Question 1

   Question: Let $y = f(x)$ be a differentiable function such that $f(0) = 0$ and $\frac{dy}{dx} + \frac{xy}{x^2 - 1} = \frac{x^6 + 4x}{\sqrt{1 - x^2}}$ for $-1 < x < 1$. If $6 \int_{-1/2}^{1/2} f(x) dx = 2\pi - \alpha$, then $\alpha^2$ is equal to?

   Options: (None provided in the source)

   Correct Answer: 27

   Year: JEE Main 2025 (Online) 22nd January Evening Shift

   Solution: The problem is a linear differential equation where the Integrating Factor (I.F.) is $e^{-\frac{1}{2}\int \frac{2x}{1-x^2}dx} = \sqrt{1-x^2}$. Multiplying the equation by the I.F. and integrating leads to $y \sqrt{1-x^2} = \frac{x^7}{7} + 2x^2 + c$. Given $y(0)=0$, $c$ becomes 0. Solving for $y$ and integrating $f(x)$ over $[-1/2, 1/2]$ removes the odd term ($x^7$), leaving the integral of $\frac{2x^2}{\sqrt{1-x^2}}$, which evaluates to $2\pi - 3\sqrt{3}$. Comparing this to $2\pi - \alpha$ gives $\alpha = 3\sqrt{3}$, hence $\alpha^2 = 27$.

   Step Solution:

    1.  Find the I.F.: Identify $P(x) = \frac{x}{x^2-1}$ and calculate $I.F. = e^{\int \frac{x}{x^2-1}dx} = \sqrt{1-x^2}$ (note: the solution uses $1-x^2$ to keep the term inside the root positive).

    2.  General Solution: Set up $y \cdot \sqrt{1-x^2} = \int \frac{x^6+4x}{\sqrt{1-x^2}} \cdot \sqrt{1-x^2} dx$, which simplifies to $\int (x^6+4x) dx = \frac{x^7}{7} + 2x^2 + c$.

    3.  Find Constant: Use $f(0)=0$ to determine $c=0$, yielding $f(x) = \frac{x^7/7 + 2x^2}{\sqrt{1-x^2}}$.

    4.  Definite Integral: Evaluate $6 \int_{-1/2}^{1/2} \frac{x^7/7 + 2x^2}{\sqrt{1-x^2}} dx$; the $x^7$ term integrates to 0 because it is an odd function over a symmetric interval.

    5.  Final Calculation: Use substitution $x = \sin\theta$ to solve $24 \int_0^{1/2} \frac{x^2}{\sqrt{1-x^2}} dx$, resulting in $2\pi - 3\sqrt{3}$. Thus $\alpha^2 = (3\sqrt{3})^2 = 27$.

   Difficulty Level: Medium

   Concept Name: First Order Linear Differential Equation, Even and Odd Functions in Integration.

   Shortcut Solution: Recognize $x^7/\sqrt{1-x^2}$ as an odd function immediately to skip its integration over the symmetric limits $[-1/2, 1/2]$.

Question 4

   Question: If $y = y(x)$ is the solution of the differential equation $\sqrt{4-x^2} \frac{dy}{dx} = \left( \left( \sin^{-1} \left( \frac{x}{2} \right) \right)^2 - y \right) \sin^{-1} \left( \frac{x}{2} \right)$ for $-2 \leq x \leq 2$, with $y(2) = \frac{\pi^2}{4} - 2$, then $y^2(0)$ is equal to?

   Options: (None provided in the source)

   Correct Answer: 4

   Year: JEE Main 2025 (Online) 28th January Evening Shift

   Solution: Rewriting the equation into linear form $\frac{dy}{dx} + \frac{\sin^{-1}(x/2)}{\sqrt{4-x^2}} y = \frac{(\sin^{-1}(x/2))^3}{\sqrt{4-x^2}}$. The Integrating Factor is $e^{\frac{1}{2}(\sin^{-1}(x/2))^2}$. Solving the integral leads to $y = (\sin^{-1}(x/2))^2 - 2 + c \cdot e^{-\frac{1}{2}(\sin^{-1}(x/2))^2}$. Applying the condition $y(2) = \frac{\pi^2}{4}-2$ gives $c=0$. Evaluating at $x=0$ results in $y(0) = -2$, so $y^2(0) = 4$.

   Step Solution:

    1.  Standardize DE: Divide by $\sqrt{4-x^2}$ to get $\frac{dy}{dx} + \frac{\sin^{-1}(x/2)}{\sqrt{4-x^2}} y = \frac{(\sin^{-1}(x/2))^3}{\sqrt{4-x^2}}$.

    2.  Integrating Factor: Calculate $I.F. = e^{\int \frac{\sin^{-1}(x/2)}{\sqrt{4-x^2}} dx} = e^{\frac{1}{2}(\sin^{-1}(x/2))^2}$.

    3.  Solve Integral: Integrate $y \cdot I.F. = \int \frac{(\sin^{-1}(x/2))^3}{\sqrt{4-x^2}} e^{\frac{1}{2}(\sin^{-1}(x/2))^2} dx$ using substitution $t = \sin^{-1}(x/2)$.

    4.  Find c: Using $y(2) = \frac{\pi^2}{4}-2$, we find $c=0$.

    5.  Evaluate Answer: Find $y(0) = 0 - 2 + 0 = -2$. Therefore $y^2(0) = 4$.

   Difficulty Level: Medium

   Concept Name: First Order Linear Differential Equation, Substitution Method.

   Shortcut Solution: Notice that the derivative of the Integrating Factor's exponent is present in the $Q(x)$ term, facilitating quick integration.

 Question 5

   Question: Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} + 2y\sec^2 x = 2\sec^2 x + 3\tan x \cdot \sec^2 x$ such that $y(0) = \frac{5}{4}$. Then $12(y(\frac{\pi}{4}) - e^{-2})$ is equal to?

   Options: (None provided in the source)

   Correct Answer: 21

   Year: JEE Main 2025 (Online) 2nd April Evening Shift

   Solution: This is a linear differential equation with $P(x) = 2\sec^2 x$ and $Q(x) = (2 + 3\tan x)\sec^2 x$. The I.F. is $e^{2\tan x}$. Integration by substitution ($u = \tan x$) gives $y e^{2\tan x} = e^{2\tan x} + 3[\frac{\tan x e^{2\tan x}}{2} - \frac{e^{2\tan x}}{4}] + c$. Using $y(0)=5/4$ results in $c=1$. Evaluating $y(\pi/4)$ and substituting into the target expression gives the result 21.

   Step Solution:

    1.  Identify I.F.: $I.F. = e^{\int 2\sec^2 x dx} = e^{2\tan x}$.

    2.  General Solution: $y \cdot e^{2\tan x} = \int e^{2\tan x}(2 + 3\tan x)\sec^2 x dx$.

    3.  Substitution: Let $u = \tan x$, $du = \sec^2 x dx$. Solve $\int e^{2u}(2+3u)du = e^{2u} + 3[\frac{u e^{2u}}{2} - \frac{e^{2u}}{4}] + c$.

    4.  Find c: Substitute $y(0)=5/4$ and $\tan(0)=0$ to find $5/4 = 1 - 3/4 + c$, which gives $c=1$.

    5.  Compute Final Value: Solve for $y(\pi/4)$ to get $7/4 + e^{-2}$. The expression $12(y(\pi/4) - e^{-2})$ simplifies to $12(7/4) = 21$.

   Difficulty Level: Medium

   Concept Name: First Order Linear Differential Equation, Integration by Parts.

   Shortcut Solution: During the integral $\int e^{2u}(2+3u)du$, split it into $\int 2e^{2u} du$ and $3\int u e^{2u} du$ for easier manipulation.

 Question 8

   Question: Let $x = x(y)$ be the solution of the differential equation $y^2 dx + \left(x - \frac{1}{y}\right) dy = 0$. If $x(1) = 1$, then $x\left(\frac{1}{2}\right)$ is:

   Options: 

    A. $\frac{3}{2} + e$ 

    B. $\frac{1}{2} + e$ 

    C. $3 + e$ 

    D. $3 - e$

   Correct Answer: D

   Year: JEE Main 2025 (Online) 22nd January Morning Shift

   Solution: The equation is rearranged to $\frac{dx}{dy} + \frac{x}{y^2} = \frac{1}{y^3}$, which is a linear differential equation in $x$. The Integrating Factor (I.F.) is $e^{-1/y}$. Multiplying and integrating gives $x e^{-1/y} = \int e^{-1/y} \frac{1}{y^3} dy + C$. Using substitution $t = -1/y$, the integral evaluates to $-e^t(t-1)$. Applying the initial condition $x(1)=1$ gives $C = -e^{-1}$. Finally, substituting $y=1/2$ into the general solution yields $x = 3 - e$.

   Step Solution:

    1.  Standardize DE: Divide by $y^2 dy$ to get $\frac{dx}{dy} + \frac{1}{y^2}x = \frac{1}{y^3}$.

    2.  Integrating Factor: $I.F. = e^{\int \frac{1}{y^2} dy} = e^{-1/y}$.

    3.  Integrate: Solve $x \cdot e^{-1/y} = \int e^{-1/y} \cdot \frac{1}{y^3} dy$. Let $t = -1/y \implies dt = \frac{1}{y^2} dy$. The integral becomes $\int -t e^t dt = -e^t(t-1) = e^{-1/y}(\frac{1}{y} + 1) + C$.

    4.  Find C: Substitute $x(1)=1 \implies 1 \cdot e^{-1} = e^{-1}(1+1) + C \implies C = -e^{-1}$.

    5.  Calculate $x(1/2)$: At $y=1/2$, $x \cdot e^{-2} = e^{-2}(2+1) - e^{-1} \implies x = 3 - e$.

   Difficulty Level: Medium

   Concept Name: First Order Linear Differential Equation (in $x$).

   Shortcut Solution: Notice that the integral $\int e^{-1/y} \frac{1}{y^3} dy$ is simply a standard $t e^t$ form after substitution, which integrates to $(t-1)e^t$.

 Question 9

   Question: If $x = f(y)$ is the solution of the differential equation $(1 + y^2) + (x - 2e^{\tan^{-1} y}) \frac{dy}{dx} = 0, y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ with $f(0) = 1$, then $f\left(\frac{1}{\sqrt{3}}\right)$ is equal to:

   Options: 

    A. $e^{\pi/4}$ 

    B. $e^{\pi/12}$ 

    C. $e^{\pi/6}$ 

    D. $e^{\pi/3}$

   Correct Answer: C

   Year: JEE Main 2025 (Online) 22nd January Evening Shift

   Solution: Converting the equation to $\frac{dx}{dy} + \frac{1}{1+y^2}x = \frac{2e^{\tan^{-1} y}}{1+y^2}$, we find it is linear in $x$. The I.F. is $e^{\tan^{-1} y}$. Integrating the right side leads to $x e^{\tan^{-1} y} = e^{2\tan^{-1} y} + C$. Given $f(0)=1$, $C$ is 0. Evaluating for $y = 1/\sqrt{3}$ gives $x = e^{\pi/6}$.

   Step Solution:

    1.  Standardize DE: Rearrange to $\frac{dx}{dy} + \frac{1}{1+y^2}x = \frac{2e^{\tan^{-1} y}}{1+y^2}$.

    2.  Integrating Factor: $I.F. = e^{\int \frac{1}{1+y^2} dy} = e^{\tan^{-1} y}$.

    3.  Integrate: $x \cdot e^{\tan^{-1} y} = \int \frac{2e^{\tan^{-1} y}}{1+y^2} \cdot e^{\tan^{-1} y} dy = \int \frac{2e^{2\tan^{-1} y}}{1+y^2} dy = e^{2\tan^{-1} y} + C$.

    4.  Find C: Using $f(0)=1$, $1 \cdot e^0 = e^0 + C \implies C = 0$.

    5.  Final Result: $f(1/\sqrt{3}) = e^{2\tan^{-1}(1/\sqrt{3})} / e^{\tan^{-1}(1/\sqrt{3})} = e^{\tan^{-1}(1/\sqrt{3})} = e^{\pi/6}$.

   Difficulty Level: Medium

   Concept Name: First Order Linear Differential Equation.

   Shortcut Solution: Since the result is $x = e^{\tan^{-1} y}$, simply evaluate $e^{\tan^{-1}(1/\sqrt{3})}$ directly.

Question 10

   Question: Let a curve $y = f(x)$ pass through the points $(0, 5)$ and $(\log_e 2, k)$. If the curve satisfies the differential equation $2(3+y)e^{2x} dx - (7+e^{2x}) dy = 0$, then $k$ is equal to?

   Options: 

    A. 32 

    B. 8 

    C. 4 

    D. 16

   Correct Answer: B

   Year: JEE Main 2025 (Online) 23rd January Morning Shift

   Solution: The equation is solved using the Integrating Factor method by writing it as $\frac{dy}{dx} - \frac{2e^{2x}}{7+e^{2x}}y = \frac{6e^{2x}}{7+e^{2x}}$. The I.F. is $\frac{1}{7+e^{2x}}$. Solving the equation gives $y = e^{2x} + 4$. Substituting $x = \ln 2$ into this equation gives $y = 8$, so $k=8$.

   Step Solution:

    1.  Separate Variables: $\frac{dy}{3+y} = \frac{2e^{2x}}{7+e^{2x}} dx$. (Note: while the source uses a linear form, variable separation is also shown).

    2.  Integrate: $\int \frac{1}{3+y} dy = \int \frac{2e^{2x}}{7+e^{2x}} dx \implies \ln(3+y) = \ln(7+e^{2x}) + C$.

    3.  Find C: Curve passes through $(0, 5) \implies \ln(8) = \ln(8) + C \implies C = 0$.

    4.  Solve for y: $3+y = 7+e^{2x} \implies y = e^{2x} + 4$.

    5.  Evaluate k: At $x = \ln 2$, $k = e^{2 \ln 2} + 4 = 4 + 4 = 8$.

   Difficulty Level: Easy

   Concept Name: Variable Separable Method.

   Shortcut Solution: Directly integrate both sides as they are already in $f'(g)/f(g)$ form, leading immediately to $3+y = C(7+e^{2x})$.

Question 12

   Question: Let $y = y(x)$ be the solution of the differential equation $\left( xy - 5x^2 \sqrt{1 + x^2} \right) dx + \left( 1 + x^2 \right) dy = 0$, $y(0) = 0$. Then $y(\sqrt{3})$ is equal to?

   Options: 

    A. $5\sqrt{3}$

    B. $\sqrt{\frac{15}{2}}$

    C. $\sqrt{\frac{14}{3}}$

    D. $2\sqrt{2}$

   Correct Answer: A

   Year: JEE Main 2025 (Online) 24th January Morning Shift

   Solution: Rewriting the equation as $\frac{dy}{dx} + \frac{x}{1+x^2} y = \frac{5x^2}{\sqrt{1+x^2}}$. This is a linear differential equation. The Integrating Factor (I.F.) is $\sqrt{1+x^2}$. Multiplying the equation by the I.F. and integrating gives $y \sqrt{1+x^2} = \int 5x^2 dx = \frac{5x^3}{3} + C$. Given $y(0)=0$, $C$ becomes 0. For $x=\sqrt{3}$, the equation yields $y \cdot 2 = \frac{5 \cdot 3\sqrt{3}}{3} = 5\sqrt{3}$.

   Step Solution:

    1.  Rearrange to Linear Form: Divide by $(1+x^2)dx$ to get $\frac{dy}{dx} + \frac{x}{1+x^2} y = \frac{5x^2 \sqrt{1+x^2}}{1+x^2} = \frac{5x^2}{\sqrt{1+x^2}}$.

    2.  Integrating Factor: Calculate $I.F. = e^{\int \frac{x}{1+x^2} dx} = e^{\frac{1}{2} \ln(1+x^2)} = \sqrt{1+x^2}$.

    3.  General Solution: $y \cdot \sqrt{1+x^2} = \int \frac{5x^2}{\sqrt{1+x^2}} \cdot \sqrt{1+x^2} dx = \int 5x^2 dx = \frac{5x^3}{3} + C$.

    4.  Find C: Use $y(0)=0 \implies 0 \cdot 1 = 0 + C \implies C = 0$.

    5.  Evaluate Answer: At $x = \sqrt{3}$, $y \sqrt{1+3} = \frac{5(\sqrt{3})^3}{3} \implies 2y = 5\sqrt{3} \implies y = \frac{5\sqrt{3}}{2}$. (Note: The source lists the answer as A ($5\sqrt{3}$), which corresponds to the value of $2y$ or implies a coefficient variation in the original problem transcript).

   Difficulty Level: Medium

   Concept Name: First Order Linear Differential Equation.

   Shortcut Solution: Notice that the term $\frac{x}{1+x^2}$ is exactly half the derivative of $\ln(1+x^2)$, leading to a simple square root I.F.

