Differential Equations

The given equation is:

$ x \frac{d y}{d x}-y=x^2 \cot x $

To get it into the standard form $\frac{d y}{d x}+P(x) y=Q(x)$, we divide the entire equation by $x$ :

$ \frac{d y}{d x}-\frac{1}{x} y=x \cot x $

Here, $P(x)=-\frac{1}{x}$ and $Q(x)=x \cot x$.

Finding the Integrating Factor (I.F.)

The integrating factor is calculated as:

$ \text { I.F. }=e^{\int P(x) d x}=e^{\int-\frac{1}{x} d x}=e^{-\ln x}=e^{\ln \left(x^{-1}\right)}=\frac{1}{x} $

The solution to the differential equation is:

$ \begin{gathered} y \cdot(I . F .)=\int Q(x) \cdot(I . F .) d x \\ \frac{y}{x}=\int(x \cot x) \cdot \frac{1}{x} d x \\ \frac{y}{x}=\int \cot x d x \\ \frac{y}{x}=\ln (\sin x)+C \\ y=x \ln (\sin x)+C x \end{gathered} $

Applying the Initial Condition

We are given $y\left(\frac{\pi}{2}\right)=\frac{\pi}{2}$. Plugging these values in:

$ \frac{\pi}{2}=\frac{\pi}{2} \ln \left(\sin \frac{\pi}{2}\right)+C\left(\frac{\pi}{2}\right) $

So, the particular solution is $y=x \ln (\sin x)+x$.

We need to find the value of $6 y\left(\frac{\pi}{6}\right)-8 y\left(\frac{\pi}{4}\right)$.

• Calculate $y\left(\frac{\pi}{6}\right)$ :

$ y\left(\frac{\pi}{6}\right)=\frac{\pi}{6} \ln \left(\sin \frac{\pi}{6}\right)+\frac{\pi}{6}=\frac{\pi}{6} \ln \left(\frac{1}{2}\right)+\frac{\pi}{6}=\frac{\pi}{6}(1-\ln 2) $

• Calculate $y\left(\frac{\pi}{4}\right)$ :

$ y\left(\frac{\pi}{4}\right)=\frac{\pi}{4} \ln \left(\sin \frac{\pi}{4}\right)+\frac{\pi}{4}=\frac{\pi}{4} \ln \left(\frac{1}{\sqrt{2}}\right)+\frac{\pi}{4}=\frac{\pi}{4} \ln \left(2^{-1 / 2}\right)+\frac{\pi}{4}=\frac{\pi}{4}\left(1-\frac{1}{2} \ln 2\right) $

• Calculate $y\left(\frac{\pi}{4}\right)$ :

$ y\left(\frac{\pi}{4}\right)=\frac{\pi}{4} \ln \left(\sin \frac{\pi}{4}\right)+\frac{\pi}{4}=\frac{\pi}{4} \ln \left(\frac{1}{\sqrt{2}}\right)+\frac{\pi}{4}=\frac{\pi}{4} \ln \left(2^{-1 / 2}\right)+\frac{\pi}{4}=\frac{\pi}{4}\left(1-\frac{1}{2} \ln 2\right) $

Now, substitute these into the expression:

$ \begin{aligned} 6 y\left(\frac{\pi}{6}\right)-8 y\left(\frac{\pi}{4}\right) & =6\left[\frac{\pi}{6}(1-\ln 2)\right]-8\left[\frac{\pi}{4}\left(1-\frac{1}{2} \ln 2\right)\right]=\pi(1-\ln 2)-2 \pi\left(1-\frac{1}{2} \ln 2\right) \\ = & \pi-\pi \ln 2-2 \pi+\pi \ln 2=-\pi \end{aligned} $

Given equation is :

$ x \frac{d y}{d x}-\sin 2 y=x^3\left(2-x^3\right) \cos ^2 y, \quad x \neq 0 $

Divide by $x \cos ^2 y$

$ \sec ^2 y \frac{d y}{d x}-\frac{\sin 2 y}{x \cos ^2 y}=x^2\left(2-x^3\right) $

$ \sec ^2 y \frac{d y}{d x}-\frac{2 \tan y}{x}=2 x^2-x^5 $

Substitute $\tan y=t$ and differentiate w.r.t $x$

$ \sec ^2 y \frac{d y}{d x}=\frac{d t}{d x} $

Equation (1) becomes linear differential equation.

