Differential Equations

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 $

Here, $\frac{d y}{d x}=\cos ^2(3 x+y)$

On putting $3 x+y=t$

$\begin{array}{ll} 3+\frac{d y}{d x} & =\frac{d t}{d x} \\ \Rightarrow & \frac{d y}{d x}=\frac{d t}{d x}-3 \Rightarrow \frac{d t}{d x}-3=\cos ^2 t \\ \Rightarrow & \frac{d t}{d x}=\cos ^2 t+3 \Rightarrow \frac{d t}{\cos ^2 t+3}=d x \end{array}$

Integrate both side,

$\int \frac{d t}{\cos ^2 t+3}=\int d x$

$\begin{aligned} & \Rightarrow \int \frac{\sec ^2 t d t}{1+3 \sec ^2 t}=\int d x \Rightarrow \int \frac{\sec ^2 t}{1+3+3 \tan ^2 t}=\int d x \\ & \Rightarrow \int \frac{\sec ^2 t d t}{4+3 \tan ^2 t}=\int d x \end{aligned}$

$\begin{aligned} & \text { On putting tant }=m, \\ & \sec ^2 t d x=d m=\frac{1}{4} \int \frac{d m}{1+\left(\frac{\sqrt{3}}{2} m\right)^2} x+C \\ & =\frac{1}{4} \times \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3}}{2} m\right)=x+C \\ & =\frac{1}{2 \sqrt{3}} \tan ^{-1}\left[\frac{\sqrt{3}}{2} \tan t\right]=x+C \quad[\because m=\tan t] \\ & =\tan ^{-1}\left[\frac{\sqrt{3}}{2} \tan (3 x+y)=2 \sqrt{3}(x+C)\right] \quad[\therefore t=3 x+y] \end{aligned}$

So, $f(x)=2 \sqrt{3}(x+C)$

Given,

$\begin{aligned} & \cos ^2 x \frac{d y}{d x}+y=\tan x \\ \Rightarrow & \frac{d y}{d x}+y \sec ^2 x=\tan x \cdot \sec ^2 x \quad \text{... (i)} \end{aligned}$

Here, $p=\sec ^2 x$

$\begin{aligned} & \Rightarrow \quad \int p d p=\int \sec ^2 x d x=\tan x \\ & I F=e^{\tan x} \end{aligned}$

Multiplying Eq. (i) by $I F$, we get

$e^{\tan x} \frac{d y}{d x}+e^{\tan x} y \sec ^2 x=e^{\tan x} \cdot \tan x \cdot \sec ^2 x$

Integrating both sides, we get

$y e^{\tan x}=\int e^{\tan x} \tan x \cdot \sec ^2 x d x$

On putting $\tan x=t$,

$\begin{array}{ll} & \sec ^2 x d x=d t \\ \therefore & y e^t=\int t e^{\prime} d t=e^t(t-1)+C \\ \therefore & y e^{\tan x}=\tan x-1+C e^{-\tan x} \\ \text { If } & y\left(\frac{\pi}{4}\right)=1 \\ \Rightarrow & \quad 1=1-1+C e^{-1} \Rightarrow 1=C e^{-1} \\ \therefore & C=e \end{array}$

Any circle with given radius can be written as, $(x-h)^2+(y-k)^2=a^2$ where $(h, k)$ be the centre of the circle which is variable. So, in above algebraic equation, there are two arbitrary constant $h$ and $k$. Hence, order of differential equation will be second order. Hence, assertion and reason are true and reason is the correct explanation to assertion.

Any straight line which is at a constant distance $p$ from the origin is

$x \cos \alpha+y \sin \alpha=p\quad \text{.... (i)}$

Differentiating Eq. (i) on both sides w.r.t. $x$,

$\cos \alpha+\sin \alpha \frac{d y}{d x}=0 \Rightarrow \tan \alpha=-\frac{d x}{d y}=-\frac{1}{y^{\prime}}$

Now, $\sin \alpha=\frac{1}{\sqrt{1+y^{\prime 2}}}, \cos \alpha=\frac{-y^{\prime}}{\sqrt{1+y^{\prime 2}}}$

On putting Eq. (i), we get

$\frac{x\left(-y^{\prime}\right)}{\sqrt{1+y^{\prime 2}}}+\frac{y}{\sqrt{1+y^{\prime 2}}}=p$