 Question 14

   Question: Let $y = y(x)$ be the solution of the differential equation: $\cos x (\log_e(\cos x))^2 dy + (\sin x - 3y \sin x \log_e(\cos x)) dx = 0$, $x \in (0, \pi/2)$. If $y(\pi/4) = -\frac{1}{\log_e 2}$, then $y(\pi/6)$ is equal to?

   Options: 

    A. $\frac{2}{\log_e(3) - \log_e(4)}$

    B. $-\frac{1}{\log_e(4)}$

    C. $\frac{1}{\log_e(4) - \log_e(3)}$

    D. $\frac{1}{\log_e(3) - \log_e(4)}$

   Correct Answer: D

   Year: JEE Main 2025 (Online) 29th January Morning Shift

   Solution: Convert to linear form $\frac{dy}{dx} - \frac{3 \tan x}{\ln(\cos x)} y = \frac{-\tan x}{(\ln(\cos x))^2}$. The I.F. is $(\ln(\sec x))^3$. Integrating the right side leads to $y = \frac{1}{2 \ln(\cos x)} + \frac{C}{(\ln(\cos x))^3}$. Applying the initial condition gives $C=0$. Evaluating at $x=\pi/6$ results in Option D.

   Step Solution:

    1.  Standardize DE: Rearrange to $\frac{dy}{dx} - \frac{3 \sin x}{\cos x \ln(\cos x)} y = \frac{-\sin x}{\cos x (\ln \cos x)^2} \implies \frac{dy}{dx} - \frac{3\tan x}{\ln(\cos x)} y = \frac{-\tan x}{(\ln \cos x)^2}$.

    2.  Integrating Factor: $I.F. = e^{\int \frac{-3\tan x}{\ln \cos x} dx}$. Let $t = \ln \cos x \implies dt = -\tan x dx$. $I.F. = e^{\int \frac{3}{t} dt} = t^3 = (\ln \cos x)^3$.

    3.  Integrate: $y (\ln \cos x)^3 = \int \frac{-\tan x}{(\ln \cos x)^2} (\ln \cos x)^3 dx = \int -\tan x \ln \cos x dx = \frac{(\ln \cos x)^2}{2} + C$.

    4.  Find C: $y(\pi/4) = \frac{-1}{\ln 2}$. Using $\ln \cos(\pi/4) = \ln(1/\sqrt{2}) = -\frac{1}{2}\ln 2$, we find $C = 0$.

    5.  Final Value: At $x = \pi/6$, $\ln \cos(\pi/6) = \ln(\sqrt{3}/2)$. $y = \frac{1}{2 \ln(\sqrt{3}/2)} = \frac{1}{\ln 3 - \ln 4}$.

   Difficulty Level: Hard

   Concept Name: First Order Linear Differential Equation, Integration by Substitution.

   Shortcut Solution: Recognizing the logarithmic derivative relationship early allows for a very fast I.F. calculation.

 Question 15

   Question: If for the solution curve $y = f(x)$ of the differential equation $\frac{dy}{dx} + (\tan x)y = \frac{2 + \sec x}{(1 + 2\sec x)^2}$, $x \in (-\pi/2, \pi/2)$, $f(\pi/3) = \frac{\sqrt{3}}{10}$, then $f(\pi/4)$ is equal to?

   Options: 

    A. $5-\sqrt{3}$

    B. $\frac{4 - \sqrt{2}}{14}$

    C. $\frac{9\sqrt{3} + 3}{10(4 + \sqrt{3})}$

    D. $\frac{\sqrt{3} + 1}{10(4 + \sqrt{3})}$

   Correct Answer: B

   Year: JEE Main 2025 (Online) 29th January Evening Shift

   Solution: The I.F. is $\sec x$. The solution is $y \cdot \sec x = \int \frac{(2 + \sec x)\sec x}{(1 + 2\sec x)^2} dx$. Using the substitution $t = \tan(x/2)$, the integral simplifies. Solving with $f(\pi/3) = \sqrt{3}/10$ gives the integration constant $C=0$. Evaluating for $x = \pi/4$ yields $\frac{4-\sqrt{2}}{14}$.

   Step Solution:

    1.  Integrating Factor: $I.F. = e^{\int \tan x dx} = e^{\ln \sec x} = \sec x$.

    2.  General Solution: $y \cdot \sec x = \int \frac{(2 + \sec x)\sec x}{(1 + 2\sec x)^2} dx = \int \frac{2\sec x + \sec^2 x}{(1 + 2\sec x)^2} dx$.

    3.  Substitution: Let $t = \tan(x/2)$. The integral evaluates to $y \cdot \sec x = \frac{2t}{1+3t^2} + C$.

    4.  Find C: At $x = \pi/3, t = 1/\sqrt{3}$. Substituting into the solution with $y = \sqrt{3}/10$ gives $C = 0$.

    5.  Evaluate f(π/4): At $x = \pi/4, t = \sqrt{2}-1, \sec x = \sqrt{2}$. Solving for $y$ leads to $\frac{4-\sqrt{2}}{14}$.

   Difficulty Level: Hard

   Concept Name: Linear Differential Equation, Half-Angle (Weierstrass) Substitution.

   Shortcut Solution: The integral is of the form $\int \frac{d}{dx} (\text{rational function of } \tan(x/2)) dx$. Recognizing the derivative structure saves significant algebraic effort.

 Question 16

   Question: Let $g$ be a differentiable function such that $\int_0^x g(t) dt = x - \int_0^x t g(t) dt, x \geq 0$ and let $y = y(x)$ satisfy the differential equation $\frac{dy}{dx} - y \tan x = 2(x + 1) \sec x g(x)$, $x \in [0, \pi/2)$. If $y(0) = 0$, then $y(\pi/3)$ is equal to?

   Options: 

    A. $\frac{4\pi}{3}$

    B. $\frac{2\pi}{3}$

    C. $\frac{2\pi}{3\sqrt{3}}$

    D. $\frac{4\pi}{3\sqrt{3}}$

   Correct Answer: A

   Year: JEE Main 2025 (Online) 3rd April Morning Shift

   Solution: Differentiate the integral equation $\int_0^x g(t) dt = x - \int_0^x t g(t) dt$ to get $g(x) = 1 - x g(x)$, which gives $g(x) = \frac{1}{1+x}$. Substituting this into the differential equation gives $\frac{dy}{dx} - y \tan x = 2 \sec x$. The Integrating Factor is $e^{\int -\tan x dx} = \cos x$. Multiplying and integrating gives $y \cos x = 2x + c$. Given $y(0)=0$, $c=0$. Evaluating at $x = \pi/3$ yields $y(1/2) = 2\pi/3$, hence $y = 4\pi/3$.

   Step Solution:

    1.  Find g(x): Differentiate the integral equation using Newton-Leibniz: $g(x) = 1 - [x g(x)]$.

    2.  Solve for g(x): $g(x)(1+x) = 1 \implies g(x) = \frac{1}{1+x}$.

    3.  Simplify DE: Substitute $g(x)$ into the DE: $\frac{dy}{dx} - (\tan x)y = 2(x+1) \sec x \cdot \frac{1}{1+x} = 2 \sec x$.

    4.  Integrate: Using $I.F. = \cos x$, we get $\frac{d}{dx}(y \cos x) = 2 \sec x \cdot \cos x = 2$. Integrating gives $y \cos x = 2x + c$.

    5.  Final Value: Use $y(0)=0$ to find $c=0$. At $x = \pi/3$, $y \cdot \cos(\pi/3) = 2(\pi/3) \implies y \cdot (1/2) = 2\pi/3 \implies y = 4\pi/3$.

   Difficulty Level: Medium

   Concept Name: Newton-Leibniz Theorem, First Order Linear Differential Equation.

   Shortcut Solution: Multiply the entire DE by $\cos x$ immediately; the left side becomes the exact derivative $\frac{d}{dx}(y \cos x)$, and the right side becomes $2$, skipping the I.F. formula.

 Question 17

   Question: Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} + 3(\tan^2 x)y + 3y = \sec^2 x, y(0) = \frac{1}{3} + e^3$. Then $y(\pi/4)$ is equal to?

   Options: 

    A. $\frac{4}{3}$

    B. $\frac{2}{3} + e^3$

    C. $\frac{4}{3} + e^3$

    D. $\frac{2}{3}$

   Correct Answer: A

   Year: JEE Main 2025 (Online) 3rd April Evening Shift

   Solution: The equation is rewritten as $\frac{dy}{dx} + 3\sec^2 x y = \sec^2 x$. The Integrating Factor (I.F.) is $e^{\int 3 \sec^2 x dx} = e^{3 \tan x}$. The solution is $y e^{3 \tan x} = \int e^{3 \tan x} \sec^2 x dx = \frac{e^{3 \tan x}}{3} + c$. Using $y(0) = 1/3 + e^3$ gives $c = e^3$. For $x = \pi/4$, the equation yields $y e^3 = \frac{e^3}{3} + e^3$, which simplifies to $y = 4/3$.

   Step Solution:

    1.  Identify Linear Form: Group $y$ terms: $\frac{dy}{dx} + 3(\tan^2 x + 1)y = \sec^2 x \implies \frac{dy}{dx} + 3\sec^2 x y = \sec^2 x$.

    2.  Integrating Factor: $I.F. = e^{\int 3\sec^2 x dx} = e^{3 \tan x}$.

    3.  General Solution: $y \cdot e^{3 \tan x} = \int \sec^2 x e^{3 \tan x} dx = \frac{1}{3}e^{3 \tan x} + c$.

    4.  Find c: Substitute $x=0, y=1/3+e^3 \implies (1/3+e^3) \cdot 1 = 1/3 + c \implies c = e^3$.

    5.  Evaluate: At $x=\pi/4$, $y \cdot e^3 = \frac{1}{3}e^3 + e^3 \implies y = \frac{1}{3} + 1 = \frac{4}{3}$.

   Difficulty Level: Medium

   Concept Name: First Order Linear Differential Equation, Trigonometric Substitution.

   Shortcut Solution: Notice that the equation is of the form $y' + P(x)y = \frac{1}{3}P(x)$. The solution for such forms is always $y = \frac{1}{3} + c \cdot e^{-\int P(x)dx}$.

 Question 19

   Question: Let $y = y(x)$ be the solution curve of the differential equation $x(x^2 + e^x) dy + (e^x(x - 2)y - x^3) dx = 0, x > 0$, passing through the point $(1, 0)$. Then $y(2)$ is equal to?

   Options: 

    A. $\frac{2}{2 + e^2}$

    B. $\frac{4}{4 - e^2}$

    C. $\frac{4}{4 + e^2}$

    D. $\frac{2}{2 - e^2}$

   Correct Answer: C

   Year: JEE Main 2025 (Online) 7th April Morning Shift

   Solution: The equation is arranged into linear form $\frac{dy}{dx} + \frac{e^x(x-2)}{x(x^2+e^x)} y = \frac{x^2}{x^2+e^x}$. The Integrating Factor is found to be $\frac{x^2+e^x}{x^2}$. Multiplying and integrating leads to $y(\frac{e^x+x^2}{x^2}) = x + c$. Given $(1, 0)$, $c$ is $-1$. Finally, $y(2)$ is calculated as $4/(e^2+4)$.

   Step Solution:

    1.  Standardize DE: Divide by $x(x^2+e^x)dx$: $\frac{dy}{dx} + \frac{e^x(x-2)}{x(x^2+e^x)}y = \frac{x^3}{x(x^2+e^x)} = \frac{x^2}{x^2+e^x}$.

    2.  Integrating Factor: Calculate $I.F. = \exp\left(\int \frac{e^x(x-2)}{x(x^2+e^x)} dx\right) = \frac{x^2+e^x}{x^2}$.

    3.  Integrate: $y \cdot \frac{x^2+e^x}{x^2} = \int \frac{x^2}{x^2+e^x} \cdot \frac{x^2+e^x}{x^2} dx = \int 1 dx = x + c$.

    4.  Constant: Pass through $(1, 0) \implies 0 = 1 + c \implies c = -1$.

    5.  Calculate y(2): $y \cdot \frac{4+e^2}{4} = 2 - 1 = 1 \implies y(2) = \frac{4}{4+e^2}$.

   Difficulty Level: Hard

   Concept Name: First Order Linear Differential Equation, Derivative of a Quotient.

   Shortcut Solution: Recognize that $\frac{d}{dx}(\frac{x^2+e^x}{x^2}) = \frac{x^2(e^x+2x)-2x(x^2+e^x)}{x^4} = \frac{xe^x-2e^x}{x^3} = \frac{e^x(x-2)}{x^3}$. This matches the $P(x)$ term after dividing the DE by $x^2$, revealing the product rule structure immediately.

 Question 20

   Question: Let $y = y(x)$ be the solution of the differential equation $(x^2 + 1)y' - 2xy = (x^4 + 2x^2 + 1) \cos x$, $y(0) = 1$. Then $\int_{-3}^3 y(x) dx$ is:

   Options: 

    A. 36 

    B. 24 

    C. 18 

    D. 30

   Correct Answer: B

   Year: JEE Main 2025 (Online) 7th April Evening Shift

   Solution: The equation is a linear differential equation: $\frac{dy}{dx} - (\frac{2x}{x^2+1})y = (x^2+1)\cos x$. The Integrating Factor (I.F.) is $\frac{1}{x^2+1}$. Multiplying and integrating gives $\frac{y}{x^2+1} = \sin x + c$. Using $y(0)=1$, $c=1$. Thus $y = (x^2+1)(\sin x + 1)$. Integrating this from $-3$ to $3$ results in $24$, as the odd part of the integrand ($\sin x$ terms) evaluates to zero.

   Step Solution:

    1.  Standardize DE: Divide by $(x^2+1)$ to get $\frac{dy}{dx} - \frac{2x}{x^2+1} y = (x^2+1) \cos x$.

    2.  Find I.F.: Calculate $I.F. = e^{\int \frac{-2x}{x^2+1} dx} = e^{-\ln(x^2+1)} = \frac{1}{x^2+1}$.

    3.  General Solution: $y \cdot \frac{1}{x^2+1} = \int (x^2+1)\cos x \cdot \frac{1}{x^2+1} dx = \int \cos x dx = \sin x + c$.

    4.  Find Constant: Use $y(0)=1 \implies \frac{1}{1} = \sin(0) + c \implies c = 1$. So, $y = (x^2+1)(\sin x + 1)$.

    5.  Definite Integral: Evaluate $\int_{-3}^3 (x^2\sin x + x^2 + \sin x + 1) dx$. The odd terms ($x^2\sin x, \sin x$) vanish, leaving $2 \int_0^3 (x^2+1) dx = 2 [\frac{x^3}{3} + x]_0^3 = 2(9+3) = 24$.

   Difficulty Level: Medium

   Concept Name: First Order Linear Differential Equation, Even and Odd Functions.

   Shortcut Solution: Recognizing that $x^2\sin x$ and $\sin x$ are odd functions allows you to instantly eliminate them from the definite integral over the symmetric interval $[-3, 3]$.

Question 21

   Question: Let $f(x) = x - 1$ and $g(x) = e^x$ for $x \in \mathbb{R}$. If $\frac{dy}{dx} = (e^{-2\sqrt{x}} g(f(f(x))) - \frac{y}{\sqrt{x}}), y(0) = 0$, then $y(1)$ is:

   Options: 

    A. $\frac{1 - e^3}{e^4}$ 

    B. $\frac{e - 1}{e^4}$ 

    C. $1 - e^2$ 

    D. $\frac{2e - 1}{e^3}$

   Correct Answer: B

   Year: JEE Main 2025 (Online) 8th April Evening Shift

   Solution: First, $f(f(x)) = x-2$, so $g(f(f(x))) = e^{x-2}$. The DE becomes $\frac{dy}{dx} + \frac{1}{\sqrt{x}}y = e^{x-2\sqrt{x}-2}$. The I.F. is $e^{2\sqrt{x}}$. Solving gives $ye^{2\sqrt{x}} = e^{x-2} + c$. Using $y(0)=0$, $c = -e^{-2}$. For $x=1$, $y(1) = \frac{e-1}{e^4}$.

   Step Solution:

    1.  Simplify Composites: $f(f(x)) = (x-1)-1 = x-2 \implies g(f(f(x))) = e^{x-2}$.

    2.  Linear Form: Rewrite DE as $\frac{dy}{dx} + \frac{1}{\sqrt{x}}y = e^{-2\sqrt{x}} \cdot e^{x-2} = e^{x-2\sqrt{x}-2}$.

    3.  Find I.F.: $I.F. = e^{\int \frac{1}{\sqrt{x}} dx} = e^{2\sqrt{x}}$.

    4.  Solve DE: $y \cdot e^{2\sqrt{x}} = \int e^{2\sqrt{x}} \cdot e^{x-2\sqrt{x}-2} dx = \int e^{x-2} dx = e^{x-2} + c$.

    5.  Compute y(1): Use $y(0)=0 \implies 0 = e^{-2} + c \implies c = -e^{-2}$. At $x=1$, $y \cdot e^2 = e^{-1} - e^{-2} \implies y = \frac{1/e - 1/e^2}{e^2} = \frac{e-1}{e^4}$.

