Differential Equations

To solve the differential equation $(x + 2y^3) \frac{dy}{dx} = y$, we begin by rearranging terms:

$ \frac{x + 2y^3}{y} = \frac{dx}{dy} $

This can be rewritten as:

$ \frac{dx}{dy} = \frac{x}{y} + 2y^2 $

This gives us a form where:

$ \frac{dx}{dy} - \frac{x}{y} = 2y^2 $

We recognize this as a linear first-order differential equation, which can be solved using an integrating factor. The integrating factor ($\text{IF}$) is derived as follows:

$ \text{IF} = e^{\int -\frac{1}{y} \, dy} = e^{-\ln|y|} = \frac{1}{y} $

Apply the integrating factor to both sides of the differential equation:

$ \frac{x}{y} = \int \left(2y^2 \cdot \frac{1}{y}\right) \, dy $

This integral becomes:

$ \frac{x}{y} = 2 \int y \, dy $

Calculate the integral:

$ \frac{x}{y} = 2 \left(\frac{y^2}{2}\right) + C $

Simplifying gives:

$ \frac{x}{y} = y^2 + C $

Therefore, multiplying both sides by $y$ results in:

$ x = y(y^2 + C) $

Thus, the solution to the differential equation is:

$ x = y(y^2 + C) $

Given differential equation is

$ \begin{aligned} \frac{d^3 y}{d x^3}=\left[1+\left(\frac{d y}{d x}\right)^2\right]^{\frac{5}{2}} \\ \Rightarrow \quad\left(\frac{d^3 y}{d x^3}\right)^2=\left[1+\left(\frac{d y}{d x}\right)^2\right]^5 \end{aligned} $

Highest derivative is third order and power of highest order derivative is 2 .

Hence, order is 3 and degree is 2 .

To find the integrating factor of the given differential equation, we start with the equation:

$ \sin x \frac{d y}{d x} - y \cos x = 1 $

Rewriting the equation, we have:

$ \frac{d y}{d x} - \frac{\cos x}{\sin x} y = \frac{1}{\sin x} $

This simplifies to:

$ \frac{d y}{d x} - \cot x \cdot y = \operatorname{cosec} x $

The standard form for integrating factors is:

$ I F = e^{\int -\cot x \, dx} $

Calculating the integral, we find:

$ I F = e^{-\log |\sin x|} $

This can further be expressed as:

$ I F = e^{\log (\sin x)^{-1}} = \frac{1}{\sin x} $

Therefore, the integrating factor for the given differential equation is:

$ \frac{1}{\sin x} = \operatorname{cosec} x $

To solve the given differential equation:

$ \left(x \sin \frac{y}{x}\right) d y = \left(y \sin \frac{y}{x} - x\right) d x $

we start by expressing it in differential form:

$ \frac{d y}{d x} = \frac{y \sin \frac{y}{x} - x}{x \sin \frac{y}{x}} $

Introduce a new variable $ v $ such that $ y = vx $. This implies:

$ \frac{d y}{d x} = v + x \frac{d v}{d x} $

Substitute $ y = vx $ into the differential equation:

$ v + x \frac{d v}{d x} = \frac{vx \sin v - x}{x \sin v} $

Simplify:

$ x \frac{d v}{d x} = \frac{vx \sin v - x - vx \sin v}{x \sin v} = \frac{-x}{x \sin v} = \frac{-1}{\sin v} $

This results in the following separable equation:

$ \sin v \, d v = -\frac{d x}{x} $

Integrate both sides:

$ \int \sin v \, d v = -\int \frac{d x}{x} $

The integration yields:

$ -\cos v = -\log x + C $

Rearranging gives:

$ \log x - \cos v = C $

Substituting back $ v = \frac{y}{x} $, we arrive at the solution:

$ \cos \left(\frac{y}{x}\right) = \log_e x + C $

To determine the sum of the order and degree of the given differential equation:

$ \frac{d^4 y}{d x^4} = \left\{C + \left(\frac{d y}{d x}\right)^2\right\}^{3 / 2} $

First, rewrite the equation in a form where the degree is clear:

$ \left(\frac{d^4 y}{d x^4}\right)^2 = \left(C + \left(\frac{d y}{d x}\right)^2\right)^3 $

From this, we observe:

The order of the differential equation is the highest derivative present, which is 4.

The degree of the differential equation is the power of the highest derivative when the equation is free of fractions and any radical signs. Here, the degree is 2 (as the highest derivative, $ \frac{d^4 y}{d x^4} $, is squared).

Therefore, the sum of the order and degree is $ 4 + 2 = 6 $.