$\begin{aligned} & \Rightarrow\left(y-x y^{\prime 2}\right)^2=p^2\left(l+y^{\prime 2}\right) \\ & \text { Order }(l)=1 \\ & \text { Degree }(m)=2 \\ & \therefore lm^2+l^2 m=(1 \cdot 2)^2+(1)^2 \cdot 2=4+2=6 \end{aligned}$

Given, $\frac{d y}{d x}=\frac{2 x-4 y-5}{x-2 y+2}$

Let $x-2=v$

$\therefore 1-2 \frac{d y}{d x}=\frac{d v}{d x} \Rightarrow \frac{d y}{d x}=\frac{\left(1-\frac{d v}{d x}\right)}{2}$

$\therefore \frac{\left(1-\frac{d v}{d x}\right)}{2}=\frac{2 v-5}{v+2}$

$\begin{aligned} & \Rightarrow 1-\frac{d v}{d x}=\frac{4 v-10}{v+2} \Rightarrow 1-\frac{4 v-10}{v+2}=\frac{d v}{d x} \\ & \Rightarrow \frac{d v}{d x}=\frac{12-3 v}{v+2} \Rightarrow \int \frac{(v+2)}{4-v} d v=3 \int d x+C \\ & \Rightarrow-1 \int \frac{-v+4-6}{(4-v)} d v=3 x+C \\ & \Rightarrow-\int d v+6 \int \frac{d v}{4-v}=3 x+C \end{aligned}$

$\begin{aligned} & \Rightarrow-(v)-6 \log |4-v|=3 x+C \\ & \Rightarrow-(x-2 y)-6 \log |4-x+2 y|=3 x+C \\ & \Rightarrow(4 x-2 y)+6 \log |4-x+2 y|=-C \\ & \Rightarrow(2 x-y)+3 \log |x-2 y-4|=k \quad\left(k=-\frac{C}{2}\right) \end{aligned}$

Thus, C = 3

Given, $y=(a+b) \sin (x+c)-d e^{x+e+f}\quad \text{... (i)}$

$ \begin{aligned} & \frac{d y}{d x}=(a+b) \cos (x+c)-d e^{x+c+f} \quad \text{... (ii)}\\ & \frac{d^2 y}{d x^2}=-(a+b) \sin (x+c)-d e^{x+e+f} \quad \text{... (iii)}\\ & =\frac{d^3 y}{d x^3}=-(a+b) \cos (x+c)-d e^{x+c+f} \quad \text{... (iv)} \end{aligned}$

On adding Eqs. (i) and (iii), we get

$\left(y+\frac{d^2 y}{d x^2}\right)=-2 d e^{x+e+f}\quad \text{... (v)}$

Adding Eqs (ii) and (iv), we get

$\left(\frac{d y}{d x}+\frac{d^3 y}{d x^3}\right)=-2 d e^{x+e+s} \quad \text{... (vi)}$

On equating Eqs. (v) and (vi) we get

$\begin{aligned} & \frac{d^3 y}{d x^3}+\frac{d y}{d x}=\frac{d^2 y}{d x^2}+y \\ & \Rightarrow \frac{d^3 y}{d x^3}-\frac{d^2 y}{d x^2}+\frac{d y}{d x}-y=0 \end{aligned}$

Order of above differential equation is 3.

$\begin{aligned} & \frac{d y}{d x}-y \log _c 0.5=0 \\ & \Rightarrow \frac{d y}{y}=\log 0.5 d x \\ & \Rightarrow \quad \int \frac{d y}{y}=\int \log 0.5 d x \\ & \Rightarrow \log y=(\log 0.5) x+C \quad \text{... (i)}\\ & \because \quad y(0)=1 \text { i.e at } x=0, y=1 \\ & \therefore \log (1)=0+C \\ & \Rightarrow \quad C=0 \\ \end{aligned}$

Therefore, Eq. (i) becomes $\log y=(\log 0.5) x$

$\begin{aligned} & \Rightarrow \quad \log y=\log (0.5)^x \\ & \Rightarrow \quad y=(0.5)^x \end{aligned}$

When $x \rightarrow \infty, y \rightarrow k$

$\begin{array}{ll} \therefore & k=(0.9)^{\infty} \\ \Rightarrow & k=\left(\frac{1}{2}\right)^{\infty}=\frac{1}{2^{\infty}}=\frac{1}{\infty}=0 \end{array}$