   Difficulty Level: Medium

   Concept Name: First Order Linear Differential Equation, Composite Functions.

   Shortcut Solution: Notice that the I.F. $e^{2\sqrt{x}}$ perfectly cancels the $e^{-2\sqrt{x}}$ term in the exponent of the source term, making the integration trivial.

Question 22

   Question: Let $x = x(t)$ and $y = y(t)$ be solutions of the differential equations $\frac{dx}{dt} + ax = 0$ and $\frac{dy}{dt} + by = 0$ respectively, $a, b \in \mathbb{R}$. Given that $x(0) = 2; y(0) = 1$ and $3y(1) = 2x(1)$, the value of $t$, for which $x(t) = y(t)$, is:

   Options: 

    A. $\log_{2/3} 2$ 

    B. $\log_4 3$ 

    C. $\log_3 4$ 

    D. $\log_{4/3} 2$

   Correct Answer: D

   Year: 27-Jan-2024 Shift 1

   Solution: Solving the separate variable equations gives $x(t) = 2e^{-at}$ and $y(t) = e^{-bt}$. The condition $3y(1) = 2x(1)$ implies $3e^{-b} = 2(2e^{-a})$, which simplifies to $e^{a-b} = \frac{4}{3}$. Setting $x(t) = y(t)$ leads to $2e^{-at} = e^{-bt} \implies 2 = e^{(a-b)t}$. Taking the logarithm base $4/3$ yields $t = \log_{4/3} 2$.

   Step Solution:

    1.  Solve for x(t): $\frac{dx}{x} = -a dt \implies \ln x = -at + c$. Given $x(0)=2 \implies x(t) = 2e^{-at}$.

    2.  Solve for y(t): $\frac{dy}{y} = -b dt \implies \ln y = -bt + c'$. Given $y(0)=1 \implies y(t) = e^{-bt}$.

    3.  Relate a and b: Use $3y(1) = 2x(1) \implies 3e^{-b} = 4e^{-a} \implies e^{a-b} = \frac{4}{3}$.

    4.  Set Equations Equal: For $x(t) = y(t) \implies 2e^{-at} = e^{-bt} \implies 2 = e^{(a-b)t}$.

    5.  Final Value: $2 = (e^{a-b})^t \implies 2 = (4/3)^t \implies t = \log_{4/3} 2$.

   Difficulty Level: Easy

   Concept Name: Variable Separable Method, Exponential Functions.

   Shortcut Solution: Directly compare the ratios: since $x(1)/y(1) = 3/2$ and $x(0)/y(0) = 2$, use the exponential nature to solve for $t$ when the ratio $x(t)/y(t)$ becomes $1$.

Question 26

   Question: A function $y = f(x)$ satisfies $f(x) \sin 2x + \sin x - (1 + \cos^2 x) f'(x) = 0$ with condition $f(0) = 0$. Then $f(\frac{\pi}{2})$ is equal to:

   Options: A. 1, B. 0, C. -1, D. 2

   Correct Answer: A

   Year: 29-Jan-2024 Shift 1

   Solution: $\frac{dy}{dx} - \left( \frac{\sin 2x}{1 + \cos^2 x} \right) y = \frac{\sin x}{1 + \cos^2 x}$. The Integrating Factor (I.F.) is $1 + \cos^2 x$. Multiplying and integrating gives $y \cdot (1 + \cos^2 x) = \int \sin x \, dx = -\cos x + C$. Given $f(0) = 0$, $C$ becomes 1. Thus $f(x) = \frac{1 - \cos x}{1 + \cos^2 x}$. Evaluating at $x = \frac{\pi}{2}$ gives $y = \frac{1 - 0}{1 + 0} = 1$.

   Step Solution:

    1.  Rearrange to Standard Form: Rewrite as $\frac{dy}{dx} + \left( \frac{-\sin 2x}{1 + \cos^2 x} \right)y = \frac{\sin x}{1 + \cos^2 x}$.

    2.  Find I.F.: $I.F. = e^{\int \frac{-\sin 2x}{1 + \cos^2 x} dx}$. Let $1 + \cos^2 x = u$, then $-\sin 2x \, dx = du$. $I.F. = e^{\ln(1 + \cos^2 x)} = 1 + \cos^2 x$.

    3.  General Solution: $y(1 + \cos^2 x) = \int \frac{\sin x}{1 + \cos^2 x} (1 + \cos^2 x) dx = \int \sin x \, dx = -\cos x + C$.

    4.  Determine Constant: Substitute $x = 0, y = 0 \implies 0(2) = -1 + C \implies C = 1$.

    5.  Final Value: At $x = \frac{\pi}{2}, y(1 + 0) = -0 + 1 \implies y = 1$.

   Difficulty Level: Medium

   Concept Name: First Order Linear Differential Equation

   Shortcut Solution: Notice the expression is $(1 + \cos^2 x) dy - y(\sin 2x) dx = \sin x \, dx$. The left side is almost the derivative of a product; dividing by $(1 + \cos^2 x)^2$ or simply recognizing the $I.F.$ speeds up the process.

Question 29

   Question: Let $y = y(x)$ be the solution of the differential equation $(1 - x^2) dy = [xy + (x^3 + 2)\sqrt{3(1 - x^2)}] dx, -1 < x < 1, y(0) = 0$. If $y(\frac{1}{2}) = \frac{m}{n}$, where $m$ and $n$ are coprime numbers, then $m + n$ is equal to:

   Options: (None provided in source)

   Correct Answer: 97

   Year: 30-Jan-2024 Shift 1

   Solution: $\frac{dy}{dx} - \frac{x}{1 - x^2} y = \frac{(x^3 + 2)\sqrt{3}}{\sqrt{1 - x^2}}$. I.F. is $\sqrt{1 - x^2}$. $y\sqrt{1 - x^2} = \int \sqrt{3}(x^3 + 2) dx$. With $y(0) = 0$, the constant $C = 0$. Solving for $x = 1/2$ gives $y = \frac{65}{32}$, so $m + n = 65 + 32 = 97$.

   Step Solution:

    1.  Linear Form: $\frac{dy}{dx} - \frac{x}{1 - x^2} y = \frac{(x^3 + 2)\sqrt{3(1 - x^2)}}{1 - x^2} = \frac{(x^3 + 2)\sqrt{3}}{\sqrt{1 - x^2}}$.

    2.  Integrating Factor: $I.F. = e^{\int \frac{-x}{1 - x^2} dx} = e^{\frac{1}{2}\ln(1 - x^2)} = \sqrt{1 - x^2}$.

    3.  Integrate: $y\sqrt{1 - x^2} = \int \frac{\sqrt{3}(x^3 + 2)}{\sqrt{1 - x^2}} \sqrt{1 - x^2} dx = \sqrt{3} \left[ \frac{x^4}{4} + 2x \right] + C$.

    4.  Solve for C: $y(0) = 0 \implies 0 = 0 + C \implies C = 0$.

    5.  Calculate $y(1/2)$: $y\sqrt{3/4} = \sqrt{3} \left[ \frac{1}{64} + 1 \right] \implies y \frac{\sqrt{3}}{2} = \sqrt{3} \frac{65}{64} \implies y = \frac{65}{32}$. $m + n = 65 + 32 = 97$.

   Difficulty Level: Hard

   Concept Name: First Order Linear Differential Equation

   Shortcut Solution: The cancellation of $\sqrt{1-x^2}$ after multiplying by the $I.F.$ is a common JEE pattern; focus on the polynomial integration $\int (x^3+2)dx$ immediately.

Question 30

   Question: Let $y = y(x)$ be the solution of the differential equation $\sec x \, dy + \{2(1 - x) \tan x + x(2 - x)\} dx = 0$ such that $y(0) = 2$. Then $y(2)$ is equal to:

   Options: A. 2, B. $2\{1 - \sin(2)\}$, C. $2\{\sin(2) + 1\}$, D. 1

   Correct Answer: A

   Year: 30-Jan-2024 Shift 1

   Solution: $\frac{dy}{dx} = - \cos x [2(1 - x) \tan x + x(2 - x)] = 2(x - 1) \sin x + (x^2 - 2x) \cos x$. This is the derivative of $(x^2 - 2x) \sin x$. Integrating gives $y(x) = (x^2 - 2x) \sin x + C$. Using $y(0) = 2$, we find $C = 2$. Thus $y(2) = (4 - 4) \sin(2) + 2 = 2$.

   Step Solution:

    1.  Isolate $dy/dx$: $\frac{dy}{dx} = - \frac{2(1 - x) \tan x + x(2 - x)}{\sec x} = 2(x - 1) \sin x + (x^2 - 2x) \cos x$.

    2.  Observe Derivative: Recognize that the RHS is of the form $u'v + uv'$, where $u = x^2 - 2x$ and $v = \sin x$.

    3.  Integrate: $y = \int \frac{d}{dx} [(x^2 - 2x) \sin x] dx = (x^2 - 2x) \sin x + C$.

    4.  Constant: $y(0) = 2 \implies 2 = (0) \sin(0) + C \implies C = 2$.

    5.  Solve for $y(2)$: $y(2) = (2^2 - 2(2)) \sin(2) + 2 = 0 + 2 = 2$.

   Difficulty Level: Medium

   Concept Name: Direct Integration (Exact Derivative)

   Shortcut Solution: Notice that at $x=2$, the term $(x^2 - 2x)$ becomes zero. Since $y = (x^2 - 2x) \sin x + C$, the value $y(2)$ must equal the constant $C$, which you already found to be $y(0)=2$.

 Question 32

   Question: Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} = \frac{\sin x + y \cos x}{\sin x (\sec x - \sin x \tan x)}$, $x \in (0, \pi/2)$ satisfying the condition $y(\pi/4) = 2$. Then, $y(\pi/3)$ is equal to?

   Options: 

    A. $\sqrt{3}(2 + \log_e \sqrt{3})$

    B. $\frac{\sqrt{3}}{2}(2 + \log_e 3)$

    C. $\sqrt{3}(1 + 2 \log_e 3)$

    D. $\sqrt{3}(2 + \log_e 3)$

   Correct Answer: A

   Year: 31-Jan-2024 Shift 1

   Solution: The denominator simplifies to $\sin x \cos x$ since $(\sec x - \sin x \tan x) = \frac{1 - \sin^2 x}{\cos x} = \cos x$. The equation becomes $\frac{dy}{dx} = \sec x + y \cot x$, which is a linear differential equation $\frac{dy}{dx} - y \cot x = \sec x$. Using the Integrating Factor $I.F. = \frac{1}{\sin x}$, the solution is $y \csc x = \ln |\tan x| + c$. With $y(\pi/4) = 2$, we find $c = 2$. Substituting $x = \pi/3$ gives $y = \sqrt{3}(\ln \sqrt{3} + 2)$.

   Step Solution:

    1.  Simplify Denominator: Observe $(\sec x - \sin x \tan x) = \frac{1}{\cos x} - \frac{\sin^2 x}{\cos x} = \cos x$. The DE becomes $\frac{dy}{dx} = \frac{\sin x + y \cos x}{\sin x \cos x}$.

    2.  Standard Linear Form: Rewrite as $\frac{dy}{dx} = \sec x + y \cot x \implies \frac{dy}{dx} - (\cot x)y = \sec x$.

    3.  Integrating Factor: Calculate $I.F. = e^{\int -\cot x dx} = e^{-\ln \sin x} = \csc x$.

    4.  Solve DE: $y \cdot \csc x = \int \sec x \csc x dx = \int \frac{2}{\sin 2x} dx = \ln |\tan x| + c$.

    5.  Find $c$ and Evaluate: $y(\pi/4)=2 \implies 2(\sqrt{2}) = \ln(1) + c$ (Note: the source uses a slightly different simplification path in leading to $y = \tan x (\ln \tan x + 2)$). Evaluating at $x = \pi/3$: $y = \sqrt{3}(\ln \sqrt{3} + 2)$.

   Difficulty Level: Medium

   Concept Name: First Order Linear Differential Equation

   Short cut solution: Simplify the trigonometric denominator first to realize it is $\frac{1}{2} \sin 2x$, making the division into the numerator terms much faster.

 Question 34

   Question: Let $y = y(x)$ satisfy the differential equation $\sec^2 x \frac{dx}{dy} + e^{2y} \tan^2 x + \tan x = 0$ for $0 < x < \pi/2$ with $y(\pi/4) = 0$. If $y(\pi/6) = a$, then $e^{8a}$ is equal to?

   Options: (None provided in source)

   Correct Answer: 9

   Year: 31-Jan-2024 Shift 2

   Solution: Let $\tan x = t$, then the equation becomes $\frac{dt}{dy} + t = -e^{2y} t^2$, which is Bernoulli's equation. Substituting $u = 1/t$ converts it to the linear form $\frac{du}{dy} - u = e^{2y}$. The Integrating Factor is $e^{-y}$, leading to the solution $\frac{e^{-y}}{\tan x} = e^y + c$. Using the initial condition $y(\pi/4)=0$ gives $c=0$. At $x=\pi/6$, $e^{2a} = \sqrt{3}$, so $e^{8a} = 9$.

   Step Solution:

    1.  Substitution: Let $t = \tan x \implies \sec^2 x \frac{dx}{dy} = \frac{dt}{dy}$. The DE is $\frac{dt}{dy} + t = -e^{2y} t^2$.

    2.  Bernoulli Transform: Divide by $t^2$ and let $u = t^{-1}$, resulting in the linear DE: $\frac{du}{dy} - u = e^{2y}$.

    3.  Integrating Factor: $I.F. = e^{\int -1 dy} = e^{-y}$.

    4.  General Solution: $u \cdot e^{-y} = \int e^{2y} \cdot e^{-y} dy = e^y + c \implies \frac{1}{\tan x e^y} = e^y + c$.

    5.  Final Value: $y(\pi/4)=0 \implies 1 = 1 + c \implies c = 0$. Thus $\frac{1}{\tan x} = e^{2y}$. At $x = \pi/6$, $\sqrt{3} = e^{2a}$. Squaring twice gives $e^{8a} = (\sqrt{3})^4 = 9$.

   Difficulty Level: Hard

   Concept Name: Bernoulli’s Differential Equation

   Short cut solution: Recognize $c=0$ immediately from the initial condition $y=0, \tan x=1$ to get the simple relation $\cot x = e^{2y}$.

 Question 36

   Question: If $x = x(t)$ is the solution of the differential equation $(t + 1) dx = (2x + (t + 1)^4) dt$, $x(0) = 2$, then $x(1)$ equals?

   Options: (None provided in source)

   Correct Answer: 14

   Year: 1-Feb-2024 Shift 1

   Solution: Rewrite the equation in the linear form $\frac{dx}{dt} - \frac{2x}{t+1} = (t+1)^3$. The Integrating Factor is $\frac{1}{(t+1)^2}$. The general solution is $\frac{x}{(t+1)^2} = \frac{(t+1)^2}{2} + c$. Using $x(0)=2$ results in $c = 3/2$. Substituting $t=1$ yields $x(1) = 14$.

   Step Solution:

    1.  Linear Form: Rearrange to $\frac{dx}{dt} - \frac{2}{t+1}x = (t+1)^3$.

    2.  Integrating Factor: Calculate $I.F. = e^{\int \frac{-2}{t+1} dt} = (t+1)^{-2} = \frac{1}{(t+1)^2}$.

    3.  Integrate: $x \cdot \frac{1}{(t+1)^2} = \int (t+1)^3 \cdot \frac{1}{(t+1)^2} dt = \int (t+1) dt = \frac{(t+1)^2}{2} + c$.

    4.  Solve for $c$: $x(0)=2 \implies \frac{2}{1} = \frac{1}{2} + c \implies c = \frac{3}{2}$.

    5.  Compute $x(1)$: $x = (t+1)^2 [\frac{(t+1)^2}{2} + \frac{3}{2}]$. At $t=1$, $x = 4 [2 + 1.5] = 4(3.5) = 14$.

   Difficulty Level: Medium

   Concept Name: First Order Linear Differential Equation

   Short cut solution: After finding the general form $x = \frac{(t+1)^4}{2} + c(t+1)^2$, use the initial condition to find $c$ and evaluate $x(1)$ by noting it involves powers of 2.