To solve the differential equation:

$ (x+y) y \, dx + (y-x) x \, dy = 0 $

we can start by rewriting the equation:

$ \frac{dy}{dx} = \frac{(x+y) y}{(x-y) x} $

Substitute $ y = vx $, where $ v $ is a function of $ x $. Then, the derivative $ \frac{dy}{dx} $ becomes:

$ \frac{dy}{dx} = v + x \frac{dv}{dx} $

Substitute $ y = vx $ into the differential equation:

$ v + x\frac{dv}{dx} = \frac{(x + vx) vx}{(x - vx) x} $

Simplify the equation:

$ v + x\frac{dv}{dx} = \frac{v + v^2}{1-v} $

Rearranging gives:

$ x\frac{dv}{dx} = \frac{v + v^2 - v + v^2}{1-v} $

Next, separate variables and integrate:

$ \int \left(\frac{1-v}{v^2}\right) dv = \int 2x \, dx $

This can be rewritten as:

$ \int \left(\frac{1}{v^2} - \frac{1}{v}\right) dv = \int 2x \, dx $

Integrate both sides:

$ -\frac{1}{v} - \log v = 2 \log x + \log c $

Substitute back $ v = \frac{y}{x} $:

$ -\frac{x}{y} = \log \frac{y}{x} + \log x^2 + \log c $

Which simplifies to the equation:

$ -\frac{x}{y} = \log (cxy) $

Thus, the general solution is:

$ x + y \log (cxy) = 0 $

We are given the differential equation:

$ \left(y^2 + x + 1\right) \, dy = (y + 1) \, dx $

This can be rewritten as:

$ \frac{dx}{dy} - \frac{1}{y+1} x = \frac{y^2 + 1}{y+1} $

This is a linear differential equation. To solve it, we first find the integrating factor (IF):

$ \mathrm{IF} = e^{\int P \, dy} = e^{\int \frac{-1}{y+1} \, dy} = e^{-\log(y+1)} = \frac{1}{y+1} $

Multiplying through by the integrating factor gives:

$ x \cdot \frac{1}{y+1} = \int \frac{1}{y+1} \cdot \frac{y^2 + 1}{y+1} \, dy = \int \frac{y^2 + 1}{y^2 + 2y + 1} \, dy $

This simplifies as:

$ = \int \left(1 - \frac{2y}{y^2 + 2y + 1}\right) \, dy + C $

Carrying out the integration:

$ = y - \int \left[\frac{2y + 2}{y^2 + 2y + 1} - \frac{2}{(y+1)^2}\right] \, dy + C $

$ = y - \log(y+1)^2 + 2 \int \frac{1}{(y+1)^2} \, dy + C $

$ = y - \log(y+1)^2 - \frac{2}{y+1} + C $

Thus, the general solution is:

$ \frac{x+2}{y+1} + \log(y+1)^2 = y + C $

We have,

$ \left(\frac{d^2 y}{d x^2}\right)^{\frac{-7}{2}}\left(\frac{d^3 y}{d x^3}\right)^2-\left(\frac{d^2 y}{d x^2}\right)^{\frac{-5}{2}}\left(\frac{d^4 y}{d x^4}\right)=0 $

$ \Rightarrow\left(\frac{d^3 y}{d x^3}\right)^2\left(\frac{d^2 y}{d x^2}\right)^{\frac{5}{2}}=\left(\frac{d^4 y}{d x^4}\right)\left(\frac{d^2 y}{d x^2}\right)^{\frac{7}{2}} $

On squaring both sides, we get

$ \Rightarrow\left(\frac{d^3 y}{d x^3}\right)^4\left(\frac{d^2 y}{d x^2}\right)^5=\left(\frac{d^4 y}{d x^4}\right)^2\left(\frac{d^2 y}{d x^2}\right)^7 $

$\therefore$ Order is 4 and degree is 2 .

Hence, required difference $=2$

To solve the given differential equation:

$ x \, dy + \left(y + y^2 x \right) \, dx = 0 $

Rearrange it as:

$ \frac{dy}{dx} + \frac{y}{x} + y^2 = 0 $

Let's make a substitution to simplify this equation. Set $ \frac{1}{y} = t $, which implies $ -\frac{1}{y^2} \frac{dy}{dx} = \frac{dt}{dx} $.

This transforms the equation into:

$ \frac{dt}{dx} - \frac{1}{x} t = 1 $

This is a first-order linear differential equation of the form:

$ \frac{dt}{dx} -\frac{1}{x} t = 1 $

To solve this, we use an integrating factor (IF):

$ \mathrm{IF} = e^{\int -\frac{1}{x} \, dx} = \frac{1}{x} $

Multiplying through by the integrating factor gives:

$ \frac{1}{x} t = \int \frac{1}{x} \, dx + C $

Integrating the right side:

$ \frac{1}{x} t = \log x + C $

Substituting back in terms of $ y $:

$ \frac{1}{xy} = \log x + C $

Given the initial condition $ y = 1 $ when $ x = 1 $:

$ \frac{1}{1 \cdot 1} = \log 1 + C \Rightarrow 1 = 0 + C \Rightarrow C = 1 $

So the equation becomes:

$ \frac{1}{xy} = \log x + 1 $

Rearranging, we find:

$ xy(1 + \log x) = 1 $

Finally, solving for $ y $:

$ y = \frac{1}{x(1 + \log x)} $

To solve the differential equation $ x \, dy - y \, dx = \sqrt{x^2 + y^2} \, dx $, with the initial condition $ y(\sqrt{3}) = 1 $, follow these steps:

Start by rewriting the equation:

$ \frac{dy}{dx} = \frac{\sqrt{x^2 + y^2} + y}{x} $

Apply the substitution $ y = vx $, where $ v $ is a function of $ x $. Then, differentiate to get:

$ \frac{dy}{dx} = v + x \frac{dv}{dx} $

Substitute back into the equation:

$ v + x \frac{dv}{dx} = \frac{\sqrt{x^2 + (vx)^2} + vx}{x} $

Simplify:

$ v + x \frac{dv}{dx} = \frac{\sqrt{x^2(1 + v^2)} + vx}{x} $

$ x \frac{dv}{dx} = \sqrt{1 + v^2} $

Separate the variables:

$ \frac{1}{\sqrt{1 + v^2}} dv = \frac{1}{x} dx $

Integrate both sides:

$ \log|v + \sqrt{1 + v^2}| = \log x + \log c $

Combine the logs:

$ |v + \sqrt{1 + v^2}| = cx $

Substituting back for $ v = \frac{y}{x} $:

$ \frac{y}{x} + \frac{\sqrt{x^2 + y^2}}{x} = cx $

$ y + \sqrt{x^2 + y^2} = cx^2 $

Use the initial condition $ y(\sqrt{3}) = 1 $:

$ 1 + \sqrt{(\sqrt{3})^2 + 1^2} = c (\sqrt{3})^2 $

$ 1 + 2 = 3c $

$ 3 = 3c \implies c = 1 $

The required solution is:

$ y + \sqrt{x^2 + y^2} = x^2 $

To find the differential equation representing a family of circles with centers on the Y-axis, let's consider the circle with center $(0, K)$ and radius $r$. The general equation for such a circle is:

$ x^2 + (y - K)^2 = r^2 \quad \ldots \text{(i)} $

Differentiate equation (i) with respect to $x$:

$ 2x + 2(y - K)\frac{dy}{dx} = 0 $

Which simplifies to:

$ y - K = \frac{-x}{\frac{dy}{dx}} \quad \text{or} \quad K = y + \frac{x}{\frac{dy}{dx}} $

Let's denote $\frac{dy}{dx}$ as $y_1$. So, from the above, we have:

$ y - K = \frac{-x}{y_1} \quad \Rightarrow \quad K = y + \frac{x}{y_1} $

Now, differentiate this equation with respect to $x$ again:

$ 0 = 1 + \left(y - K\right) \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 $

Substituting $y - K = \frac{-x}{y_1}$ into this, we get:

$ 1 + \left(-\frac{x}{y_1}\right)\frac{d^2y}{dx^2} + y_1^2 = 0 $

This can be rearranged to:

$ 1 - \frac{x \frac{d^2y}{dx^2}}{y_1} + y_1^2 = 0 $

$ x \frac{d^2y}{dx^2} = y_1(y_1^2 + 1) $

So, the differential equation is:

$ x \frac{d^2y}{dx^2} = y_1(y_1^2 + 1) $

To solve the given differential equation:

$\left(\sin y \cos^2 y - x \sec^2 y\right) dy = (\tan y) dx,$

we start by rewriting it in the standard linear form:

$\frac{dx}{dy} + \frac{x}{\sin y \cos y} = \cos^3 y.$

Recognizing this as a linear differential equation in $x$, we determine the integrating factor (IF):

Compute the Integrating Factor (IF):

$\mathrm{IF} = e^{\int \frac{1}{\sin y \cos y} \, dy}.$

Solve the Integral:

$\int \frac{1}{\sin y \cos y} \, dy = 2 \int \csc 2y \, dy.$

Solving this integral gives:

$e^{\int \csc 2y \, dy} = e^{\log |\csc 2y - \cot 2y|} = \csc 2y - \cot 2y.$

Thus, the integrating factor simplifies to:

$\mathrm{IF} = \frac{1 - \cos 2y}{\sin 2y} = \frac{2 \sin^2 y}{2 \sin y \cos y} = \tan y.$

Apply the Integrating Factor:

Multiply the entire linear differential equation by this integrating factor:

$x \times \tan y = \int \cos^3 y \times \tan y \, dy.$

Integrate the Right-Hand Side:

Substitute $\cos^2 y = 1 - \sin^2 y$. Then:

$\int \cos^3 y \tan y \, dy = \int \cos^2 y \sin y \, dy.$

Perform the integration:

$= -\frac{\cos^3 y}{3} + C.$

Resulting General Solution:

Therefore, the general solution to the differential equation is:

$3x \tan y + \cos^3 y = C.$

To solve the given differential equation, we start by analyzing the equation:

$ (x-y-1) dy = (x+y+1) dx $

This can be rewritten in differential form as:

$ \frac{dy}{dx} = \frac{x+y+1}{x-y-1} $

This expression can also be represented as:

$ \frac{dy}{dx} = \frac{x+(y+1)}{x-(y+1)} $

Introduce a substitution where $ y+1 = z $. Thus, we have:

$ \frac{dy}{dx} = \frac{dz}{dx} $

After substitution, the equation becomes:

$ \frac{dz}{dx} = \frac{x+z}{x-z} $

Next, we use another substitution $ z = vx $. Therefore, we differentiate $ z $ with respect to $ x $:

$ \frac{dz}{dx} = v + x \frac{dv}{dx} $

Substituting this back into our equation results in:

$ v + x \frac{dv}{dx} = \frac{1+v}{1-v} $

Reorganize terms and separate variables:

$ x \frac{dv}{dx} = \frac{1+v-v+v^2}{1-v} $

This simplifies to:

$ \left(\frac{1-v}{1+v^2}\right) dv = \frac{dx}{x} $

Integrating both sides, we have:

$ \int \left( \frac{1}{1+v^2} - \frac{v}{1+v^2} \right) dv = \int \frac{dx}{x} $

The integrals yield:

$ \tan^{-1} v - \frac{1}{2} \log \left| 1+v^2 \right| = \log x + C $

Substitute back $ v = \frac{z}{x} $:

$ \tan^{-1} \frac{z}{x} - \frac{1}{2} \log \left| \frac{z^2}{x^2} \right| = \log x + C $

Simplifying using $ z = y+1 $:

$ \tan^{-1} \left(\frac{y+1}{x}\right) - \frac{1}{2} \log \left| \frac{x^2 + (y+1)^2}{x^2} \right| - \log x = C $

Finally, this simplifies to:

$ \tan^{-1} \left(\frac{y+1}{x}\right) - \frac{1}{2} \log \left| x^2 + y^2 + 2y + 1 \right| = C $

To solve the given differential equation

$ (x+1) \frac{d y}{d x} - y = e^{3x}(x+1)^2 $

we start by rewriting it in the standard linear form:

$ \frac{d y}{d x} - \frac{y}{x+1} = e^{3x}(x+1) $

Here, it's a first-order linear differential equation of the form

$ \frac{d y}{d x} + P(x)y = Q(x) $

where $ P(x) = -\frac{1}{x+1} $ and $ Q(x) = e^{3x}(x+1) $.

Step 1: Calculate the Integrating Factor (IF)

The integrating factor is given by:

$ \text{IF} = e^{\int P(x) \, dx} = e^{\int -\frac{1}{x+1} \, dx} = e^{-\ln(x+1)} = \frac{1}{x+1} $

Step 2: Determine the Solution

Using the integrating factor, the solution of the differential equation is:

$ y \cdot \text{IF} = \int Q(x) \cdot \text{IF} \, dx $

Substituting the known values:

$ y \cdot \frac{1}{x+1} = \int e^{3x}(x+1) \cdot \frac{1}{x+1} \, dx $

This simplifies to:

$ \frac{y}{x+1} = \frac{e^{3x}}{3} + C' $

Therefore, the complete solution is:

$ \frac{3y}{x+1} = e^{3x} + C' $

where $ C' = C/3 $ representing an integration constant.

The given differential equation is

$\frac{d y}{d x} = y + 7$

This is a first order linear differential equation and can be solved using an integrating factor.

Rearrange the equation to the standard form of a linear differential equation :

$\frac{d y}{d x} - y = 7$

The integrating factor is $e^{-\int dx} = e^{-x}$.

Multiplying each side of the equation by the integrating factor gives :

$e^{-x} \frac{d y}{d x} - e^{-x} y = 7e^{-x}$

The left-hand side of the equation is the derivative of $(e^{-x}y)$ with respect to $x$. So we can write the equation as :

$\frac{d}{d x}(e^{-x}y) = 7e^{-x}$

Integrate both sides with respect to $x$ :

$e^{-x}y = -7e^{-x} + C$

Multiply both sides by $e^{x}$ to isolate $y$ :

$y = -7 + Ce^{x}$

So, the general solution to the differential equation is $y = -7 + Ce^{x}$.

Now, let's apply the initial conditions to find the particular solutions :

For $y_{1}(0)=0$, we substitute into the general solution and solve for $C$ :

$0 = -7 + C$

So, $C = 7$, and the solution for $y_{1}$ is $y_{1}(x) = 7e^{x} - 7$.

For $y_{2}(0)=1$, again substitute into the general solution:

$1 = -7 + C$

So, $C = 8$, and the solution for $y_{2}$ is $y_{2}(x) = 8e^{x} - 7$.

The two curves intersect when $y_{1}(x) = y_{2}(x)$. Setting these equal and solving for $x$ gives :

$7e^{x} - 7 = 8e^{x} - 7$

$e^{x} = 0$

But $e^{x} = 0$ has no solution, because the exponential function never equals zero.

So, the curves $y=y_{1}(x)$ and $y=y_{2}(x)$ do not intersect at any point.