$y=A e^x+B e^{-2 x}$

On differentiating w.r.t. $x$, we get

$\frac{d y}{d x}=A e^x-2 B e^{-2 x} \quad \text{... (i)}$

Again on differentiating w.r.t. $x$, we get

$\frac{d^2 y}{d x^2}=A e^x+4 B e^{-2 x} \quad \text{.... (ii)}$

Adding Eqs. (ii) and (iii), we get

$\begin{aligned} & \frac{d^2 y}{d x^2}+\frac{d y}{d x}=2 A e^x+2 B e^{-2 x}=2 y \\ \Rightarrow & \frac{d^2 y}{d x^2}+\frac{d y}{d x}-2 y=0 \end{aligned}$

$\begin{aligned} & y=\sin (\sin x), \text { then } y^{\prime}=\cos (\sin x) \cdot \cos x \\ & y^{\prime \prime}=-\sin x \cos (\sin x)-\cos ^2 x \sin (\sin x) \\ &=-\sin x \cos (\sin x)-\left(\cos ^2 x\right) y \\ &=-\sin x\left(\frac{y^{\prime}}{\cos x}\right)-\left(\cos ^2 x\right) y \\ & \Rightarrow y^{\prime \prime}+(\tan x) y^{\prime}+\left(\cos ^2 x\right) y=0 \\ & \Rightarrow \quad \quad f(x)=\tan x, g(x) \cos ^2 x \\ & \Rightarrow f(x) \cdot g(x)=\sin x \cos x=\frac{1}{2}(2 \sin x \cos x) \\ & \quad f(x) \cdot g(x)=\frac{1}{2} \sin 2 x\end{aligned}$

$\left.\left(e^x \tan y\right) d x+\left(1+e^x\right) \sec ^2 y\right) d y=0$ ...... (i)

Its of the form $M d x+N d y=0$

Now, $\frac{\partial M}{\partial y}=e^x \sec ^2 y$ and $\frac{\partial N}{\partial x}=e^x \sec ^2 y$

$\Rightarrow$ Eq. (i) is exact.

Thus, its solution will be

$\int M d x+\int N d y=C$

(treat $y$-constant) No terms of $x$ included

$\begin{aligned} \int e^x \tan y d x+\int \sec ^2 y d y & =C \\ e^x \tan y+\tan y & =C \\ \tan y\left(1+e^x\right) & =C \quad \text{......(ii)} \end{aligned}$

Since, Eq. (ii) passes through $(0, \pi / 4)$

$\tan \frac{\pi}{4}(1+1)=C \Rightarrow C=2$

From Eq. (ii), $\tan y\left(1+e^x\right)=2$

$\begin{aligned} 2 x \frac{d y}{d x}-y & =4 \\ \frac{d y}{d x}-\left(\frac{y}{2 x}\right) & =\left(\frac{4}{2 x}\right) \\ \mathrm{IF} & =e^{\int \frac{-1}{2 x} d x}=e^{\frac{\log x}{-2}}=\frac{1}{\sqrt{x}} \\ \left(\frac{1}{\sqrt{x}} y\right) & =\int \frac{2}{x \sqrt{x}} d x \Rightarrow \frac{y}{\sqrt{x}}=2\left(\frac{x^{-\frac{3}{2}+1}}{\frac{-3}{2}+1}\right)+C \\ \frac{y}{\sqrt{x}} & =2\left(\frac{x^{-\frac{1}{2}}}{-\frac{1}{2}}\right)+C \\ y & =C \sqrt{x}-4 \Rightarrow(y+4)^2=C^2 x \end{aligned}$

$\Rightarrow$ Represents a parabola.

Given, differential equation $\frac{d^2 y}{d x^2}+y=0$

A solution is function of $x$ which satisfies given differential equation.

$\begin{aligned} & \text { Option (a) } y=3 \sin x+4 \cos x \\ & \qquad \begin{aligned} \frac{d y}{d x} & =3 \cos x-4 \sin x \\ \frac{d^2 y}{d x^2} & =-3 \sin x-4 \cos x=-y \\ \Rightarrow \frac{d^2 y}{d x^2}+y & =0 \end{aligned} \end{aligned}$