 Question 37

   Question: Let $\alpha$ be a non-zero real number. Suppose $f : \mathbb{R} \to \mathbb{R}$ is a differentiable function such that $f(0) = 2$ and $\lim_{x \to -\infty} f(x) = 1$. If $f'(x) = \alpha f(x) + 3$, for all $x \in \mathbb{R}$, then $f(-\log_e 2)$ is equal to?

   Options: 

    A. 3

    B. 5

    C. 9

    D. 7

   Correct Answer: A

   Year: 1-Feb-2024 Shift 2

   Solution: The equation is a linear differential equation $f'(x) - \alpha f(x) = 3$. The general solution is $f(x) = Ce^{\alpha x} - \frac{3}{\alpha}$. Given the limit $\lim_{x \to -\infty} f(x) = 1$, for a finite value to exist as $x \to -\infty$, $\alpha$ must be positive, making $e^{\alpha x} \to 0$. Thus, $1 = -\frac{3}{\alpha} \implies \alpha = -3$ (Note: there is a discrepancy in the source's sign convention, but it concludes with the result 3). Using $f(0)=2$ to find $C$ and evaluating at $x = -\ln 2$ yields the answer.

   Step Solution:

    1.  Linear Form: Rewrite the DE as $\frac{dy}{dx} - \alpha y = 3$.

    2.  General Solution: The solution for this standard form is $y = Ce^{\alpha x} - \frac{3}{\alpha}$.

    3.  Identify Constants: Apply the limit $\lim_{x \to -\infty} f(x) = 1$. This implies the steady-state term $-\frac{3}{\alpha} = 1$, so $\alpha = -3$.

    4.  Find C: Use $f(0) = 2 \implies 2 = C e^0 + 1 \implies C = 1$. Thus, $f(x) = 1 + e^{-3x}$.

    5.  Final Calculation: Evaluate $f(-\ln 2) = 1 + e^{-3(-\ln 2)} = 1 + e^{\ln 8} = 1 + 8 = 9$. (Note: while the mathematical calculation leads to 9, the source explicitly marks the answer as 3, likely due to a typo in the original exam's constant term or limit direction).

   Difficulty Level: Hard

   Concept Name: First Order Linear Differential Equation, Limits at Infinity.

   Shortcut Solution: Recognize that the limit as $x \to -\infty$ directly gives the value of the constant term in the general solution, bypassing complex integration.

 Question 38

   Question: If $\frac{dx}{dy} = \frac{1 + x - y^2}{y}, x(1) = 1$, then $5x(2)$ is equal to?

   Options: (None provided in the source)

   Correct Answer: 5

   Year: 1-Feb-2024 Shift 2

   Solution: The equation is rewritten as $\frac{dx}{dy} - \frac{x}{y} = \frac{1 - y^2}{y}$, which is linear in $x$. The Integrating Factor (I.F.) is $1/y$. Integrating both sides gives $x \cdot \frac{1}{y} = -\frac{1}{y} - y + c$. Applying the condition $x(1)=1$ results in $c=3$. Substituting $y=2$ gives $x=1$, so $5x(2) = 5$.

   Step Solution:

    1.  Standardize DE: Rearrange to $\frac{dx}{dy} - \frac{1}{y}x = \frac{1}{y} - y$.

    2.  Integrating Factor: Calculate $I.F. = e^{\int -\frac{1}{y} dy} = \frac{1}{y}$.

    3.  General Solution: $x \cdot \frac{1}{y} = \int (\frac{1}{y} - y) \frac{1}{y} dy = \int (\frac{1}{y^2} - 1) dy = -\frac{1}{y} - y + c$.

    4.  Find Constant: $x(1)=1 \implies 1(1) = -1 - 1 + c \implies c = 3$.

    5.  Evaluate: At $y=2, x(\frac{1}{2}) = -\frac{1}{2} - 2 + 3 \implies \frac{x}{2} = \frac{1}{2} \implies x = 1$. Therefore, $5x(2) = 5$.

   Difficulty Level: Easy

   Concept Name: First Order Linear Differential Equation (in $x$).

   Shortcut Solution: After finding $x = 3y - y^2 - 1$, notice that at $y=2$, $x = 6 - 4 - 1 = 1$, allowing for a very quick mental calculation.

 Question 39

   Question: Let $y = y(x)$ be the solution of the differential equation $x^3 dy + (xy - 1) dx = 0, x > 0, y(\frac{1}{2}) = 3 - e$. Then $y(1)$ is equal to?

   Options: 

    A. 1

    B. e

    C. 2 - e

    D. 3

   Correct Answer: A

   Year: 24-Jan-2023 Shift 1

   Solution: The equation is converted to the linear form $\frac{dy}{dx} + \frac{y}{x^2} = \frac{1}{x^3}$. The Integrating Factor is $e^{-1/x}$. Multiplying and integrating gives $y e^{-1/x} = (1 + \frac{1}{x})e^{-1/x} + C$. Using the initial condition $y(1/2) = 3 - e$, the constant $C$ is determined to be $-e^{-1}$. Evaluating at $x=1$ yields $y=1$.

   Step Solution:

    1.  Linear Form: Divide by $x^3 dx$ to get $\frac{dy}{dx} + \frac{1}{x^2}y = \frac{1}{x^3}$.

    2.  Integrating Factor: Calculate $I.F. = e^{\int \frac{1}{x^2} dx} = e^{-1/x}$.

    3.  Integrate: $y \cdot e^{-1/x} = \int \frac{1}{x^3} e^{-1/x} dx$. Use substitution $t = -1/x \implies dt = \frac{1}{x^2} dx$. This gives $\int -t e^t dt = (1-t)e^t + C = (1 + \frac{1}{x})e^{-1/x} + C$.

    4.  Find C: Substitute $x = 1/2, y = 3-e \implies (3-e)e^{-2} = (1+2)e^{-2} + C \implies C = -e^{-1}$.

    5.  Evaluate: At $x=1$, $y e^{-1} = (1+1)e^{-1} - e^{-1} \implies y e^{-1} = e^{-1} \implies y = 1$.

   Difficulty Level: Medium

   Concept Name: First Order Linear Differential Equation, Integration by Substitution.

   Shortcut Solution: Notice that at $x=1$, the term $(1+1/x)$ becomes 2, and the $C$ term involves $e^{-1}$, simplifying the final $y(1)$ calculation significantly.

 Question 41

   Question: Let $y = y(x)$ be the solution curve of the differential equation $\frac{dy}{dx} = \frac{y}{x} (1 + xy^2(1 + \log_e x))$, $x > 0$, $y(1) = 3$. Then $\frac{y^2(x)}{9}$ is equal to :

   Options: 

    A. $\frac{x^2}{5 - 2x^3(2 + \log_e x^3)}$

    B. $\frac{x^2}{2x^3(2 + \log_e x^3) - 3}$

    C. $\frac{x^2}{3x^3(1 + \log_e x^2) - 2}$

    D. $\frac{x^2}{7 - 3x^3(2 + \log_e x^2)}$

   Correct Answer: A

   Year: 25-Jan-2023 Shift 1

   Solution: The equation is a Bernoulli's Differential Equation. Rewriting gives $\frac{dy}{dx} - \frac{y}{x} = y^3(1 + \log_e x)$. Dividing by $y^3$ and substituting $\frac{1}{y^2} = t$ transforms it into a linear differential equation: $\frac{dt}{dx} + \frac{2}{x}t = -2(1 + \log_e x)$. Using the Integrating Factor $IF = x^2$, the solution is $\frac{x^2}{y^2} = \frac{-2}{3} \left( (1 + \log_e x)x^3 - \frac{x^3}{3} \right) + C$. Using $y(1) = 3$, we find $C = \frac{5}{9}$. Simplifying for $\frac{y^2}{9}$ leads to Option A.

   Step Solution:

    1.  Transform to Bernoulli: Divide the DE by $y^3$ to get $y^{-3} \frac{dy}{dx} - \frac{1}{x} y^{-2} = 1 + \log_e x$.

    2.  Substitution: Let $t = y^{-2} \implies \frac{dt}{dx} = -2y^{-3} \frac{dy}{dx}$. The DE becomes $\frac{dt}{dx} + \frac{2}{x} t = -2(1 + \log_e x)$.

    3.  Integrating Factor: $IF = e^{\int \frac{2}{x} dx} = e^{2 \ln x} = x^2$.

    4.  Integrate: $t \cdot x^2 = \int -2x^2(1 + \log_e x) dx$. Using Integration by Parts, we get $\frac{x^2}{y^2} = -\frac{2}{3}x^3(1 + \log_e x) + \frac{2}{9}x^3 + C$.

    5.  Final Value: Use $y(1)=3$ to find $C = \frac{1}{9} + \frac{2}{3} - \frac{2}{9} = \frac{5}{9}$. Rearranging gives $\frac{y^2}{9} = \frac{x^2}{5 - 2x^3(2 + \log_e x^3)}$.

   Difficulty Level: Hard

   Concept Name: Bernoulli’s Differential Equation, Integration by Parts.

   Short cut solution: Recognize the standard Bernoulli form $y' + Py = Qy^n$ early to jump directly to the linear substitution $t = y^{1-n}$.

 Question 42

   Question: Let $y = y(t)$ be a solution of the differential equation $\frac{dy}{dt} + \alpha y = \gamma e^{-\beta t}$ where, $\alpha > 0, \beta > 0$ and $\gamma > 0$. Then $\lim_{t \to \infty} y(t)$ is:

   Options: 

    A. is 0

    B. does not exist

    C. is 1

    D. is -1

   Correct Answer: A

   Year: 25-Jan-2023 Shift 2

   Solution: This is a First Order Linear Differential Equation. The Integrating Factor is $e^{\alpha t}$. The general solution is $y e^{\alpha t} = \int \gamma e^{-\beta t} e^{\alpha t} dt$, which simplifies to $y = \frac{\gamma}{(\alpha - \beta)e^{\beta t}} + \frac{c}{e^{\alpha t}}$. Since $\alpha, \beta > 0$, as $t \to \infty$, both terms in the expression for $y$ tend to zero.

   Step Solution:

    1.  Identify IF: For $\frac{dy}{dt} + \alpha y = Q$, $IF = e^{\int \alpha dt} = e^{\alpha t}$.

    2.  General Solution: $y \cdot e^{\alpha t} = \int \gamma e^{(\alpha - \beta)t} dt$.

    3.  Integrate: $y \cdot e^{\alpha t} = \frac{\gamma}{\alpha - \beta} e^{(\alpha - \beta)t} + C$.

    4.  Isolate y: $y(t) = \frac{\gamma}{\alpha - \beta} e^{-\beta t} + C e^{-\alpha t}$.

    5.  Limit: Since $\alpha, \beta > 0$, $\lim_{t \to \infty} e^{-\beta t} = 0$ and $\lim_{t \to \infty} e^{-\alpha t} = 0$. Thus, $\lim_{t \to \infty} y(t) = 0$.

   Difficulty Level: Easy

   Concept Name: First Order Linear Differential Equation, Limits at Infinity.

   Short cut solution: In linear equations with positive constant coefficients for $y$ and decaying exponential sources, the transient and steady-state solutions both decay to zero as $t \to \infty$.

 Question 44

   Question: Let $y = y(x)$ be the solution of the differential equation $x \log_e x \frac{dy}{dx} + y = x^2 \log_e x, (x > 1)$. If $y(2) = 2$, then $y(e)$ is equal to:

   Options: 

    A. $\frac{4 + e^2}{4}$

    B. $1 + e^2$

    C. $2 + e^2$

    D. $1 + \frac{e^2}{4}$

   Correct Answer: A

   Year: 29-Jan-2023 Shift 2

   Solution: Dividing the DE by $x \ln x$ gives $\frac{dy}{dx} + \frac{y}{x \ln x} = x$. The Integrating Factor is $\ln x$. The solution is $y \ln x = \int x \ln x dx$. Integrating by parts gives $y \ln x = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C$. Using $y(2)=2$, we find $C = 1$. Evaluating at $x=e$ yields $y(e) = 1 + \frac{e^2}{4}$, which is $\frac{4+e^2}{4}$.

   Step Solution:

    1.  Standardize DE: Divide by $x \ln x$ to get $\frac{dy}{dx} + \frac{1}{x \ln x} y = x$.

    2.  Integrating Factor: $IF = e^{\int \frac{1}{x \ln x} dx} = e^{\ln(\ln x)} = \ln x$.

    3.  General Solution: $y \cdot \ln x = \int x \ln x dx$.

    4.  Integrate by Parts: $\int x \ln x dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C$.

    5.  Find C and Solve: At $x=2, y=2 \implies 2 \ln 2 = 2 \ln 2 - 1 + C \implies C = 1$. At $x=e$, $y(1) = \frac{e^2}{2} - \frac{e^2}{4} + 1 = \frac{e^2+4}{4}$.

   Difficulty Level: Medium

   Concept Name: First Order Linear Differential Equation, Integration by Parts.

   Short cut solution: Recognize that $\frac{1}{x \ln x}$ is the derivative of $\ln(\ln x)$, making the Integrating Factor calculation instantaneous.

 Question 45

   Question: Let the solution curve $y = y(x)$ of the differential equation $\frac{dy}{dx} - \frac{3x^5 \tan^{-1}(x^3)}{(1+x^6)^{3/2}}y = 2x \exp\left(\frac{x^3 - \tan^{-1}(x^3)}{\sqrt{1+x^6}}\right)$ pass through the origin. Then $y(1)$ is equal to?

   Options: 

    A. $\exp\left(\frac{4-\pi}{4\sqrt{2}}\right)$ 

    B. $\exp\left(\frac{\pi-4}{4\sqrt{2}}\right)$ 

    C. $\exp\left(\frac{1-\pi}{4\sqrt{2}}\right)$ 

    D. $\exp\left(\frac{4+\pi}{4\sqrt{2}}\right)$

   Correct Answer: A

   Year: 30-Jan-2023 Shift 1

   Solution: The equation is a linear differential equation. The Integrating Factor (I.F.) is found by integrating $P(x)$, which results in $e^{\frac{\tan^{-1}x^3 - x^3}{\sqrt{1+x^6}}}$. Multiplying the DE by this I.F. simplifies the right side to $2x$. Integrating both sides gives $y \cdot e^{\frac{\tan^{-1}x^3 - x^3}{\sqrt{1+x^6}}} = x^2 + c$. Since the curve passes through the origin, $c=0$. Evaluating at $x=1$ yields $y(1) = e^{\frac{4-\pi}{4\sqrt{2}}}$.

   Step Solution:

    1.  Identify Linear Form: The DE is $\frac{dy}{dx} + P(x)y = Q(x)$.

    2.  Calculate I.F.: $I.F. = e^{\int \frac{-3x^5 \tan^{-1}x^3}{(1+x^6)^{3/2}} dx}$. Using substitution $x^3 = \tan\theta$, this evaluates to $e^{\frac{\tan^{-1}x^3 - x^3}{\sqrt{1+x^6}}}$.

    3.  General Solution: $y \cdot e^{\frac{\tan^{-1}x^3 - x^3}{\sqrt{1+x^6}}} = \int 2x \cdot \exp\left(\frac{x^3 - \tan^{-1}x^3}{\sqrt{1+x^6}}\right) \cdot \exp\left(\frac{\tan^{-1}x^3 - x^3}{\sqrt{1+x^6}}\right) dx = \int 2x dx$.

    4.  Find constant: $y \cdot (I.F.) = x^2 + c$. At $(0,0)$, $0 = 0 + c \implies c = 0$.

    5.  Evaluate y(1): $y(1) \cdot e^{\frac{\pi/4 - 1}{\sqrt{2}}} = 1^2 \implies y(1) = e^{\frac{1 - \pi/4}{\sqrt{2}}} = e^{\frac{4-\pi}{4\sqrt{2}}}$.

   Difficulty Level: Hard

   Concept Name: First Order Linear Differential Equation, Integration by Substitution.

   Short cut solution: Observe that the exponent in $Q(x)$ is the negative of the integral of $P(x)$; therefore, the product $Q(x) \cdot I.F.$ will always simplify to the non-exponential part of $Q(x)$, which is $2x$.

 Question 48

   Question: If $y = y(x)$ is the solution curve of the differential equation $(x^2 - 4) dy - (y^2 - 3y) dx = 0$, $x > 2, y(4) = 3/2$ and the slope of the curve is never zero, then the value of $y(10)$ equals?

   Options: 

    A. $\frac{3}{1+(8)^{1/4}}$ 

    B. $\frac{3}{1+2\sqrt{2}}$ 

    C. $\frac{3}{1-2\sqrt{2}}$ 

    D. $\frac{3}{1-(8)^{1/4}}$

   Correct Answer: A

   Year: 27-Jan-2024 Shift 2

   Solution: The variables are separated to get $\int \frac{dy}{y^2-3y} = \int \frac{dx}{x^2-4}$. Integrating both sides using partial fractions results in $\frac{1}{3} \ln|\frac{y-3}{y}| = \frac{1}{4} \ln|\frac{x-2}{x+2}| + C$. Using the initial condition $y(4)=3/2$, the constant $C$ is found to be $\frac{1}{4} \ln 3$. Substituting $x=10$ and solving for $y$ leads to the final result.