$y = vx$

$v + x{{dv} \over {dx}} = - \left( {{{1 + 3{v^2}} \over {3 + {v^2}}}} \right)$

$x{{dv} \over {dx}} = - \left( {{{1 + 3{v^2}} \over {3 + {v^2}}} + v} \right)$

${{dv} \over {dx}} = - \left( {{{{{(1 + v)}^3}} \over {3 + {v^2}}}} \right)$

$ \Rightarrow {{3 + {v^2}} \over {{{(1 + v)}^3}}} = {{ - dx} \over x}$

$ \Rightarrow \ln |v + 1| + {{2v} \over {{{(v + 1)}^2}}} = C - \ln |x|$

$x = 1,v = 0 \Rightarrow C = 0$

$ \Rightarrow \ln |x + y| - \ln |x| + {{2xy} \over {{{(x + y)}^2}}} = - \ln |x|$

${{dy} \over {dx}} - {{3{x^5}{{\tan }^{ - 1}}({x^3})} \over {{{(1 + {x^6})}^{{3 \over 2}}}}}y = 2x\exp \left\{ {{{{x^3} - {{\tan }^{ - 1}}{x^3}} \over {\sqrt {1 + {x^6}} }}} \right\}$

$IF = {e^{ - \int {{{3{x^5}{{\tan }^{ - 1}}({x^3})} \over {{{(1 + {x^6})}^{{3 \over 2}}}}}dx} }}$

Let ${\tan ^{ - 1}}{x^3} = t \Rightarrow {{3{x^2}} \over {1 + {x^6}}}dx = dt$

$ \Rightarrow IF = {e^{ - \int {{{\tan t} \over {\sec t}}.t\,dt} }} = {e^{ - \int {\sin t.tdt} }} = {e^{t\cos t - \sin t}}$

$ \Rightarrow IF = {e^{{{{{\tan }^{ - 1}}({x^3})} \over {\sqrt {1 + {x^6}} }} - {{{x^3}} \over {\sqrt {1 + {x^6}} }}}}$

$\therefore$ Solution is

$y\,.\,{e^{{{{{\tan }^{ - 1}}{x^3}} \over {\sqrt {1 + {x^6}} }} - {{{x^3}} \over {\sqrt {1 + {x^6}} }}}} = \int {2x\,dx + c} $

$ \Rightarrow y\,.\,{e^{{{{{\tan }^{ - 1}}{x^3} - {x^3}} \over {\sqrt {1 + {x^6}} }}}} = {x^2} + c$

$y(0) = 0 \Rightarrow c = 0$

$x = 1$

$y\,.\,{e^{{{{\pi \over 4} - 1} \over {\sqrt 2 }}}} = 1$

$ \Rightarrow y = {e^{{{1 - {\pi \over 4}} \over {\sqrt 2 }}}}$

$ \Rightarrow y = {e^{{{4 - \pi } \over {4\sqrt 2 }}}}$

$x\ln x{{dy} \over {dx}} + y = {x^2}\ln x$

${{dy} \over {dx}} + {1 \over {x\ln x}}\,.\,y = x$

If $ = {e^{\int {{1 \over {x\ln x}}dx} }} = {e^{\int {{1 \over t}dt} }}$, where $t = \ln x$

$ = {e^{\ln t}} = t = \ln x$

$y.\ln x = \int {x\ln x = {{{x^2}} \over 2}\ln x - \int {{{{x^2}} \over 2}\,.\,{1 \over x}} } $

$y\ln x = {{{x^2}} \over 2}\ln x - {{{x^2}} \over 4} + C$ ..... (i)

$y(2) = 2 \Rightarrow C = 1$

Putting $x = e$ in (i),

$y = {{{e^2}} \over 4} + 1 = {{4 + {e^2}} \over 4}$

Given,

$y(x + 1)dx - {x^2}dy = 0$

$ \Rightarrow \left( {{{x + 1} \over {{x^2}}}} \right)dx = {{dy} \over y}$

$ \Rightarrow {1 \over x}dx + {{dx} \over {{x^2}}} = {{dy} \over y}$

Integrating both sides, we get

$\int {{{dx} \over x} + \int {{{dx} \over {{x^2}}} = \int {{{dy} \over y}} } } $

$ \Rightarrow \ln |x| - {1 \over x} = \ln |y| + C$ ..... (1)

Given $y(1) = e$

$\therefore$ $x = 1$ and $y = e$

Putting value of x and y in equation (1), we get

$\ln |1| - {1 \over 1} = \ln |e| + C$

$ \Rightarrow 0 - 1 = 1 + C$

$ \Rightarrow C = - 2$

$\therefore$ Equation (1) becomes,

$\ln |x| = - {1 \over x} = \ln |y| - 2$

$ \Rightarrow \ln |y| = \ln |x| - {1 \over x} + 2$

$ \Rightarrow y = {e^{\ln |x|}}\,.\,{e^{2 - {1 \over x}}}$

$ \Rightarrow y = x\,.\,{e^{2 - {1 \over x}}}$

Now,

$\mathop {\lim }\limits_{x \to {0^ + }} y$

$ = \mathop {\lim }\limits_{x \to {0^ + }} \left( {x\,.\,{e^{2 - {1 \over x}}}} \right)$