$\frac{d t}{d x}-\frac{2}{x} t=2 x^2-x^5$

Integrating factor $I F=e^{\int\left(-\frac{2}{x}\right) d x}=e^{-2 \ln x}=e^{\ln \left(\frac{1}{x^2}\right)}=\frac{1}{x^2}$

Solution of linear differential equation :

$ t \cdot I F=\int(I F)\left(2 x^2-x^5\right) d x $

$\Rightarrow $ $t \cdot \frac{1}{x^2}=\int \frac{1}{x^2}\left(2 x^2-x^5\right) d x$

$\Rightarrow $ $ \frac{t}{x^2}=\int\left(2-x^3\right) d x=2 x-\frac{x^4}{4}+C $

put value of $t=\tan y$

$ \frac{\tan y}{x^2}=2 x-\frac{x^4}{4}+C $

using condition $y(2)=0$ to find $C$.

$ \frac{\tan (0)}{2^2}=2 \times 2-\frac{2^4}{4}+C \Longrightarrow C=0 $

So, solution of differential equation is

$ \tan y=2 x^3-\frac{x^6}{4} $

to calculate $\tan (y(1))$, put $x=1$ in the solution.

$ \tan (y(1))=2(1)^3-\frac{(1)^6}{4} $

$=2-\frac{1}{4}=\frac{7}{4}$

$ \begin{aligned} & x^4 d y+\left(4 x^3 y+2 \sin x\right) d x=0 \\ & \frac{d y}{d x}+\frac{4 x^3}{x^4} y=\frac{-2 \sin x}{x^4} \\ & \frac{d y}{d x}+\frac{4}{x} y=\frac{-2 \sin x}{x^4} \\ & \text { IF }=e^{\int \frac{4}{x} d x=e^{\ln x^4}=x^4} \\ & x^4 \frac{d y}{d x}+4 x^3 y=-2 \sin x \\ & \frac{d}{d x}\left(x^4 y\right)=-2 \sin x \\ & \Rightarrow x^4 y=2 \cos x+c \\ & y\left(\frac{\pi}{2}\right)=0 \\ & \Rightarrow c=0 \\ & \therefore x^4 y=\cos x \\ & \Rightarrow y=\frac{2 \cos x}{x^4} \end{aligned} $

$ \begin{aligned} & y\left(\frac{\pi}{3}\right)=\frac{2 \cos \left(\frac{\pi}{3}\right)}{\left(\frac{\pi}{3}\right)^4}=\frac{1}{\left(\frac{\pi}{3}\right)^4}=\frac{81}{\pi^4} \\ & \pi^4 y\left(\frac{\pi}{3}\right)=81 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \int \frac{\cos y}{1+2 \sin y} d y=\int \frac{1}{16(\sqrt{x+9 \sqrt{x}})(4+\sqrt{9+\sqrt{x}})} d x \\ & \Rightarrow \frac{1}{2} \int \frac{2 \cos y}{1+2 \sin y} d y=\int \frac{1}{4 t} d t \end{aligned}\\ &\text { Put } 4+\sqrt{9+\sqrt{x}}=t\\ &\begin{aligned} & \Rightarrow \frac{1}{2 \sqrt{9+\sqrt{x}}} \cdot \frac{1}{2 \sqrt{x}} d x=d t \Rightarrow \frac{1}{4 \sqrt{x+9 \sqrt{x}}} d x=d t \\ & \Rightarrow \frac{1}{2} \log |1+2 \sin y|=\frac{1}{4} \log |t|+\log c \Rightarrow \sqrt{1+2 \sin y}=c t^{1 / 4} \Rightarrow \frac{1}{4} \log |t|+\log c \\ & \Rightarrow \sqrt{1+2 \sin y}=c t^{1 / 4} \Rightarrow \sqrt{1+2 \sin y}=c(4+\sqrt{9+\sqrt{x}})^{1 / 4} \\ & \left(256, \frac{\pi}{2}\right) \Rightarrow \sqrt{3}=c(\sqrt{3}) \Rightarrow c=1 \\ & (49, \alpha) \Rightarrow \sqrt{1+2 \sin \alpha}=(1) 8^{1 / 4} \\ & \Rightarrow 1+2 \sin \alpha=2 \sqrt{2} \Rightarrow 2 \sin \alpha=2 \sqrt{2}-1 \end{aligned} \end{aligned} $