   Step Solution:

    1.  Separate Variables: $\frac{dy}{y(y-3)} = \frac{dx}{(x-2)(x+2)}$.

    2.  Integrate: $\frac{1}{3} \ln|\frac{y-3}{y}| = \frac{1}{4} \ln|\frac{x-2}{x+2}| + C$.

    3.  Find C: Substitute $x=4, y=3/2 \implies \frac{1}{3} \ln|-1| = \frac{1}{4} \ln|2/6| + C \implies 0 = \frac{1}{4} \ln(1/3) + C \implies C = \frac{1}{4} \ln 3$.

    4.  Substitute x=10: $\frac{1}{3} \ln|\frac{y-3}{y}| = \frac{1}{4} \ln(8/12) + \frac{1}{4} \ln 3 = \frac{1}{4} \ln(2/3 \cdot 3) = \ln(2^{1/4})$.

    5.  Solve for y: $|\frac{y-3}{y}| = 2^{3/4} = 8^{1/4}$. Since $y \in (0,3)$, $\frac{3-y}{y} = 8^{1/4} \implies y = \frac{3}{1+8^{1/4}}$.

   Difficulty Level: Medium

   Concept Name: Variable Separable Method.

   Short cut solution: At $y=3/2$, the term $(y-3)/y = -1$. Since $\ln| -1 | = 0$, the constant $C$ is simply the negative of the RHS evaluated at $x=4$.

 Question 49

   Question: Let $y = y(x)$ be the solution of the differential equation $2x^2 y dy - (1 - xy^2) dx = 0$, $x > 0, y(2) = \sqrt{\ln 2}$. Then $xy^2 = \ln(ax)$, and if $u(x) = \exp(x^\beta y^\gamma)$, find $a + \beta - \gamma$.

   Options: 

    A. 1 

    B. -1 

    C. 0 

    D. 3

   Correct Answer: A

   Year: 1-Feb-2023 Shift 2

   Solution: Substitute $y^2 = t$ to transform the equation into a linear differential equation in $t$: $\frac{dt}{dx} + \frac{t}{x} = \frac{1}{x^2}$. The Integrating Factor is $x$. Solving gives $xy^2 = \ln x + C$. Using the condition $y(2) = \sqrt{\ln 2}$ gives $C = \ln 2$, so $xy^2 = \ln(2x)$. Comparing this with the required forms gives $a=2, \beta=1, \gamma=2$.

   Step Solution:

    1.  Linearize: Let $y^2 = t \implies 2y \frac{dy}{dx} = \frac{dt}{dx}$. The DE becomes $x^2 \frac{dt}{dx} = 1 - xt$.

    2.  Standard Form: $\frac{dt}{dx} + \frac{1}{x} t = \frac{1}{x^2}$.

    3.  I.F. and Integration: $I.F. = e^{\int 1/x dx} = x$. Then $t \cdot x = \int \frac{1}{x^2} \cdot x dx = \ln x + C$.

    4.  Find Constant: $y^2 x = \ln x + C$. At $x=2, y^2 = \ln 2 \implies 2 \ln 2 = \ln 2 + C \implies C = \ln 2$.

    5.  Final Parameters: $xy^2 = \ln(2x)$. Here $a=2$. For $u(x) = e^{x^1 y^2}$, $\beta=1, \gamma=2$. Result: $2 + 1 - 2 = 1$.

   Difficulty Level: Medium

   Concept Name: Bernoulli’s Form (Reducible to Linear).

   Short cut solution: Recognize $2x^2 y dy + xy^2 dx = dx$ as $x d(xy^2) = dx$, which integrates directly to $xy^2 = \ln x + C$.

 Question 50

   Question: Let $y = y(x)$ be a solution of the differential equation $(x \cos x) dy + (xy \sin x + y \cos x - 1) dx = 0, 0 < x < \pi/2$. If $y(\frac{\pi}{3}) = \frac{\sqrt{3}}{\pi}$, then $|\frac{\pi}{6} y''(\frac{\pi}{6}) + 2y'(\frac{\pi}{6})|$ is equal to?

   Options: (None provided in the source)

   Correct Answer: 2

   Year: 6-Apr-2023 Shift 1

   Solution: The equation is rewritten as $\frac{dy}{dx} + \left( \frac{x \sin x + \cos x}{x \cos x} \right) y = \frac{1}{x \cos x}$. The Integrating Factor (I.F.) is $x \sec x$. Multiplying by the I.F. and integrating gives $y(x \sec x) = \tan x + c$. Given the condition at $x = \pi/3$, the constant $c$ is found to be $\sqrt{3}$. The final expression evaluates to 2 after differentiation.

   Step Solution:

    1.  Standardize DE: Divide by $x \cos x$ to get $\frac{dy}{dx} + (\tan x + \frac{1}{x})y = \frac{1}{x \cos x}$.

    2.  Integrating Factor: Calculate $I.F. = e^{\int (\tan x + 1/x) dx} = e^{\ln \sec x + \ln x} = x \sec x$.

    3.  General Solution: Solve $y(x \sec x) = \int \sec^2 x dx = \tan x + c$, which simplifies to $xy = \sin x + c \cos x$.

    4.  Find Constant: Substitute $y(\frac{\pi}{3}) = \frac{\sqrt{3}}{\pi} \implies \frac{\pi}{3} \cdot \frac{\sqrt{3}}{\pi} = \frac{\sqrt{3}}{2} + c \cdot \frac{1}{2}$, giving $c = \sqrt{3}$.

    5.  Calculate Final Value: Differentiate $xy = \sin x + \sqrt{3} \cos x$ twice to find $y'$ and $y''$ at $x = \pi/6$; the required expression modulus results in 2.

   Difficulty Level: Hard

   Concept Name: First Order Linear Differential Equation

   Short cut solution: Recognize that $xy = \sin x + \sqrt{3} \cos x = 2 \sin(x + \frac{\pi}{3})$, which simplifies the derivatives significantly.

 Question 51

   Question: If the solution curve $f(x, y) = 0$ of the differential equation $(1 + \log_e x) \frac{dx}{dy} - x \log_e x = e^y, x > 0$, passes through the points (1, 0) and $(\alpha, 2)$, then $\alpha^\alpha$ is equal to?

   Options: 

    A. $e^{2e^2}$

    B. $e^{e^2}$

    C. $2e^{\sqrt{2}}$

    D. $e^{2e^2}$

   Correct Answer: D

   Year: 6-Apr-2023 Shift 2

   Solution: Let $x \ln x = t$. Then $(1 + \ln x) \frac{dx}{dy} = \frac{dt}{dy}$. The equation becomes $\frac{dt}{dy} - t = e^y$. This is a linear differential equation in $t$. The I.F. is $e^{-y}$. Solving gives $t e^{-y} = y + c$. Using the point (1, 0) gives $c = 0$. For the point $(\alpha, 2)$, we find $\alpha^\alpha = e^{2e^2}$.

   Step Solution:

    1.  Substitution: Let $t = x \ln x \implies \frac{dt}{dy} = (1 + \ln x) \frac{dx}{dy}$.

    2.  Linear Form: Rewrite the DE as $\frac{dt}{dy} - t = e^y$.

    3.  I.F. and Integration: $I.F. = e^{\int -1 dy} = e^{-y}$. Then $t e^{-y} = \int e^y e^{-y} dy = y + c$.

    4.  Find c: Pass through (1, 0): $x = 1 \implies t = 0$. Thus $0 = 0 + c \implies c = 0$.

    5.  Final Result: At $(\alpha, 2)$, $\alpha \ln \alpha \cdot e^{-2} = 2 \implies \alpha \ln \alpha = 2e^2 \implies \ln \alpha^\alpha = 2e^2 \implies \alpha^\alpha = e^{2e^2}$.

   Difficulty Level: Medium

   Concept Name: First Order Linear Differential Equation (Reducible by substitution)

   Short cut solution: The substitution $t = x \ln x$ is a standard trick for terms like $(1 + \ln x) dx$ which appears in the DE.

 Question 52

   Question: Let the solution curve $x = x(y), 0 < y < \pi/2$ of the differential equation $(\log_e(\cos y))^2 \cos y dx - (1 + 3x \log_e(\cos y)) \sin y dy = 0$ satisfy $x(\frac{\pi}{3}) = \frac{1}{2 \log_e 2}$. If $x(\frac{\pi}{6}) = \frac{1}{\log_e m - \log_e n}$, where $m$ and $n$ are co-prime, then $mn$ is equal to?

   Options: (None provided in the source)

   Correct Answer: 12

   Year: 8-Apr-2023 Shift 2

   Solution: The equation is rearranged to $\frac{dx}{dy} - \left( \frac{3 \tan y}{\ln \cos y} \right) x = \frac{\tan y}{\ln^2 \cos y}$. Using the Integrating Factor $(\ln \cos y)^3$, the solution becomes $x \ln^3 \cos y = -\frac{1}{2} \ln^2 \cos y + c$. Using the initial condition, $c = 0$. Evaluating at $y = \pi/6$ gives $x = \frac{1}{\ln(4/3)}$, leading to $mn = 12$.

   Step Solution:

    1.  Rearrange DE: $\frac{dx}{dy} = \tan y \left( \frac{3x}{\ln \cos y} + \frac{1}{\ln^2 \cos y} \right) \implies \frac{dx}{dy} - \frac{3 \tan y}{\ln \cos y} x = \frac{\tan y}{\ln^2 \cos y}$.

    2.  Integrating Factor: $I.F. = e^{\int \frac{-3 \tan y}{\ln \cos y} dy}$. Let $t = \ln \cos y \implies dt = -\tan y dy$. $I.F. = e^{\int \frac{3}{t} dt} = t^3 = \ln^3 \cos y$.

    3.  Integrate: $x \ln^3 \cos y = \int \frac{\tan y}{\ln^2 \cos y} \ln^3 \cos y dy = \int \tan y \ln \cos y dy = -\frac{\ln^2 \cos y}{2} + c$.

    4.  Find Constant: Substitute $y = \pi/3, \cos y = 1/2$. $\frac{1}{2 \ln 2} (-\ln 2)^3 = -\frac{(-\ln 2)^2}{2} + c \implies c = 0$.

    5.  Final Result: At $y = \pi/6$, $x = \frac{-1}{2 \ln \cos(\pi/6)} = \frac{-1}{2 \ln(\sqrt{3}/2)} = \frac{1}{\ln(4/3)}$. Thus $m=4, n=3, mn=12$.

   Difficulty Level: Hard

   Concept Name: First Order Linear Differential Equation, Substitution Method

   Short cut solution: Once you find $c=0$, the relation $x = \frac{-1}{2 \ln \cos y}$ is very clean for direct evaluation at any $y$.

 Question 57

   Question: Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} + \frac{5}{x(x^5 + 1)} y = \frac{(x^5 + 1)^2}{x^7}, x > 0$. If $y(1) = 2$, then $y(2)$ is equal to?

   Options: 

    A. 693/128

    B. 637/128

    C. 697/128

    D. 679/128

   Correct Answer: A

   Year: 11-Apr-2023 shift 2

   Solution: The Integrating Factor is $I.F = e^{\int \frac{5}{x(x^5+1)} dx} = e^{\int \frac{5x^{-6}}{x^{-5}+1} dx} = \frac{x^5}{1+x^5}$. The solution to the DE is $y \cdot \frac{x^5}{1+x^5} = \int \frac{(1+x^5)^2}{x^7} \cdot \frac{x^5}{1+x^5} dx = \int (x^{-2} + x^3) dx = \frac{x^4}{4} - \frac{1}{x} + c$. Using $y(1)=2$, $c$ is found. Substituting $x=2$ gives $y(2) = 693/128$.

   Step Solution:

    1.  Integrating Factor: Calculate $I.F. = e^{\int \frac{5}{x(x^5+1)} dx}$. Using partial fractions or $x^{-5}$ substitution, this simplifies to $\frac{x^5}{x^5+1}$.

    2.  General Solution: $y \cdot \frac{x^5}{x^5+1} = \int \frac{(x^5+1)^2}{x^7} \cdot \frac{x^5}{x^5+1} dx = \int \frac{x^5+1}{x^2} dx = \int (x^3 + x^{-2}) dx$.

    3.  Integrate: $y \cdot \frac{x^5}{x^5+1} = \frac{x^4}{4} - \frac{1}{x} + C$.

    4.  Find C: Substitute $y(1)=2 \implies 2(\frac{1}{2}) = \frac{1}{4} - 1 + C \implies 1 = -0.75 + C \implies C = \frac{7}{4}$.

    5.  Evaluate y(2): $y(2) \cdot \frac{32}{33} = \frac{16}{4} - \frac{1}{2} + \frac{7}{4} = \frac{21}{4} \implies y(2) = \frac{21}{4} \cdot \frac{33}{32} = \frac{693}{128}$.

   Difficulty Level: Hard

   Concept Name: First Order Linear Differential Equation

   Short cut solution: Use the substitution $x^5 = t$ early in the Integrating Factor calculation to avoid complex partial fractions.

 Question 58

   Question: Let $y = y(x), y > 0$, be a solution curve of the differential equation $(1 + x^2) dy = y(x - y) dx$. If $y(0) = 1$ and $y(2\sqrt{2}) = \beta$, then?

   Options: 

    A. $e^{3\beta^{-1}} = e(5 + \sqrt{2})$

    B. $e^{3\beta^{-1}} = e(3 + 2\sqrt{2})$

    C. $e^{\beta^{-1}} = e^{-2}(3 + 2\sqrt{2})$

    D. $e^{\beta^{-1}} = e^{-2}(5 + \sqrt{2})$

   Correct Answer: B

   Year: 12-Apr-2023 shift 1

   Solution: Rewriting the equation gives $\frac{dy}{dx} - \frac{x}{1+x^2}y = -\frac{y^2}{1+x^2}$, which is Bernoulli's equation. Substituting $1/y = t$ transforms it into the linear form $\frac{dt}{dx} + \frac{x}{1+x^2} t = \frac{1}{1+x^2}$. The I.F. is $\sqrt{1+x^2}$. Integrating leads to $t \sqrt{1+x^2} = \ln|x + \sqrt{1+x^2}| + C$. Using $y(0)=1$, $C=1$. Substituting $x=2\sqrt{2}$ gives the result for $\beta$.

   Step Solution:

    1.  Bernoulli Form: Rearrange to $\frac{dy}{dx} - \frac{x}{1+x^2}y = -\frac{y^2}{1+x^2}$.

    2.  Linearize: Substitute $t = \frac{1}{y} \implies \frac{dt}{dx} + \frac{x}{1+x^2} t = \frac{1}{1+x^2}$.

    3.  I.F. and Integration: $I.F. = \sqrt{1+x^2}$. Then $t \sqrt{1+x^2} = \int \frac{\sqrt{1+x^2}}{1+x^2} dx = \int \frac{1}{\sqrt{1+x^2}} dx = \ln(x + \sqrt{1+x^2}) + C$.

    4.  Condition: $y(0)=1 \implies t(0)=1$. Thus $1 \cdot 1 = \ln(1) + C \implies C = 1$.

    5.  Final Value: At $x=2\sqrt{2}, \sqrt{1+x^2}=3$. So $3\beta^{-1} = \ln(2\sqrt{2}+3) + 1$. Exponential form: $e^{3\beta^{-1}} = e(3+2\sqrt{2})$.

   Difficulty Level: Hard

   Concept Name: Bernoulli’s Differential Equation

   Short cut solution: Recognize that $x + \sqrt{1+x^2}$ and its reciprocal appear frequently in these types of calculus problems, allowing for faster integration of the $1/\sqrt{1+x^2}$ term.

 Question 60

   Question: If $y = y(x)$ is the solution of the differential equation $\frac{dy}{dx} + \frac{4x}{(x^2 - 1)} y = \frac{x + 2}{(x^2 - 1)^{5/2}}, x > 1$ such that $y(2) = \frac{2}{9} \log_e(2 + \sqrt{3})$ and $y(\sqrt{2}) = \alpha \log_e(\sqrt{\eta} \alpha + \beta) + \beta - \sqrt{\gamma}, \alpha, \beta, \gamma \in N$, then $\alpha\beta\gamma$ is equal to?

   Options: (None listed; numerical answer provided)

   Correct Answer: 6

   Year: 13-Apr-2023 shift 2

   Solution: This is a linear differential equation. $I.F = (x^2-1)^2$. Multiplying and integrating gives $y(x^2-1)^2 = \int \frac{x+2}{\sqrt{x^2-1}} dx = \sqrt{x^2-1} + 2\ln[x+\sqrt{x^2-1}] + C$. Given the condition at $x=2$, $C = -\sqrt{3}$. Evaluating at $x=\sqrt{2}$ gives $\alpha=2, \beta=1, \gamma=3$, so $\alpha\beta\gamma = 6$.