$ = 0\,.\,{e^{ - \alpha }}$

$ = 0$

Given,

$\left( {{x^2} - 3{y^2}} \right)dx + 3xydy = 0$

$ \Rightarrow {x^2}dx - 3{y^2}dx + 3xydy = 0$

$ \Rightarrow {{{x^2}dx} \over {3xdx}} - {{3{y^2}dx} \over {3xdx}} + {{3ydy} \over {3xdx}} = 0$

$ \Rightarrow {x \over 3} - {{{y^2}} \over x} + y{{dy} \over {dx}} = 0$

Let ${y^2} = t$

$2y{{dy} \over {dx}} = {{dt} \over {dx}}$

$\therefore$ ${1 \over 2}{{dt} \over {dx}} - {t \over x} + {x \over 3} = 0$

$ \Rightarrow {{dt} \over {dx}} - {{2t} \over x} = - {{2x} \over 3}$

This is linear Differential Equation.

$ \text { I.F. }=e^{\int-\frac{2}{x} d x}=e^{-2 \ln |x|}=e^{\ln \frac{1}{x^2}}=\frac{1}{x^2} $

So, $t \cdot \frac{1}{x^2}=\int \frac{1}{x^2}\left(\frac{-2 x}{3}\right) d x $

$\Rightarrow \frac{y^2}{x^2}=\frac{-2}{3} \ln |x|+C$

When, $x=1, y=1$

$ 1=\frac{-2}{3} \ln (1)+C \Rightarrow C=1 $

$ \therefore \frac{y^2}{x^2}=\frac{-2}{3} \ln |x|+1 $

At $x=e, \frac{y^2(e)}{e^2}=\frac{-2}{3}+1 $

$\Rightarrow y^2(e)=\frac{e^2}{3} $

$\Rightarrow 6 y^2(e)=2 e^2$

$ \begin{aligned} & \text { } y=x a \cos (\log x)+x b \sin (\log x) \\ & \frac{d y}{d x}=a \cos (\log x)+a x \frac{(-\sin (\log x))}{x} +b \sin (\log x)+b x \frac{\cos (\log x)}{x} \end{aligned} $

$ \begin{array}{r} \Rightarrow \frac{d y}{d x}=\frac{y}{x}+(-a \sin (\log g x))+b \cos (\log x) \\ \Rightarrow \frac{d^2 y}{d x^2}=\frac{d y}{d x} \cdot \frac{1}{x}+y\left(\frac{-1}{x^2}\right) -\frac{a \cos (\log x)}{x}-\frac{b \sin (\log x)}{x} \end{array} $

$ \begin{aligned} & \Rightarrow \frac{d^2 y}{d x^2}=\frac{d y}{d x} \cdot \frac{1}{x}-\frac{y}{x^2}-\frac{1}{x^2}(y) \\ & \Rightarrow x^2 \frac{d^2 y}{d x^2}-x \frac{d y}{d x}+2 y=0 \end{aligned} $

The given differential equation is

$ \begin{aligned} & y^2 d x+\left(x^2-x y-y^2\right) d y=0 \\ \Rightarrow \quad & \frac{d y}{d x}=\frac{-y^2}{x^2-x y-y^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

above equation is in the form of homogeneous equation.

So, we put $y=v x$

$ \begin{aligned} & \Rightarrow \quad \frac{d y}{d x}=v+x \frac{d v}{d x} \text { in Eq. (i), We get, } \\ & v+x \frac{d v}{d x}=\frac{-v^2 x^2}{x^2-x(v x)-v^2 x^2} \\ & \Rightarrow x \frac{d v}{d x}=\frac{-v^2}{1-v-v^2}-v \end{aligned} $

$ \Rightarrow x \frac{d v}{d x}=\frac{-v\left(v^2-1\right)}{v^2+v-1}-\left[\frac{1}{v^2-1}+\frac{1}{v}\right] d v=\frac{d x}{x} $

On integrating both sides, we get

$ \begin{aligned} & -\left[\frac{1}{2 \times 1} \log \left|\frac{v-1}{v+1}\right|+\log v\right]=(\log x)+\log C \\ & \Rightarrow-\log \left(\sqrt{\frac{v-1}{v+1}} \cdot v\right)=\log C x \\ & \Rightarrow-\log \sqrt{\frac{y-x}{y+x}} \cdot \frac{y}{x}=\log C x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

On putting $x=2, y=1$

$ \begin{aligned} -\log \sqrt{\frac{1}{3}} \cdot \frac{1}{2} & =\log 2 C \\ \Rightarrow \quad C & =\sqrt{3} \end{aligned} $

So, solution of above differential equation is (from Eq. (ii)

$ \begin{aligned} & \sqrt{\frac{y-x}{y+x}} \cdot \frac{y}{x}=\frac{1}{\sqrt{3} x} \\ & 3 y^2(y-x)=y+x \\ & \quad-3 y^2(x-y)=x+y \end{aligned} $

On comparing it with given solution, we get

$ k=-3 $

Given, differential equation

$ \frac{d y}{d x}+\frac{y}{x}=x^2 $

This is linear differential equation

$ \begin{aligned} & P=1 / x, Q=x^2 \\ & \mathrm{IF}=e^{\int \frac{1}{x} d x}=e^{\ln x}=x \end{aligned} $