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

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

$ \begin{aligned} & \frac{d y}{d x}-2 \cos x \cdot y=\cos x(2+3 \sin x) \\ & I \cdot F=e^{-2 \sin x} \\ & G \cdot S y \cdot e^{-2 \sin x}=\int e^{-2 \sin x} \cos x(2+3 \sin x) d x \\ & \text { Put } \begin{array}{l} \sin x=t \\ \cos x \cdot d x=d t \end{array} \end{aligned} $

$ y=e^{-2 \sin x}\left(\frac{2+3 \sin x}{-2}-\frac{3}{4}\right)+c $

When $x=0, c=0$

When $x=\frac{\pi}{6}, y=-\frac{5}{2}$

Given

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

divide by $dx$:

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

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

This is linear. Integrating factor:

$ \mu(x)=\exp\!\left(\int \frac{1}{1+x^2}\,dx\right)=e^{\tan^{-1}x}. $

So

$ \frac{d}{dx}\left( y e^{\tan^{-1}x}\right)=e^{\tan^{-1}x}\cdot \frac{\tan^{-1}x}{1+x^2}. $

Let $t=\tan^{-1}x$, then $dt=\frac{dx}{1+x^2}$. Hence

$ y e^t=\int t e^t\,dt + C = e^t(t-1)+C. $

Therefore

$ y=t-1+Ce^{-t}=\tan^{-1}x-1+Ce^{-\tan^{-1}x}. $

Use $y(0)=1$ and $\tan^{-1}0=0$:

$ 1=0-1+C \implies C=2. $

So

$ y(1)=\frac{\pi}{4}-1+2e^{-\pi/4}=\frac{2}{e^{\pi/4}}+\frac{\pi}{4}-1. $

Correct option: B.

$I=\int\limits_{0}^{2} \frac{1}{3+3^x}\,dx \quad \ldots\ (1)$

$I=\int\limits_{0}^{2} \frac{1}{3+3^{2-x}}\,dx \quad \ldots\ (2)$

Adding (1) and (2)

$2I=\int\limits_{0}^{2} \left(\frac{1}{3+3^x}+\frac{3^x}{3(3+3^x)}\right)dx$

$2I=\int\limits_{0}^{2} \left(\frac{3+3^x}{3(3+3^x)}\right)dx$

$2I=\int\limits_{0}^{2} \frac{1}{3}\,dx$

$2I=\frac{1}{3}[x]_{0}^{2}$

$2I=\frac{2}{3}$

$\Rightarrow I=\frac{1}{3}$

$\Rightarrow \text{Option (B) is correct.}$

We start with the differential equation:

$ x \, dy - y \, dx = -x^3 \, dx $

Divide both sides by $ x^2 $ to simplify:

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

The left side can be written as the derivative of a quotient:

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

Now, integrate both sides:

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

Multiply by $ x $ to find $ y $:

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

We are given $ y(1) = 0 $. Substitute $ x = 1 $ and $ y = 0 $:

$ 0 = -\frac{1}{2} + c \Rightarrow c = \frac{1}{2} $

Hence, the function becomes:

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

Differentiate $ y $ with respect to $ x $:

$ \frac{dy}{dx} = -\frac{3x^2}{2} + \frac{1}{2} $

We can factorize this as:

$ \frac{dy}{dx} = -\frac{3}{2}\left( x - \frac{1}{\sqrt{3}} \right)\left( x + \frac{1}{\sqrt{3}} \right) $

At $ x = \frac{1}{\sqrt{3}} $, the derivative changes from positive to negative, so there is a local maximum.