   Step Solution:

    1.  Find I.F: $I.F. = e^{\int \frac{4x}{x^2-1} dx} = e^{2\ln(x^2-1)} = (x^2-1)^2$.

    2.  Integrate: $y(x^2-1)^2 = \int \frac{x+2}{(x^2-1)^{5/2}} (x^2-1)^2 dx = \int \frac{x+2}{\sqrt{x^2-1}} dx$.

    3.  Solve Integral: $y(x^2-1)^2 = \sqrt{x^2-1} + 2\ln|x+\sqrt{x^2-1}| + C$.

    4.  Find Constant: $y(2) = \frac{2}{9}\ln(2+\sqrt{3}) \implies \frac{2}{9}\ln(2+\sqrt{3}) \cdot 9 = \sqrt{3} + 2\ln(2+\sqrt{3}) + C \implies C = -\sqrt{3}$.

    5.  Compute Product: At $x = \sqrt{2}$, $y(1)^2 = 1 + 2\ln(\sqrt{2}+1) - \sqrt{3}$. Thus $\alpha=2, \beta=1, \gamma=3$. $\alpha\beta\gamma = 6$.

   Difficulty Level: Hard

   Concept Name: First Order Linear Differential Equation

   Short cut solution: Directly split the integral $\int \frac{x+2}{\sqrt{x^2-1}} dx$ into the derivative form $\frac{x}{\sqrt{x^2-1}}$ and the standard log form $\frac{2}{\sqrt{x^2-1}}$ to save time.

 Question 61

   Question: Let $x = x(y)$ be the solution of the differential equation $2(y+2)\log_e(y+2) dx + (x+4-2\log_e(y+2)) dy = 0, y > -1$ with $x(e^4-2) = 1$. Then $x(e^9-2)$ is equal to?

   Options: 

    A. $4/9$

    B. $32/9$

    C. $10/3$

    D. $3$

   Correct Answer: B

   Year: 15-Apr-2023 shift 1

   Solution: The equation is rearranged into a linear differential equation in $x$: $\frac{dx}{dy} + \frac{1}{2(y+2)\ln(y+2)}x = \frac{1}{y+2} - \frac{2}{(y+2)\ln(y+2)}$. The Integrating Factor (I.F.) is $\sqrt{\ln(y+2)}$. Multiplying and integrating leads to $x\sqrt{\ln(y+2)} = \frac{2}{3}(\ln(y+2))^{3/2} - 4\sqrt{\ln(y+2)} + C$. Using the initial condition $x(e^4-2)=1$, $C$ is found to be $14/3$. Evaluating at $y = e^9-2$ gives $x = 32/9$.

   Step Solution:

    1.  Standardize DE: Rearrange as $\frac{dx}{dy} + \frac{x}{2(y+2)\ln(y+2)} = \frac{2\ln(y+2)-4}{2(y+2)\ln(y+2)} = \frac{1}{y+2} - \frac{2}{(y+2)\ln(y+2)}$.

    2.  Integrating Factor: Calculate $I.F. = e^{\int \frac{1}{2(y+2)\ln(y+2)} dy} = \sqrt{\ln(y+2)}$.

    3.  General Solution: Integrate $x \sqrt{\ln(y+2)} = \int [ \frac{\sqrt{\ln(y+2)}}{y+2} - \frac{2}{(y+2)\sqrt{\ln(y+2)}} ] dy$ using substitution $t = \ln(y+2)$ to get $\frac{2}{3}t^{3/2} - 4\sqrt{t} + C$.

    4.  Find C: Substitute $x=1, y=e^4-2 \implies 1\sqrt{4} = \frac{2}{3}(4)^{3/2} - 4\sqrt{4} + C \implies 2 = \frac{16}{3} - 8 + C \implies C = \frac{14}{3}$.

    5.  Evaluate: For $y = e^9-2$ ($\ln(y+2)=9$), $x\sqrt{9} = \frac{2}{3}(27) - 4(3) + \frac{14}{3} \implies 3x = 18 - 12 + \frac{14}{3} \implies x = \frac{32}{9}$.

   Difficulty Level: Hard

   Concept Name: First Order Linear Differential Equation (in $x$).

   Shortcut Solution: Identify the substitution $t = \ln(y+2)$ immediately to convert the DE into the simple form $2t \frac{dx}{dt} + x = 2t - 4$.

 Question 63

   Question: Let $y = y(x)$ be the solution of the differential equation $(x + 1)y' - y = e^{3x}(x + 1)^2$, with $y(0) = \frac{1}{3}$. Then, the point $x = -\frac{4}{3}$ for the curve $y = y(x)$ is?

   Options: 

    A. not a critical point 

    B. a point of local minima 

    C. a point of local maxima

    D. a point of inflection

   Correct Answer: B

   Year: 25-Jun-2022-Shift-1

   Solution: Dividing by $(x+1)$ gives the linear DE $\frac{dy}{dx} - \frac{1}{x+1}y = e^{3x}(x+1)$. The I.F. is $\frac{1}{x+1}$. Integrating both sides leads to $\frac{y}{x+1} = \frac{e^{3x}}{3} + C$. Using $y(0)=1/3$ gives $C=0$. The function is $y = \frac{x+1}{3}e^{3x}$. Its derivative $y'$ is zero at $x = -4/3$, and the second derivative $y''$ is positive at that point, indicating a local minima.

   Step Solution:

    1.  Linear Form: Rewrite as $\frac{dy}{dx} - \frac{1}{x+1} y = e^{3x}(x+1)$.

    2.  Integrating Factor: $I.F. = e^{\int \frac{-1}{x+1} dx} = e^{-\ln(x+1)} = \frac{1}{x+1}$.

    3.  General Solution: $y \cdot \frac{1}{x+1} = \int e^{3x}(x+1) \cdot \frac{1}{x+1} dx = \int e^{3x} dx = \frac{e^{3x}}{3} + C$.

    4.  Find C: Substitute $y(0)=1/3 \implies \frac{1/3}{1} = \frac{1}{3} + C \implies C = 0$. So $y = \frac{x+1}{3}e^{3x}$.

    5.  Analyze Point: $y' = \frac{e^{3x}}{3} + (x+1)e^{3x} = e^{3x}(x + \frac{4}{3})$. At $x = -4/3, y' = 0$. $y'' = e^{3x}(3x+5)$; at $x = -4/3, y'' = e^{-4}(1) > 0$. Therefore, it is a local minima.

   Difficulty Level: Medium

   Concept Name: First Order Linear Differential Equation, Maxima and Minima.

   Shortcut Solution: Notice that dividing the DE by $(x+1)^2$ transforms the LHS into the exact derivative $\frac{d}{dx}(\frac{y}{x+1})$, making integration instantaneous.

 Question 66

   Question: Let the solution curve $y = y(x)$ of the differential equation $(4 + x^2) dy - 2x(x^2 + 3y + 4) dx = 0$ pass through the origin. Then $y(2)$ is equal to?

   Options: (None provided in the source)

   Correct Answer: 12

   Year: 26-Jun-2022-Shift-1

   Solution: The DE is rewritten as $\frac{dy}{dx} - \frac{6x}{x^2+4}y = 2x$. The Integrating Factor is $\frac{1}{(x^2+4)^3}$. Multiplying and integrating gives $\frac{y}{(x^2+4)^3} = \int \frac{2x}{(x^2+4)^3} dx = \frac{-1}{2(x^2+4)^2} + C$. Using the origin $(0,0)$ gives $C = 1/32$. Evaluating at $x=2$ results in $y=12$.

   Step Solution:

    1.  Linear Form: Rearrange to $\frac{dy}{dx} = \frac{2x(x^2+4) + 6xy}{x^2+4} \implies \frac{dy}{dx} - \frac{6x}{x^2+4}y = 2x$.

    2.  Integrating Factor: $I.F. = e^{\int \frac{-6x}{x^2+4} dx} = e^{-3 \ln(x^2+4)} = \frac{1}{(x^2+4)^3}$.

    3.  General Solution: $y \cdot \frac{1}{(x^2+4)^3} = \int \frac{2x}{(x^2+4)^3} dx = -\frac{1}{2}(x^2+4)^{-2} + C$.

    4.  Find C: At $(0,0)$, $0 = -\frac{1}{2(16)} + C \implies C = \frac{1}{32}$.

    5.  Evaluate: At $x=2, y \cdot \frac{1}{8^3} = -\frac{1}{2(8^2)} + \frac{1}{32} \implies \frac{y}{512} = -\frac{1}{128} + \frac{4}{128} = \frac{3}{128} \implies y = \frac{3 \cdot 512}{128} = 12$.

   Difficulty Level: Hard

   Concept Name: First Order Linear Differential Equation.

   Shortcut Solution: Recognize that the term $\frac{2x}{(x^2+4)}$ is the derivative of $\ln(x^2+4)$, making the power rule for the Integrating Factor integration very fast.

 Question 68

   Question: Let $y = y(x)$ satisfy the differential equation $\sec^2 x \frac{dx}{dy} + e^{2y} \tan^2 x + \tan x = 0$ for $0 < x < \frac{\pi}{2}$ with $y(\frac{\pi}{4}) = 0$. If $y(\frac{\pi}{6}) = a$, then $e^{8a}$ is equal to?

   Options: (None provided in the source)

   Correct Answer: 9

   Year: 31-Jan-2024 Shift 2

   Solution: Let $\tan x = t$. Then the equation becomes $\frac{dt}{dy} + t = -e^{2y} t^2$, which is Bernoulli's equation. Substituting $u = 1/t$ converts it to the linear form $\frac{du}{dy} - u = e^{2y}$. The Integrating Factor is $e^{-y}$, leading to the solution $\frac{e^{-y}}{\tan x} = e^y + c$. Using $y(\frac{\pi}{4})=0$ gives $c=0$. At $x=\frac{\pi}{6}$, $e^{2a} = \sqrt{3}$, so $e^{8a} = 9$.

   Step Solution:

    1.  Substitution: Let $t = \tan x \implies \sec^2 x \frac{dx}{dy} = \frac{dt}{dy}$. The DE becomes $\frac{dt}{dy} + t = -t^2 e^{2y}$.

    2.  Linearize: Divide by $t^2$ and let $u = t^{-1}$, yielding the linear DE: $\frac{du}{dy} - u = e^{2y}$.

    3.  Integrating Factor: $I.F. = e^{\int -1 dy} = e^{-y}$.

    4.  Integrate: $u \cdot e^{-y} = \int e^{2y} \cdot e^{-y} dy = e^y + c \implies \frac{1}{\tan x \cdot e^y} = e^y + c$.

    5.  Final Value: $y(\frac{\pi}{4})=0 \implies 1 = 1 + c \implies c = 0$. At $x = \frac{\pi}{6}$, $\sqrt{3} = e^{2a}$. Squaring twice gives $e^{8a} = 9$.

   Difficulty Level: Hard

   Concept Name: Bernoulli’s Differential Equation

   Shortcut Solution: Recognize $c=0$ immediately from the initial condition to get the clean relationship $\cot x = e^{2y}$.

 Question 72

   Question: Let $y = y(x)$ be the solution of the differential equation $(1 - x^2) dy = (xy + (x^3 + 2)\sqrt{1 - x^2}) dx, -1 < x < 1, y(0) = 0$. If $\int_{-1/2}^{1/2} \sqrt{1 - x^2} y(x) dx = k$, then $k^{-1}$ is equal to?

   Options: (None provided in the source)

   Correct Answer: 320

   Year: 27-Jun-2022-Shift-2

   Solution: The equation is rearranged to $\frac{dy}{dx} - \frac{x}{1-x^2} y = \frac{x^3 + 2}{\sqrt{1-x^2}}$. Using the Integrating Factor $\sqrt{1-x^2}$, the general solution becomes $y \sqrt{1-x^2} = \frac{x^4}{4} + 2x + c$. With $y(0)=0$, $c=0$. Integrating the function over $[-1/2, 1/2]$ gives $k = 1/320$, hence $k^{-1} = 320$.

   Step Solution:

    1.  Linear Form: Rewrite as $\frac{dy}{dx} - \frac{x}{1-x^2} y = \frac{x^3 + 2}{\sqrt{1-x^2}}$.

    2.  Integrating Factor: $I.F. = e^{\int \frac{-x}{1-x^2} dx} = e^{\frac{1}{2} \ln(1-x^2)} = \sqrt{1-x^2}$.

    3.  General Solution: $y \cdot \sqrt{1-x^2} = \int (x^3 + 2) dx = \frac{x^4}{4} + 2x + c$.

    4.  Determine c: $y(0)=0 \implies 0 \cdot 1 = 0 + c \implies c = 0$.

    5.  Evaluate k: $\int_{-1/2}^{1/2} (\frac{x^4}{4} + 2x) dx = [\frac{x^5}{20} + x^2]_{-1/2}^{1/2} = \frac{1/32}{20} - \frac{-1/32}{20} = \frac{1}{320}$.

   Difficulty Level: Hard

   Concept Name: First Order Linear Differential Equation, Definite Integral Properties.

   Shortcut Solution: Recognize $2x$ as an odd function to immediately set its integral over the symmetric interval $[-1/2, 1/2]$ to zero.

Question 74

   Question: Let $y = y(x)$ be the solution of the differential equation $x(1 - x^2) \frac{dy}{dx} + (3x^2y - y - 4x^3) = 0, x > 1$, with $y(2) = -2$. Then $y(3)$ is equal to?

   Options: A. −18, B. −12, C. −6, D. −3

   Correct Answer: A

   Year: 28-Jun-2022-Shift-1

   Solution: The equation is standardized to $\frac{dy}{dx} + \frac{3x^2-1}{x(1-x^2)} y = \frac{4x^2}{1-x^2}$. The Integrating Factor is found to be $\frac{1}{x^3-x}$. Integrating leads to $\frac{y}{x^3-x} = \frac{2}{x^2-1} + c$. Using the condition $y(2)=-2$ gives $c = -1$. Evaluating for $x=3$ yields $y = -18$.

   Step Solution:

    1.  Rearrange DE: $\frac{dy}{dx} + \frac{3x^2-1}{x(1-x^2)}y = \frac{4x^3}{x(1-x^2)} \implies \frac{dy}{dx} - \frac{3x^2-1}{x(x^2-1)}y = \frac{-4x^2}{x^2-1}$.

    2.  Integrating Factor: $I.F. = \exp(\int \frac{1-3x^2}{x(x^2-1)} dx) = \frac{1}{x(x^2-1)} = \frac{1}{x^3-x}$.

    3.  Integrate: $\frac{y}{x^3-x} = \int \frac{-4x^2}{x^2-1} \cdot \frac{1}{x(x^2-1)} dx = \int \frac{-4x}{(x^2-1)^2} dx = \frac{2}{x^2-1} + c$.

    4.  Find c: $y(2)=-2 \implies \frac{-2}{6} = \frac{2}{3} + c \implies c = -1$.

    5.  Evaluate: At $x=3$, $\frac{y}{24} = \frac{2}{8} - 1 = -\frac{3}{4} \implies y = 24(-\frac{3}{4}) = -18$.

   Difficulty Level: Hard

   Concept Name: First Order Linear Differential Equation, Integration by Substitution.

   Shortcut Solution: Dividing the original equation by $(x^3-x)^2$ allows one to identify the exact derivative $\frac{d}{dx}(\frac{y}{x^3-x})$.

 Question 77

   Question: Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} + \frac{\sqrt{2}y}{2\cos^4 x - \cos^2 x} = xe^{\tan^{-1}(\sqrt{2}\cot 2x)}, 0 < x < \frac{\pi}{2}$ with $y\left(\frac{\pi}{4}\right) = \frac{\pi^2}{32}$. If $y\left(\frac{\pi}{3}\right) = \frac{\pi^2}{18}e^{-\tan^{-1}(\alpha)}$, then the value of $3\alpha^2$ is equal to?

   Options: (None provided in the source)

   Correct Answer: 2

   Year: 29-Jun-2022-Shift-1

   Solution: The equation is a linear differential equation. The Integrating Factor (I.F.) is found by integrating $P(x)$, resulting in $e^{-\tan^{-1}(\sqrt{2}\cos 2x)}$. Multiplying and integrating both sides leads to $y \cdot e^{-\tan^{-1}(\sqrt{2}\cos 2x)} = \frac{x^2}{2} + c$. Using $y(\frac{\pi}{4}) = \frac{\pi^2}{32}$, the constant $c$ is found to be 0. Evaluating at $x = \frac{\pi}{3}$ yields $\alpha = \sqrt{2/3}$, so $3\alpha^2 = 2$.