Solution $y \cdot x=\int x^2 \cdot x d x+C$

$ \Rightarrow \quad x y=\frac{x^4}{4}+C $

$y^2=4 a(x-a)$

Differentiating, $2 y \frac{d y}{d x}=4 a \Rightarrow y \frac{d y}{d x}=2 a$

Again differentiating,

$ y \frac{d^2 y}{d x^2}+\frac{d y}{d x} \cdot \frac{d y}{d x}=0 \Rightarrow y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2=0 $

Order $m=1$

Degree $n=2$

$ \therefore \quad m+n^2=1+(2)^2=1+4=5 $

$\frac{d y}{d x}=\frac{2 x+3 y}{3 x-2 y}$

On putting, $y=v x$

$ \begin{aligned} & \frac{d y}{d x}=v+x \frac{d v}{d x} \\ \Rightarrow & v+x \frac{d v}{d x}=\frac{2 x+3 v x}{2 x-2 v x} \\ \Rightarrow & v+x \frac{d v}{d x}=\frac{2+3 v}{3-2 v} \\ \Rightarrow & v-\frac{2+3 v}{3-2 v}=\frac{-x d v}{d x} \\ \Rightarrow & \frac{3 v-2 v^2-2-3 v}{3-2 v}=-x \frac{d v}{d x} \\ \Rightarrow & \frac{-2 v^2-2}{3-2 v}=-x \frac{d v}{d x} \\ \Rightarrow & \frac{-d x}{x}=\frac{3-2 v}{-2 v^2-2} d v \end{aligned} $

$ \begin{aligned} &\text { On integrating both sides, we get }\\ &\begin{aligned} &-\int \frac{d x}{x}=\int \frac{2 v-3}{2 v^2+2} d v \\ & \Rightarrow-\log |x|=\int \frac{2 v}{2 v^2+2} d v-\int \frac{3}{2 v^2+2} d v \\ & \Rightarrow-\log |x|=\frac{1}{2} \log \left|v^2+1\right|-\frac{3}{2} \tan ^{-1} v+C \\ & \Rightarrow-\log |x|=\frac{1}{2} \log \left|(y / x)^2+1\right| -\frac{3}{2} \tan ^{-1}(y / x)+C \\ & \Rightarrow \frac{3}{2} \tan ^{-1}(y / x)=\log \left(\frac{y^2+x^2}{x^2}\right)^{1 / 2}+\log x+C \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \frac{3}{2} \tan ^{-1}(y / x)=\log \left(y^2+x^2\right)^{1 / 2} -\log x+\log x+C \\ & \Rightarrow 3 \tan ^{-1}(y / x)=2 \log \left(y^2+x^2\right)^{1 / 2} \\ & \Rightarrow \quad 3 \tan ^{-1}(y / x)=\log \left(y^2+x^2\right)^{\frac{1}{2} \times 2} \\ & \Rightarrow \quad \tan ^{-1}(y / x)=\frac{1}{3} \log \left(y^2+x^2\right) \\ & \Rightarrow \quad y / x=\tan \left[\frac{1}{3} \log \left(y^2+x^2\right)\right] \\ & \Rightarrow \quad y=x \tan \left[\frac{1}{3} \log \left(y^2+x^2\right)\right] \end{aligned}\\ &\text { Thus, } f(x)=\frac{1}{3} \log \left(x^2+y^2\right) \end{aligned} $

$\left(x^2+2\right) d y+2 x y d x=e^x\left(x^2+2\right) d x$

Dividing on both sides by ' $d x^{\prime}$ ',

$ \begin{aligned} & \left(x^2+2\right) \frac{d y}{d x}+2 x y=e^x\left(x^2+2\right) \\ \Rightarrow & \left(x^2+2\right) \cdot \frac{d y}{d x}+y \cdot \frac{d}{d x}\left(x^2+2\right)=e^x\left(x^2+2\right) \\ \Rightarrow & \frac{d}{d x}\left[y \cdot\left(x^2+2\right)\right]=e^x\left(x^2+2\right) \end{aligned} $

Integrating on both sides, we get

$ \begin{aligned} & \int \frac{d}{d x}\left[y \cdot\left(x^2+2\right)\right] \cdot d x=\int e^x \cdot\left(x^2+2\right) \cdot d x \\ & \begin{array}{r} \Rightarrow \quad\left(x^2+2\right) y=\left(x^2+2\right) \\ \quad \int e^x \cdot d x-\int\left[\frac{d}{d x}\left(x^2+2\right) \cdot \int e^x \cdot d x\right] d x \end{array} \\ & =\left(x^2+2\right) e^x-\int 2 x \cdot e^x \cdot d x \\ & =\left(x^2+2\right) e^x-2 \int x \cdot e^x d x \\ & =\left(x^2+2\right) e^x-2\left[x \cdot \int e^x d x\right. \left.\quad-\left[\int \frac{d}{d x}(x) \cdot \int e^x d x\right] d x\right] \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\left(x^2+2\right) e^x-2\left[x \cdot e^x-\int e^x \cdot d x\right] \\ & =\left(x^2+2\right) e^x-2\left[e^x(x-1)\right] \\ & =e^x\left[x^2+2-2 x+2\right] \\ & =e^x\left(x^2-2 x+4\right) \end{aligned}\\ &\text { Hence, }\left(x^2+2\right) y=e^x\left(x^2-2 x+4\right)+C \end{aligned} $