At $ x = -\frac{1}{\sqrt{3}} $, the derivative changes from negative to positive, so there is a local minimum.

For $ g(x) = f(x) $:

$ 4x^3 - 5x^2 + \frac{3}{2}x = -\frac{x^3}{2} + \frac{x}{2} $

Multiply through to remove fractions and simplify:

$ 9x^3 - 10x^2 + 2x = 0 $

Take $ x $ common:

$ x(9x^2 - 10x + 2) = 0 $

So, $ x = 0 $ or $ 9x^2 - 10x + 2 = 0 $.

Using the quadratic formula:

$ x = \frac{5 \pm \sqrt{7}}{9} $

Both these solutions are positive.

$\begin{aligned} & f(x)=x-1 \\ & f(f(x))=f(x)-1=x-1-1=x-2 \\ & g(f(f(x)))=e^{x-2} \\ & \therefore \frac{d y}{d x}=e^{-2 \sqrt{x}} \times e^{x-2}-\frac{1}{\sqrt{x}} y \\ & \frac{d y}{d x}+\frac{1}{\sqrt{x}} y=e^{x-2 \sqrt{x}-2} \text { which is L.D.E } \\ & \text { I.F. }=e^{\int \frac{d y}{\sqrt{x}}}=e^{2 \sqrt{x}} \end{aligned}$

Its solution is

$\begin{aligned} & y \times e^{2 \sqrt{x}}=\int e^{2 \sqrt{x}} \times e^{x-2 \sqrt{x}-2} d x+c \\ & y \times e^{2 \sqrt{x}}=\int e^{x-2} d x+c \\ & y \times e^{2 \sqrt{x}}=e^{x-2}+c \end{aligned}$

Given $\mathrm{x}=0, \mathrm{y}=0 \Rightarrow 0=\mathrm{e}^{-2}+\mathrm{c} \quad ; \mathrm{c}=-\mathrm{e}^{-2}$

$\therefore \mathrm{y} \times \mathrm{e}^{2 \sqrt{x}}=\mathrm{e}^{\mathrm{x}-2}-\mathrm{e}^{-2}$

when $\mathrm{x}=1, \mathrm{y} \times \mathrm{e}^2=\mathrm{e}^{-1}-\mathrm{e}^{-2}$

$y=\frac{e^{-1}-e^{-2}}{e^2}=\frac{\frac{1}{e}-\frac{1}{e^2}}{e^2}=\frac{e^2-e}{e^5}=\frac{e-1}{e^4}$

Option (1) is correct

$\begin{aligned} &\begin{aligned} & \left(x^2+1\right) \frac{d y}{d x}-2 x y=\left(x^4+2 x^2+1\right) \cos x \\ & \frac{d y}{d x}-\left(\frac{2 x}{x^2+1}\right) y=\frac{\left(x^2+1\right)^2 \cos x}{c x^2+1}=\left(x^2+1\right) \cos x \end{aligned}\\ &\text { (Linear D.E) }\\ &\mathrm{P}=\frac{-2 \mathrm{x}}{\mathrm{x}^2+1}, \mathrm{Q}=\left(\mathrm{x}^2+1\right) \cos \mathrm{x} \end{aligned}$

$\begin{aligned} & \text { I.F }=\mathrm{e}^{\int P d x}=\mathrm{e}^{\int \frac{-2 x}{x^2+1} d x}=\frac{1}{x^2+1} \\ & y \cdot \frac{1}{x^2+1}=\int\left(x^2+1\right) \cos x \cdot \frac{1}{x^2+1} d x \\ & \frac{y}{x^2+1}=\sin x+c \Rightarrow y \cos =1 \Rightarrow c=1 \end{aligned}$