   Step Solution:

    1.  Identify I.F.: Calculate $I.F. = \exp\left(\int \frac{\sqrt{2}}{2\cos^4 x - \cos^2 x} dx\right) = e^{-\tan^{-1}(\sqrt{2}\cos 2x)}$.

    2.  General Solution: $y \cdot e^{-\tan^{-1}(\sqrt{2}\cos 2x)} = \int x e^{\tan^{-1}(\sqrt{2}\cot 2x)} \cdot e^{-\tan^{-1}(\sqrt{2}\cos 2x)} dx = \int x dx$.

    3.  Integrate: $y \cdot e^{-\tan^{-1}(\sqrt{2}\cos 2x)} = \frac{x^2}{2} + c$.

    4.  Find c: Substitute $x = \frac{\pi}{4}, y = \frac{\pi^2}{32} \implies \frac{\pi^2}{32} \cdot e^0 = \frac{\pi^2}{32} + c \implies c = 0$.

    5.  Calculate $\alpha$: At $x = \frac{\pi}{3}$, $\cos 2x = -1/2$. Thus $y = \frac{\pi^2}{18} e^{-\tan^{-1}(\sqrt{2/3})}$. Comparing gives $\alpha = \sqrt{2/3}$, hence $3\alpha^2 = 2$.

   Difficulty Level: Hard

   Concept Name: First Order Linear Differential Equation

   Shortcut Solution: Recognize that the exponent in $Q(x)$ and the $I.F.$ are designed to cancel out, leaving only a simple polynomial integration of $x$.

 Question 79

   Question: Let $y = y(x), x > 1$, be the solution of the differential equation $(x - 1) \frac{dy}{dx} + 2xy = \frac{1}{x-1}$, with $y(2) = \frac{1 + e^4}{2e^4}$. If $y(3) = \frac{e^a + 1}{\beta e^a}$, then the value of $a + \beta$ is equal to?

   Options: (None provided in the source)

   Correct Answer: 14

   Year: 29-Jun-2022-Shift-2

   Solution: Rewriting the DE as $\frac{dy}{dx} + \frac{2x}{x-1}y = \frac{1}{(x-1)^2}$. The Integrating Factor is $(x-1)^2 e^{2x}$. Solving the DE gives $y(x-1)^2 e^{2x} = \frac{e^{2x}}{2} + c$. Using the initial condition at $x=2$, we find $c = 1/2$. Evaluating for $x=3$ yields $y(3) = \frac{e^6+1}{8e^6}$, identifying $a=6$ and $\beta=8$, thus $a+\beta = 14$.

   Step Solution:

    1.  Standard Form: Divide by $(x-1)$ to get $\frac{dy}{dx} + \frac{2x}{x-1}y = \frac{1}{(x-1)^2}$.

    2.  Integrating Factor: Calculate $I.F. = e^{\int \frac{2x}{x-1} dx} = e^{\int (2 + \frac{2}{x-1}) dx} = (x-1)^2 e^{2x}$.

    3.  General Solution: $y(x-1)^2 e^{2x} = \int \frac{1}{(x-1)^2} (x-1)^2 e^{2x} dx = \int e^{2x} dx = \frac{e^{2x}}{2} + c$.

    4.  Find c: $y(2) = \frac{1+e^4}{2e^4} \implies \frac{1+e^4}{2e^4}(1)^2 e^4 = \frac{e^4}{2} + c \implies \frac{1+e^4}{2} = \frac{e^4}{2} + c \implies c = \frac{1}{2}$.

    5.  Evaluate: At $x=3$, $y(4)e^6 = \frac{e^6}{2} + \frac{1}{2} = \frac{e^6+1}{2} \implies y = \frac{e^6+1}{8e^6}$. So $a=6, \beta=8$. Sum is 14.

   Difficulty Level: Hard

   Concept Name: First Order Linear Differential Equation

   Shortcut Solution: Use the substitution $x-1 = t$ to simplify the $P(x)$ term and the Integrating Factor integral significantly.

 Question 80

   Question: If $x = x(y)$ is the solution of the differential equation $y \frac{dx}{dy} = 2x + y^3(y + 1)e^y$, $x(1) = 0$; then $x(e)$ is equal to?

   Options: 

    A. $e^3(e^e - 1)$

    B. $e^e(e^3 - 1)$

    C. $e^2(e^e + 1)$

    D. $e^e(e^2 - 1)$

   Correct Answer: A

   Year: 24-Jun-2022-Shift-1

   Solution: The equation is a linear differential equation in $x$: $\frac{dx}{dy} - \frac{2x}{y} = y^2(y+1)e^y$. The Integrating Factor is $1/y^2$. Multiplying and integrating gives $\frac{x}{y^2} = ye^y + c$. Applying $x(1)=0$ results in $c = -e$. Evaluating at $y=e$ gives $x = e^3(e^e-1)$.

   Step Solution:

    1.  Linear Form: Rearrange to $\frac{dx}{dy} - \frac{2}{y}x = y^2(y+1)e^y$.

    2.  Integrating Factor: $I.F. = e^{\int -\frac{2}{y} dy} = e^{-2 \ln y} = \frac{1}{y^2}$.

    3.  General Solution: $\frac{x}{y^2} = \int y^2(y+1)e^y \cdot \frac{1}{y^2} dy = \int (y+1)e^y dy = ye^y + c$.

    4.  Find c: $x(1)=0 \implies 0/1 = 1e^1 + c \implies c = -e$. So, $x = y^2(ye^y - e)$.

    5.  Final Value: At $y=e$, $x(e) = e^2(e \cdot e^e - e) = e^3(e^e - 1)$.

   Difficulty Level: Medium

   Concept Name: First Order Linear Differential Equation (in $x$)

   Shortcut Solution: Recognize that the integral of $(y+1)e^y$ is the standard form for the derivative of $ye^y$, skipping the need for manual integration by parts.

 Question 86

   Question: If $\frac{dy}{dx} + 2y \tan x = \sin x, 0 < x < \frac{\pi}{2}$ and $y\left( \frac{\pi}{3} \right) = 0$, then the maximum value of $y(x)$ is :

   Options:

    A. $\frac{1}{8}$

    B. $\frac{3}{4}$

    C. $\frac{1}{4}$

    D. $\frac{3}{8}$

   Correct Answer: A

   Year: 26-Jul-2022-Shift-1

   Solution: $\frac{dy}{dx} + 2y \tan x = \sin x$ is a first order linear differential equation. Integrating factor (I.F.) = $e^{\int 2 \tan x dx} = e^{2 \ln |\sec x|} = \sec^2 x$. The solution is $y \cdot \sec^2 x = \int \sin x \cdot \sec^2 x dx = \int \sec x \cdot \tan x dx = \sec x + C$. Given $y(\pi/3) = 0$, we find $0 = \sec(\pi/3) + C \implies C = -2$. Thus $y = \frac{\sec x - 2}{\sec^2 x} = \cos x - 2 \cos^2 x$. Completing the square gives $y = \frac{1}{8} - 2(\cos x - \frac{1}{4})^2$. The maximum value is $\frac{1}{8}$.

   Step Solution:

    1.  Integrating Factor: Calculate $I.F. = e^{\int 2 \tan x dx} = \sec^2 x$.

    2.  General Solution: $y \cdot \sec^2 x = \int \sin x \sec^2 x dx = \int \sec x \tan x dx = \sec x + C$.

    3.  Determine Constant: $y(\pi/3) = 0 \implies 0 \cdot (2)^2 = 2 + C \implies C = -2$.

    4.  Function Definition: $y = \frac{\sec x - 2}{\sec^2 x} = \cos x - 2 \cos^2 x$.

    5.  Maximize: Let $u = \cos x$; maximize $f(u) = u - 2u^2$. $f'(u) = 1 - 4u = 0 \implies u = 1/4$. $y_{max} = 1/4 - 2(1/16) = 1/8$.

   Difficulty Level: Medium

   Concept Name: First Order Linear Differential Equation, Maxima and Minima.

   Shortcut Solution: Once you reach $y = \cos x - 2 \cos^2 x$, recognize it as a downward parabola in terms of $\cos x$. The vertex occurs at $\cos x = -b/2a = -1/(-4) = 1/4$, leading directly to the maximum value.

 Question 87

   Question: Let a curve $y = y(x)$ pass through the point (3, 3) and the area of the region under this curve, above the X-axis and between the abscissae 3 and $x (> 3)$ be $(\frac{y}{x})^3$. If this curve also passes through the point $(\alpha, 6\sqrt{10})$ in the first quadrant, then $\alpha$ is equal to:

   Options: (None provided in the source)

   Correct Answer: 6

   Year: 26-Jul-2022-Shift-1

   Solution: The given area condition is $\int_3^x y dt = (\frac{y}{x})^3$, which implies $x^3 \int_3^x y dt = y^3$. Differentiating both sides w.r.t. $x$ gives $x^3 y + 3x^2 \int_3^x y dt = 3y^2 \frac{dy}{dx}$. Substituting the initial area expression back in gives $x^3 y + \frac{3y^3}{x} = 3y^2 \frac{dy}{dx} \implies \frac{dy}{dx} - \frac{y}{x} = \frac{x^3}{3y}$. Letting $y^2 = t$ converts it to the linear DE $\frac{dt}{dx} - \frac{2}{x}t = \frac{2}{3}x^3$. Solving with $I.F. = 1/x^2$ yields $y^2 = \frac{x^4}{3} + Cx^2$. Using (3,3) gives $C = -2$. For $y = 6\sqrt{10}$, we get $\alpha^4 - 6\alpha^2 - 1080 = 0$, giving $\alpha = 6$.

   Step Solution:

    1.  Differentiate Area Equation: From $x^3 \int_3^x y dt = y^3$, differentiate to get $x^3 y + 3x^2 \int_3^x y dt = 3y^2 y'$.

    2.  Standardize DE: Substitute original area to get $x^3 y + \frac{3y^3}{x} = 3y^2 y' \implies y' = \frac{x^3}{3y} + \frac{y}{x}$.

    3.  Linearize: Substitute $y^2 = t$ to get $\frac{dt}{dx} - \frac{2}{x} t = \frac{2x^3}{3}$.

    4.  Find General Form: $I.F. = 1/x^2 \implies \frac{t}{x^2} = \int \frac{2x}{3} dx \implies y^2 = \frac{x^4}{3} + Cx^2$.

    5.  Evaluate $\alpha$: Use $(3,3)$ to find $C = -2$. Solve $360 = \frac{\alpha^4}{3} - 2\alpha^2$ to find $\alpha = 6$.

   Difficulty Level: Hard

   Concept Name: Integral Equation, Newton-Leibniz Theorem, First Order Linear Differential Equation.

   Shortcut Solution: Recognizing the transformation to a linear form $t' + Pt = Q$ through $y^2 = t$ is the critical step that simplifies the higher-order powers of $y$.

 Question 88

   Question: Let the solution curve $y = f(x)$ of the differential equation $\frac{dy}{dx} + \frac{xy}{x^2 - 1} = \frac{x^4 + 2x}{\sqrt{1 - x^2}}, x \in (-1, 1)$ pass through the origin. Then $\int_{-\sqrt{3}/2}^{\sqrt{3}/2} f(x) dx$ is equal to?

   Options:

    A. $\frac{\pi}{3} - \frac{1}{4}$

    B. $\frac{\pi}{3} - \frac{\sqrt{3}}{4}$

    C. $\frac{\pi}{6} - \frac{\sqrt{3}}{4}$

    D. $\frac{\pi}{6} - \frac{\sqrt{3}}{2}$

   Correct Answer: B

   Year: 26-Jul-2022-Shift-2

   Solution: This is a first order linear differential equation. The Integrating Factor is $e^{\int \frac{x}{x^2-1} dx} = \sqrt{1-x^2}$ (considering $1-x^2$ for the interval). The solution is $y \sqrt{1-x^2} = \int (x^4 + 2x) dx = \frac{x^5}{5} + x^2 + c$. Since it passes through the origin, $c = 0$. The function is $f(x) = \frac{x^5/5 + x^2}{\sqrt{1-x^2}}$. Integrating $f(x)$ from $-\sqrt{3}/2$ to $\sqrt{3}/2$, the odd part ($x^5$ term) vanishes. The remaining even part evaluates to $\frac{\pi}{3} - \frac{\sqrt{3}}{4}$.

   Step Solution:

    1.  Find I.F.: Calculate $I.F. = e^{\int \frac{x}{x^2-1} dx} = e^{\frac{1}{2} \ln|x^2-1|} = \sqrt{1-x^2}$.

    2.  General Solution: $y \sqrt{1-x^2} = \int \frac{x^4+2x}{\sqrt{1-x^2}} \sqrt{1-x^2} dx = \frac{x^5}{5} + x^2 + c$.

    3.  Identify Function: $y(0)=0 \implies c=0$. Thus $f(x) = \frac{x^5/5 + x^2}{\sqrt{1-x^2}}$.

    4.  Simplify Definite Integral: Over $[-\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}]$, the integral of the odd term ($x^5$) is zero. Evaluate $2 \int_0^{\sqrt{3}/2} \frac{x^2}{\sqrt{1-x^2}} dx$.

    5.  Final Integration: Substitute $x = \sin\theta$ to get $\int_0^{\pi/3} (1 - \cos 2\theta) d\theta = [\theta - \frac{\sin 2\theta}{2}]_0^{\pi/3} = \frac{\pi}{3} - \frac{\sqrt{3}}{4}$.

   Difficulty Level: Medium

   Concept Name: First Order Linear Differential Equation, Even and Odd Functions in Definite Integration.

   Shortcut Solution: Immediately recognize $\frac{x^5}{5\sqrt{1-x^2}}$ as an odd function to skip its integration over the symmetric limits, focusing only on the $x^2$ term.

 Question 89

   Question: Suppose $y = y(x)$ be the solution curve to the differential equation $\frac{dy}{dx} - y = 2 - e^{-x}$ such that $\lim_{x \to \infty} y(x)$ is finite. If $a$ and $b$ are respectively the $x$ and $y$-intercepts of the tangent to the curve at $x = 0$, then the value of $a - 4b$ is equal to?

   Options: (None provided in the source)

   Correct Answer: 3

   Year: 26-Jul-2022-Shift-2

   Solution: $I.F. = e^{-x}$. $y \cdot e^{-x} = -2e^{-x} + \frac{e^{-2x}}{2} + C$. $\Rightarrow y = -2 + e^{-x} + Ce^x$. Since $\lim_{x \to \infty} y(x)$ is finite, $C$ must be 0. Thus, $y = -2 + e^{-x}$. At $x=0$, $y=-1$ and $y'=-1$. Tangent equation is $x + y = -1$. Intercepts are $a=-1, b=-1$. So $a-4b = 3$.

   Step Solution:

    1.  Find I.F.: Identify $P(x) = -1$, so $I.F. = e^{\int -1 dx} = e^{-x}$.

    2.  General Solution: $y \cdot e^{-x} = \int (2 - e^{-x})e^{-x} dx = \int (2e^{-x} - e^{-2x}) dx = -2e^{-x} + \frac{1}{2}e^{-2x} + C$.

    3.  Apply Limit: Isolate $y = -2 + \frac{1}{2}e^{-x} + Ce^x$. For $y$ to be finite as $x \to \infty$, the coefficient $C$ of the growing exponential $e^x$ must be 0.

    4.  Tangent Equation: At $x=0$, $y = -2 + 1 = -1$ and $\frac{dy}{dx} = -e^0 = -1$. Tangent is $(y + 1) = -1(x - 0) \implies x + y = -1$.

    5.  Final Calculation: $x$-intercept ($a$) is $-1$, $y$-intercept ($b$) is $-1$. $a - 4b = -1 - 4(-1) = 3$.

   Difficulty Level: Hard

   Concept Name: First Order Linear Differential Equation, Limits at Infinity, Tangents and Intercepts.

   Short cut solution: Immediately set the constant $C$ of the $e^x$ term to zero to satisfy the "finite at infinity" condition.

 Question 91

   Question: Let the solution curve $y = y(x)$ of the differential equation $\sin(2x^2) \log_e(\tan x^2) dy + (4xy - 4\sqrt{2}x \sin(x^2 - \pi/4)) dx = 0, 0 < x < \sqrt{\pi/2}$, which passes through the point $(\sqrt{\pi/6}, 1)$. Then $y(\sqrt{\pi/3})$ is equal to?

   Options: (None provided in the source)

   Correct Answer: 1

   Year: 27-Jul-2022-Shift-1

   Solution: The equation is rearranged to a linear form in $y$. Using substitution $x^2 = t$, the Integrating Factor is found to be $\ln(\tan x^2)$. Integrating the remaining terms and applying the initial point $(\sqrt{\pi/6}, 1)$ allows solving for the constant and eventually finding that $y(\sqrt{\pi/3}) = 1$.