Given,

$ \begin{aligned} & (3 x-4 y)(d x-3 d y)+(6 d x-4 d y)=0 \\ & \Rightarrow(3 x-4 y+6) d x+(-9 x+12 y-4) d y=0 \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{3 x-4 y+6}{9 x-12 y+4} \end{aligned} $

Let $\quad 3 x-4 y=z$

$ \therefore \quad 3-4 \frac{d y}{d x}=\frac{d z}{d x} \Rightarrow \frac{d y}{d x}=\frac{1}{4}\left(3-\frac{d z}{d x}\right) $

Now, $\frac{1}{4}\left(3-\frac{d z}{d x}\right)=\frac{z+6}{3 z+4}$

$ \begin{aligned} \Rightarrow \quad 3-\frac{d z}{d x} & =\frac{4 z+24}{3 z+4} \\ \Rightarrow \quad \frac{d z}{d x} & =3-\frac{4 z+24}{3 z+4} \\ & =\frac{9 z+12-4 z-24}{3 z+4} \\ & =\frac{5 z-12}{3 z+4} \end{aligned} $

$ \begin{aligned} &\Rightarrow \frac{3 z+4}{5 z-12} d z=d x \Rightarrow \frac{1}{5}\left[3+\frac{56}{5 z-12}\right] d z=d x\\ &\text { On integrating both sides, we get }\\ &\frac{1}{5}\left[3 z+\frac{56}{5} \ln |5 z-12|\right]=x+A \end{aligned} $

$ \begin{array}{ll} \Rightarrow {\left[3(3 x-4 y)+\frac{56}{5} \ln |5(3 x-4 y)-12|\right]} \\ =5 x+5 A \\ \Rightarrow 9 x-12 y+\frac{56}{5} \ln |15 x-20 y-12| \\ =5 x+5 A \\ \Rightarrow 45 x-60 y+56 \ln |15 x-20 y-12| \\ \Rightarrow \begin{array}{l} 25 x+25 A \\ 20 x-60 y+56 \ln |15 x-20 y-12| \\ =25 A \end{array} \\ \Rightarrow 5 x-15 y+14 \ln |15 x-20 y-12|=C \\ {\left[\text { where, } C=\frac{25}{4} A\right]} \end{array} $

$ \begin{aligned} & \text { Given, }(\sec x+\tan x) \frac{d y}{d x}+\left(\sec ^2 x\right. \\ & +\sec x \tan x) y=1 \\ & \Rightarrow \quad(\sec x+\tan x)\left[\frac{d y}{d x}+y \cdot \sec x\right]=1 \\ & \Rightarrow \quad \frac{d y}{d x}+y \cdot \sec x=\frac{1}{(\sec x+\tan x)} \end{aligned} $

Eq. (i) is the linear differential equation of the form $\frac{d y}{d x}+P y=Q$.

where $P=\sec x$ and $Q=\frac{1}{\sec x+\tan x}$

$ \text { } \begin{array}{rlrl} & \text { Now, IF } =e^{\int P d x}=e^{\int \sec x d x} \\ & = e^{\log |\sec x+\tan x|} \\ \Rightarrow & \text { IF } =\sec x+\tan x \end{array} $

∴ Solution is $y \times$ IF $=\int$ IF $\times Q d x+C$

$ \begin{aligned} & \Rightarrow y(\sec x+\tan x)=\int 1 d x+C \\ & \Rightarrow y(\sec x+\tan x)=x+C \end{aligned} $

we have, $y=A e^x+B \sin 2 x$

$ \begin{aligned} d y / d x & =A e^x+2 B \cos 2 x \\ d^2 y / d x^2 & =A e^x-4 B \sin 2 x \end{aligned} $

On substituting these values in

$ \begin{aligned} & \text { option (a) } \\ & \begin{array}{r} \text { LHS }=(\cos 2 x-\sin 2 x)\left(A e^x-4 B \sin 2 x\right) \\ +4 \sin 2 x\left(A e^x+2 B \cos 2 x\right) \\ -4(\sin 2 x+\cos 2 x)\left(A e^x+B \sin 2 x\right) \\ =A e^x(\cos 2 x)-4 B \sin 2 x \cos 2 x-A e^x \sin 2 x \\ +4 B \sin ^2 2 x+4 A e^x \sin 2 x+8 B \sin 2 x \cos 2 x \\ -4 A e^x \sin 2 x-4 B \sin ^2 2 x-4 A e^x \cos 2 x \\ -4 B \sin 2 x \cos 2 x \neq 0 \end{array} \end{aligned} $

No other option is also satisfying the equation.