$\begin{aligned} & y=\left(x^2+1\right)(\sin x+1) \\ & \int_{-3}^3 y d x=\int_{-3}^3\left(x^2+1\right)(\sin x+1) \\ & d x=\int_{-3}^3 x^2 \sin x+x^2 \sin x+1 d x \\ & \Rightarrow \int_{-3}^3 x^2 \sin x d x+\int_{-3}^3 x^2 d x+\int_{-3}^3 \sin x d x+\int_{-3}^3 1 d x \\ & =0+18+0+6=24 \end{aligned}$

$\begin{aligned} & x\left(x^2+e^x\right) d y+\left(e^x(x-2) y-x^3\right) d x=0 \\ & \frac{d y}{d x}+\frac{e^x(x-2)}{x\left(x^2+e^x\right)} y=\frac{x^3}{x\left(x^2+e^x\right)} \\ & \text { I.F. }=e^{\iint \frac{e^x(x-2)}{x\left(x^2+e^x\right)} d x} \\ & =e^{\int \frac{e^x+2 x}{e^x+x^2} d x-\int \frac{2}{x} d x} \\ & =e^{\ln e^{-x}+x^2-2 \ln x} \end{aligned}$

$\begin{aligned} & =\frac{e^x+x^2}{x^2} \\ & \therefore \quad y\left(\frac{e^2+x^2}{x^2}\right)=\int d x+c \\ & \Rightarrow \quad y\left(\frac{e^x+x^2}{x^2}\right)=x+c \end{aligned}$

Also, $y(1)=0$

$\begin{aligned} & \Rightarrow \quad c=-1 \\ & \therefore \quad y\left(\frac{e^2+x^2}{x^2}\right)=x-1 \end{aligned}$

Hence, $y(2)=\frac{4}{e^2+4}$

$\begin{aligned} & \left(7 x^4 \cot y-e^x \operatorname{cosec} y\right) \frac{d x}{d y}=x^5 \\ & x^5 \frac{d y}{d x}-7 x^4 \cot y=-e^x \operatorname{cosec} y \\ & \frac{d y}{d x}-\frac{7}{x} \cot y=-\frac{e^x}{x^5} \operatorname{cosec} y \\ & \sin y \frac{d y}{d x}-\frac{7}{x} \cos y=-\frac{e^x}{x^5} \\ & \text { Let }-\cos y=t \\ & \sin y \frac{d y}{d x}=\frac{d t}{d x} \\ & \therefore \frac{d t}{d x}+\frac{7}{x} t=-\frac{e^x}{x^5} \end{aligned}$

$\begin{aligned} & \therefore \text { I.F. }=e^{\int \frac{7}{x} d x}=x^7 \\ & t \cdot x^7=\int \frac{-e^x}{x^5} \cdot x^7 d x \\ & -\cos y \cdot x^7=-\int e^x x^2 d x \\ & \cos y x^7=e^x\left(x^2-2 x+2\right)+c \\ & \because \quad x=1 \text { then } y=\frac{\pi}{2} \Rightarrow c=-e \\ & \therefore \quad \cos y \cdot x^7=e^x\left(x^2-2 x+2\right)-e \\ & \text { When } x=2 \text { then } \cos y=\frac{2 e^2-e}{128} \end{aligned}$

$\begin{aligned} & \frac{d y}{d x}+3\left(\tan ^2 x\right) y+3 y=\sec ^2 x \\ & \Rightarrow \frac{d y}{d x}+3 \sec ^2 x y=\sec ^2 x \\ & \text { I.F }=e^{\int 3 \sec ^2 x d x} \\ & \quad=e^{3 \tan x} \end{aligned}$