   Step Solution:

    1.  Substitution: Let $x^2 = t \implies 2x dx = dt$. The DE becomes $\sin(2t) \ln(\tan t) \frac{dy}{dt} + 2y = 2\sqrt{2} \sin(t - \pi/4)$.

    2.  Standard Linear Form: $\frac{dy}{dt} + \frac{2}{\sin 2t \ln(\tan t)} y = \frac{2\sqrt{2} \sin(t - \pi/4)}{\sin 2t \ln(\tan t)}$.

    3.  Integrating Factor: $I.F. = e^{\int \frac{2 dt}{\sin 2t \ln(\tan t)}} = \ln(\tan t)$.

    4.  Integrate: $y \ln(\tan t) = \int \frac{2\sqrt{2} \sin(t - \pi/4)}{\sin 2t} dt$. Using trigonometric expansion, this leads to a logarithmic expression plus constant $C$.

    5.  Evaluate: Applying initial conditions and substituting $t = \pi/3$, the terms simplify to yield $y = 1$.

   Difficulty Level: Hard

   Concept Name: First Order Linear Differential Equation, Substitution Method, Trigonometric Integration.

   Short cut solution: Recognize that the derivative of $\ln(\tan t)$ is exactly $\frac{2}{\sin 2t}$, making the I.F. calculation direct.

 Question 93

   Question: If $y = y(x), x \in (0, \pi/2)$ be the solution curve of the differential equation $(\sin^2 2x) \frac{dy}{dx} + (8 \sin^2 2x + 2 \sin 4x)y = 2e^{-4x}(2 \sin 2x + \cos 2x)$, with $y(\pi/4) = e^{-\pi}$, then $y(\pi/6)$ is equal to?

   Options: A. $\frac{2}{\sqrt{3}}e^{-2\pi/3}$, B. $\frac{2}{\sqrt{3}}e^{2\pi/3}$, C. $\frac{1}{\sqrt{3}}e^{-2\pi/3}$, D. $\frac{1}{\sqrt{3}}e^{2\pi/3}$

   Correct Answer: A

   Year: 28-Jul-2022-Shift-1

   Solution: Divide by $\sin^2 2x$ to get $\frac{dy}{dx} + (8 + 4\cot 2x)y = 2e^{-4x} \frac{2\sin 2x + \cos 2x}{\sin^2 2x}$. $I.F. = e^{8x} \sin^2 2x$. Solution is $y \cdot e^{8x} \sin^2 2x = e^{4x} \sin 2x + C$. Using $y(\pi/4) = e^{-\pi}$ gives $C=0$. At $x=\pi/6$, $y = \frac{2}{\sqrt{3}}e^{-2\pi/3}$.

   Step Solution:

    1.  Linearize: Divide the DE by $\sin^2 2x$ to obtain $\frac{dy}{dx} + (8 + 4\cot 2x)y = \frac{2e^{-4x}(2 \sin 2x + \cos 2x)}{\sin^2 2x}$.

    2.  Integrating Factor: $I.F. = e^{\int (8 + 4\cot 2x) dx} = e^{8x + 2\ln(\sin 2x)} = e^{8x} \sin^2 2x$.

    3.  General Solution: $y(e^{8x} \sin^2 2x) = \int 2e^{4x}(2 \sin 2x + \cos 2x) dx$.

    4.  Find C: The integral on the RHS is $e^{4x} \sin 2x + C$. Using $y(\pi/4) = e^{-\pi}$, we find $e^{-\pi} e^{2\pi} (1) = e^\pi (1) + C \implies C = 0$.

    5.  Final Result: $y = \frac{e^{4x} \sin 2x}{e^{8x} \sin^2 2x} = \frac{e^{-4x}}{\sin 2x}$. At $x = \pi/6$, $y = \frac{e^{-2\pi/3}}{\sqrt{3}/2} = \frac{2}{\sqrt{3}} e^{-2\pi/3}$.

   Difficulty Level: Hard

   Concept Name: First Order Linear Differential Equation, Product Rule in Integration.

   Short cut solution: Recognize that the RHS integral $\int 2e^{4x}(2 \sin 2x + \cos 2x) dx$ is the exact derivative of $e^{4x} \sin 2x$.

 Question 94

   Question: Let $y = y(x)$ be the solution curve of the differential equation $\frac{dy}{dx} + \frac{1}{x^2 - 1} y = \left( \frac{x - 1}{x + 1} \right)^{1/2}, x > 1$ passing through the point $\left( 2, \sqrt{\frac{1}{3}} \right)$. Then $\sqrt{7} y(8)$ is equal to?

   Options: 

    A. $11 + 6 \log_e 3$

    B. 19

    C. $12 - 2 \log_e 3$

    D. $19 - 6 \log_e 3$

   Correct Answer: B (Note: The source's answer key lists B, though the provided mathematical derivation in the source concludes with the value in Option D)

   Year: 28-Jul-2022-Shift-2

   Solution: This is a first order linear differential equation. The Integrating Factor is found to be $\sqrt{\frac{x-1}{x+1}}$. Multiplying and integrating both sides leads to the equation $y \sqrt{\frac{x-1}{x+1}} = x - 2 \ln|x+1| + C$. Using the initial condition at $x=2$, the constant $C$ is determined. Finally, substituting $x=8$ gives the required value.

   Step Solution:

    1.  Identify Linear Form: The equation is already in the form $\frac{dy}{dx} + P(x)y = Q(x)$ with $P(x) = \frac{1}{x^2-1}$ and $Q(x) = \sqrt{\frac{x-1}{x+1}}$.

    2.  Integrating Factor: Calculate $I.F. = e^{\int \frac{1}{x^2-1} dx} = e^{\frac{1}{2} \ln \left| \frac{x-1}{x+1} \right|} = \sqrt{\frac{x-1}{x+1}}$.

    3.  General Solution: $y \cdot \sqrt{\frac{x-1}{x+1}} = \int \sqrt{\frac{x-1}{x+1}} \cdot \sqrt{\frac{x-1}{x+1}} dx = \int \frac{x-1}{x+1} dx = \int (1 - \frac{2}{x+1}) dx = x - 2 \ln|x+1| + C$.

    4.  Find C: Substitute $x = 2, y = \sqrt{1/3} \implies \sqrt{1/3} \cdot \sqrt{1/3} = 2 - 2 \ln 3 + C \implies C = 2 \ln 3 - \frac{5}{3}$.

    5.  Evaluate: At $x=8$, $y(8) \cdot \sqrt{7/9} = 8 - 2 \ln 9 + 2 \ln 3 - \frac{5}{3} \implies \sqrt{7} y(8) = 19 - 6 \ln 3$.

   Difficulty Level: Hard

   Concept Name: First Order Linear Differential Equation

   Shortcut Solution: Recognize that $\int \frac{x-1}{x+1} dx$ can be written as $\int \frac{x+1-2}{x+1} dx$ to immediately obtain $x - 2\ln(x+1)$, saving time on polynomial division.

 Question 96

   Question: Let the solution curve $y = y(x)$ of the differential equation $(1 + e^{2x}) (\frac{dy}{dx} + y) = 1$ pass through the point $(0, \frac{\pi}{2})$. Then, $\lim_{x \to \infty} e^x y(x)$ is equal to?

   Options: 

    A. $\pi/4$

    B. $3\pi/4$

    C. $\pi/2$

    D. $3\pi/2$

   Correct Answer: B

   Year: 29-Jul-2022-Shift-1

   Solution: Rewriting the equation gives $\frac{dy}{dx} + y = \frac{1}{1+e^{2x}}$. The Integrating Factor is $e^x$. Multiplying the DE by the I.F. and integrating results in $e^x y = \tan^{-1}(e^x) + C$. Using the point $(0, \pi/2)$, the constant is found to be $\pi/4$. Taking the limit as $x \to \infty$ of the expression $e^x y$ yields $3\pi/4$.

   Step Solution:

    1.  Standardize DE: Divide by $(1+e^{2x})$ to get $\frac{dy}{dx} + y = \frac{1}{1+e^{2x}}$.

    2.  Integrating Factor: Calculate $I.F. = e^{\int 1 dx} = e^x$.

    3.  General Solution: $y \cdot e^x = \int \frac{1}{1+e^{2x}} \cdot e^x dx$. Use substitution $t = e^x \implies e^x y = \int \frac{1}{1+t^2} dt = \tan^{-1}(e^x) + C$.

    4.  Find C: Substitute $(0, \pi/2) \implies \frac{\pi}{2} \cdot 1 = \tan^{-1}(1) + C \implies C = \frac{\pi}{4}$.

    5.  Calculate Limit: $\lim_{x \to \infty} e^x y = \lim_{x \to \infty} (\tan^{-1}(e^x) + \frac{\pi}{4}) = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$.

   Difficulty Level: Medium

   Concept Name: First Order Linear Differential Equation, Limits at Infinity.

   Shortcut Solution: Once the form $e^x y = \tan^{-1}(e^x) + C$ is found, notice that as $x \to \infty$, $e^x \to \infty$, and $\tan^{-1}(\infty) = \pi/2$. Simply adding your constant $C$ to $\pi/2$ gives the answer immediately.

 Question 98

   Question: Let $y = y(x)$ be the solution curve of the differential equation $\frac{dy}{dx} + \left( \frac{2x^2 + 11x + 13}{x^3 + 6x^2 + 11x + 6} \right) y = \frac{x + 3}{x + 1}, x > -1$, which passes through the point (0, 1). Then y(1) is equal to?

   Options: 

    A. 1/2

    B. 3/2

    C. 5/2

    D. 7/2

   Correct Answer: B

   Year: 29-Jul-2022-Shift-2

   Solution: The $P(x)$ term is decomposed into partial fractions: $\frac{2}{x+1} + \frac{1}{x+2} - \frac{1}{x+3}$. This gives an Integrating Factor of $\frac{(x+1)^2(x+2)}{x+3}$. Multiplying the DE by the I.F. and integrating simplifies the right side to a basic quadratic. Solving for the constant using $(0, 1)$ and evaluating at $x=1$ yields $y(1) = 3/2$.

   Step Solution:

    1.  Partial Fractions: Decompose $P(x) = \frac{2x^2 + 11x + 13}{(x+1)(x+2)(x+3)} = \frac{2}{x+1} + \frac{1}{x+2} - \frac{1}{x+3}$.

    2.  Integrating Factor: Calculate $I.F. = e^{\int (\frac{2}{x+1} + \frac{1}{x+2} - \frac{1}{x+3}) dx} = \frac{(x+1)^2(x+2)}{x+3}$.

    3.  General Solution: $y \cdot \frac{(x+1)^2(x+2)}{x+3} = \int \frac{x+3}{x+1} \cdot \frac{(x+1)^2(x+2)}{x+3} dx = \int (x+1)(x+2) dx = \frac{x^3}{3} + \frac{3x^2}{2} + 2x + C$.

    4.  Find C: Substitute $(0, 1) \implies 1 \cdot \frac{1 \cdot 2}{3} = 0 + C \implies C = \frac{2}{3}$.

    5.  Evaluate y(1): $y(1) \cdot \frac{4 \cdot 3}{4} = \frac{1}{3} + \frac{3}{2} + 2 + \frac{2}{3} \implies 3y(1) = 1 + \frac{3}{2} + 2 = \frac{9}{2} \implies y(1) = \frac{3}{2}$.

   Difficulty Level: Hard

   Concept Name: First Order Linear Differential Equation, Partial Fractions.

   Shortcut Solution: After finding the $I.F.$, recognize that multiplying it by $Q(x)$ effectively cancels the "unfriendly" denominator terms, leaving a standard polynomial to integrate.

 Question 102

   Question: Let slope of the tangent line to a curve at any point $P(x, y)$ be given by $\frac{xy^2 + y}{x}$. If the curve intersects the line $x + 2y = 4$ at $x = -2$, then the value of $y$, for which the point $(3, y)$ lies on the curve, is?

   Options: 

    A. $\frac{18}{35}$ 

    B. $-\frac{4}{3}$ 

    C. $-\frac{18}{19}$ 

    D. $-\frac{18}{11}$

   Correct Answer: C

   Year: 26 Feb 2021 Shift 2

   Solution: The slope is given by $\frac{dy}{dx} = \frac{xy^2 + y}{x}$, which simplifies to $\frac{dy}{dx} - \frac{y}{x} = y^2$. This is a Bernoulli equation. Dividing by $y^2$ and substituting $t = \frac{1}{y}$ transforms it into a linear equation $\frac{dt}{dx} + \frac{1}{x}t = -1$. The Integrating Factor is $x$, leading to the general solution $\frac{x}{y} = -\frac{x^2}{2} + C$. The curve intersects $x+2y=4$ at $x=-2$, which gives the point $(-2, 3)$. Solving for $C$ gives $C = \frac{4}{3}$. Substituting $x=3$ into the curve equation results in $y = -\frac{18}{19}$.

   Step Solution:

    1.  Form the Equation: From the slope, $\frac{dy}{dx} = y^2 + \frac{y}{x} \implies \frac{x dy - y dx}{x} = y^2 dx \implies \frac{x dy - y dx}{y^2} = x dx$.

    2.  Integrate: Recognize the left side as $-d(\frac{x}{y})$. Integrating both sides: $-\frac{x}{y} = \frac{x^2}{2} + C$.

    3.  Find Point of Intersection: At $x = -2$ on the line $x + 2y = 4$, we have $-2 + 2y = 4 \implies y = 3$. The point is $(-2, 3)$.

    4.  Find Constant: Substitute $(-2, 3)$ into the solution: $-\frac{-2}{3} = \frac{(-2)^2}{2} + C \implies \frac{2}{3} = 2 + C \implies C = -\frac{4}{3}$.

    5.  Evaluate for $x=3$: $-\frac{3}{y} = \frac{3^2}{2} - \frac{4}{3} = \frac{9}{2} - \frac{4}{3} = \frac{19}{6} \implies y = -\frac{18}{19}$.

   Difficulty Level: Medium

   Concept Name: Bernoulli’s Differential Equation / Exact Differential Form.

   Shortcut Solution: Recognize the differential form $\frac{x dy - y dx}{y^2}$ as $-d(\frac{x}{y})$ immediately to skip the Bernoulli substitution steps.

 Question 105

   Question: If $y = y(x)$ is the solution of the equation $\sin y \cos y \frac{dy}{dx} + e^{\sin y} \cos x = \cos x, y(0) = 0$, then $1 + y(\frac{\pi}{6}) + \frac{\sqrt{3}}{2} y(\frac{\pi}{3}) + \frac{1}{\sqrt{2}} y(\frac{\pi}{4})$ is equal to?

   Options: (None provided in the source)

   Correct Answer: 1

   Year: 26 Feb 2021 Shift 1

   Solution: Multiply the equation by $e^{\sin y}$ to facilitate substitution. Let $t = e^{\sin y}$, then $\frac{dt}{dx} = e^{\sin y} \cos y \frac{dy}{dx}$. The equation becomes $\frac{dt}{dx} + t \cos x = \cos x$. This is a linear differential equation with Integrating Factor $e^{\sin x}$. Solving gives $e^{\sin y} \cdot e^{\sin x} = e^{\sin x} + C$. Using $y(0)=0$ results in $C=0$. This implies $e^{\sin y} = 1$, hence $y=0$ for all values of $x$. The sum therefore evaluates to $1 + 0 + 0 + 0 = 1$.

   Step Solution:

    1.  Substitution: Let $e^{\sin y} = t$, then $\frac{dt}{dx} = e^{\sin y} \cos y \frac{dy}{dx}$. The DE (multiplied by $e^{\sin y}$) becomes $\frac{dt}{dx} + t \cos x = \cos x$.

    2.  Integrating Factor: $I.F. = e^{\int \cos x dx} = e^{\sin x}$.

    3.  General Solution: $t \cdot e^{\sin x} = \int e^{\sin x} \cos x dx + C \implies e^{\sin y} \cdot e^{\sin x} = e^{\sin x} + C$.

    4.  Find C: Use $y(0)=0 \implies e^0 \cdot e^0 = e^0 + C \implies 1 = 1 + C \implies C = 0$.

    5.  Solve for y: $e^{\sin y} e^{\sin x} = e^{\sin x} \implies e^{\sin y} = 1 \implies \sin y = 0 \implies y = 0$. Thus all $y$ terms in the final expression are 0.

   Difficulty Level: Medium

   Concept Name: First Order Linear Differential Equation (Substitution Method).

   Shortcut Solution: Once $C=0$ is determined, the relationship $e^{\sin y} e^{\sin x} = e^{\sin x}$ forces $e^{\sin y} = 1$, which makes $y(x) = 0$ for all $x$, allowing you to instantly evaluate the sum as 1.