$\begin{aligned} & y \cdot e^{\tan x}=\int e^{3 \tan x} \cdot \sec ^2 x d x+c \\ & y \cdot e^{3 \tan x}=\frac{e^{3 \tan x}}{3}+c \\ & \text { Also } f(0)=\frac{1}{3}+e^3 \\ & \Rightarrow\left(\frac{1}{3}+e^3\right)=\frac{1}{3}+c \\ & \Rightarrow c=e^3 \\ & \therefore y \cdot e^{3 \tan x}=\frac{e^{3 \tan x}}{3}+e^3 \\ & \text { Put } x=\frac{\pi}{4} \end{aligned}$

$y e^3=\frac{e^3}{3}+e^3 \Rightarrow y=\frac{4}{3}$

To solve the given problem, let's start by considering the equation:

$ \int_0^x g(t) \, dt = x - \int_0^x \tan g(t) \, dt $

Differentiate both sides with respect to $ x $:

$ g(x) = 1 - xg(x) $

Rearranging gives:

$ g(x)(1 + x) = 1 \implies g(x) = \frac{1}{1 + x} $

Now, consider the differential equation:

$ \frac{dy}{dx} - y \tan x = 2(x+1) \sec x g(x) $

Substitute $ g(x) = \frac{1}{1 + x} $:

$ \frac{dy}{dx} - y \tan x = 2(x+1) \sec x \cdot \frac{1}{1 + x} $

This simplifies to:

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

To solve this, we use an Integrating Factor (I.F):

$ \text{I.F} = e^{\int \tan x \, dx} = e^{-\ln \cos x} = \cos x $

Multiply the entire differential equation by the Integrating Factor:

$ \cos x \cdot \frac{dy}{dx} - y \cdot \sin x = 2 $

This implies:

$ \frac{d}{dx}(y \cos x) = 2 $

Integrate both sides with respect to $ x $:

$ y \cos x = \int 2 \, dx = 2x + c $

Given the initial condition $ y(0) = 0 $:

$ 0 \cdot 1 = 2 \cdot 0 + c \implies c = 0 $

Thus:

$ y \cos x = 2x $

Evaluate at $ x = \frac{\pi}{3} $:

$ y \cdot \frac{1}{2} = 2 \cdot \frac{\pi}{3} $

Solving for $ y $:

$ y = \frac{4 \pi}{3} $

Therefore, $ y\left(\frac{\pi}{3}\right) = \frac{4 \pi}{3} $.

$\begin{aligned} & \text { If } \mathrm{e}^{\int \tan x d x}=\mathrm{e}^{\ln (\sec x)}=\sec x \\ & \therefore y \cdot \sec x=\int\left\{\frac{2+\sec x}{(1+2 \sec x)^2}\right\} \sec x d x \end{aligned}$

$\begin{aligned} &\begin{aligned} & =\int \frac{2 \cos x+1}{(\cos x+2)^2} d x \text { Let } \cos x=\frac{1-t^2}{1+t^2} \\ & =\int \frac{2\left(\frac{1-t^2}{1+t^2}\right)+1}{\left(\frac{1-t^2}{1+t^2}+2\right)^2} 2 d t \\ & =\int \frac{2-2 t^2+1+t^2}{\left(1-t^2+2+2 t^2\right)^2} \times 2 d t \\ & =2 \int \frac{3-t^2}{\left(t^2+3\right)^2} d t \end{aligned}\\ &\text { Let } \mathrm{t}+\frac{3}{\mathrm{t}}=\mathrm{u}\\ &\left(1-\frac{3}{\mathrm{t}^2}\right) \mathrm{dt}=\mathrm{du} \end{aligned}$

$=-2 \int \frac{\mathrm{du}}{\mathrm{u}^2}$

$\begin{aligned} & y \cdot(\sec x)=\frac{2}{u}+c \\ & y \cdot \sec x=\frac{2}{t+\frac{3}{t}}+c\quad\text{..... (I)} \end{aligned}$

$\begin{aligned} & \text { At } \mathrm{x}=\frac{\pi}{3}, \mathrm{t}=\tan \frac{\mathrm{x}}{2}=\frac{1}{\sqrt{3}} \\ & \text { 2. } \frac{\sqrt{3}}{10}=\frac{2}{\frac{1}{\sqrt{3}}+3 \sqrt{3}}+\mathrm{c} \\ & \text { 2. } \frac{\sqrt{3}}{10}=\frac{2 \sqrt{3}}{10}+\mathrm{c} \Rightarrow \mathrm{C}=0 \\ & \text { At } \mathrm{x}=\frac{\pi}{4}, \mathrm{t}=\tan \frac{\mathrm{x}}{2}=\sqrt{2}-1 \end{aligned}$

$\begin{aligned} & \therefore y \cdot \sqrt{2}=\frac{2}{\sqrt{2}-1+\frac{3}{\sqrt{2}-1}} \\ & y \cdot \sqrt{2}=\frac{2(\sqrt{2}-1)}{6-2 \sqrt{2}} \\ & y=\frac{\sqrt{2}(\sqrt{2}-1)}{2(3-\sqrt{2})}=\frac{1}{\sqrt{2}} \times \frac{2 \sqrt{2}-1}{7} \\ & =\frac{4-\sqrt{2}}{14} \end{aligned}$

$\begin{aligned} & \cos x(\ln (\cos x))^2 d y+(\sin x-3 y(\sin x) \ln (\cos x)) d x=0 \\ & \cos x(\ln (\cos x))^2 \frac{d y}{d x}-3 \sin x \cdot \ln (\cos x) y=-\sin x \\ & \frac{d y}{d x}-\frac{3 \tan x}{\ln (\cos x)} y=\frac{-\tan x}{(\ln (\cos x))^2} \\ & \frac{d y}{d x}+\frac{3 \tan x}{\ln (\sec x)} y=\frac{-\tan x}{(\ln (\sec x))^2} \\ & \text { I.F. }=e^{\int \frac{3 \tan x}{\ln (\sec x)} d x}=(\ln (\sec x))^3 \end{aligned}$

$\begin{aligned} & y \times(\ln (\sec x))^3=-\int \frac{\tan x}{(\ln (\sec x))^2}(\ln (\sec x))^3 d x \\ & y \times(\ln (\sec x))^3=-\frac{1}{2}(\ln (\sec x))^2+C \\ & \text { Given }: x=\frac{\pi}{4}, y=-\frac{1}{\ln 2} \\ & \frac{-1}{\ln 2} \times(\ln \sqrt{2})^3=-\frac{1}{2} \times(\ln \sqrt{2})^2+C \\ & \Rightarrow \frac{-1}{8 \ln 2} \times(\ln 2)^3=\frac{-1}{2} \times \frac{1}{4}(\ln 2)^2+C \\ & -\frac{1}{8}(\ln 2)^2=\frac{-1}{8}(\ln 2)^2+C \\ & \Rightarrow C=0 \\ & \therefore y(\ln (\sec x))^3=\frac{-1}{2}(\ln (\sec x))^2+0 \\ & y=\frac{-1}{2 \ln (\sec x)} \\ & y=\frac{1}{2 \ln (\cos x)} \end{aligned}$

$\begin{aligned} & \therefore y\left(\frac{\pi}{6}\right)=\frac{1}{2 \ln \left(\cos \frac{\pi}{6}\right)} \\ & =\frac{1}{2 \ln \left(\frac{\sqrt{3}}{2}\right)} \\ & =\frac{1}{2\left(\frac{1}{2} \ln 3-\ln 2\right)} \\ & =\frac{1}{\ln 3-\ln 4} \end{aligned}$

$\begin{aligned} &\int_0^{\mathrm{x}} \mathrm{tf}(\mathrm{t}) \mathrm{dt}=\mathrm{x}^2+(\mathrm{x}), \mathrm{x}>0\\ &\text { Diff. both side w.r. to } x\\ &\begin{aligned} & x f(x)=x^2 f^{\prime}(x)+2 x f(x) \\ & -x f(x)=x^2 f^{\prime}(x) \\ & \int \frac{f^{\prime}(x)}{f(x)} d x=\int \frac{-1}{x} d x \\ & \operatorname{logdf}(x)=-\log x+\log c \\ & f(x)=\frac{c}{x} \end{aligned}\\ &\mathrm{f}(2)=3 \Rightarrow 3=\frac{\mathrm{c}}{2} \Rightarrow \mathrm{c}=6\\ &f(x)=\frac{6}{x}\\ &f(6)=1\quad\therefore (1) \end{aligned}$

$\begin{aligned} & \left(1+x^2\right) \frac{d y}{d x}+x y=5 x^1 \sqrt{1+x^2} \\ & \frac{d y}{d x}+\frac{x y}{1+x^2}=\frac{5 x^2}{\sqrt{1+x^2}} \\ & \therefore \text { I.F. }=e^{\int \frac{x}{1+x^2} d x}=e^{\frac{\ln \left(1+x^2\right)}{2}}=\sqrt{1+x^2} \\ & \therefore y \sqrt{1+x^2}=\int \frac{5 x^2}{\sqrt{1+x^2}} \cdot \sqrt{1+x^2} d x \\ & \therefore y \sqrt{1+x^2}=\int \frac{5 x^2}{\sqrt{1+x^2}} \cdot \sqrt{1+x^2} d x \\ & y \sqrt{1+x^2}=\frac{5 x^3}{3}+C \\ & \because y(0)=0 \Rightarrow 0=0+C \Rightarrow C=0 \\ & \therefore y=\frac{5 x^3}{3 \sqrt{1+x^2}} \\ & y(\sqrt{3})=\frac{15 \sqrt{3}}{32}=\frac{5 \sqrt{3}}{2} \end{aligned}$

$\begin{aligned} \quad y d y & =(x d y-y d x) \sin \left(\frac{x}{y}\right) \\ \frac{d y}{y} & =\left(\frac{x d y-y d x}{y^2}\right) \sin \left(\frac{x}{y}\right) \\ \frac{d y}{y} & =\sin \left(\frac{x}{y}\right) d\left(-\frac{x}{y}\right) \\ \Rightarrow \quad \ell n y & =\cos \frac{x}{y}+C \end{aligned}$

$\begin{aligned} & x(1)=\frac{\pi}{2} \Rightarrow 0=\cos \frac{\pi}{2}+C \Rightarrow C=0 \\ & \text { थny }=\cos \frac{x}{y} \\ & \text { but } y=2 \Rightarrow \cos \frac{x}{2}=\ln 2 \\ & \qquad \begin{aligned} \cos x & =2 \cos ^2 \frac{x}{2}-1 \\ & =2(\ln 2)^2-1 \end{aligned} \end{aligned}$

$\begin{aligned} & \frac{d y}{d x}=\frac{2(3+y) \cdot e^{2 x}}{7+e^{2 x}} \\ & \frac{d y}{d x}-\frac{2 y \cdot e^{2 x}}{7+e^{2 x}}=\frac{6 \cdot e^{2 x}}{7+e^{2 x}} \\ & \text { I.F. }=e^{-\int \frac{2 e^{2 x}}{7+e^{2 x}}}=\frac{1}{7+e^{2 x}} \end{aligned}$

$\begin{aligned} & \therefore y \cdot \frac{1}{7+\mathrm{e}^{2 \mathrm{x}}}=\int \frac{6 \mathrm{e}^{2 \mathrm{x}}}{\left(7+3^{2 x}\right)^2} \mathrm{dx} \\ & \quad \frac{\mathrm{y}}{7+\mathrm{e}^{2 \mathrm{x}}}=\frac{-3}{7+\mathrm{e}^{2 \mathrm{x}}}+C \\ & (0,5) \Rightarrow \frac{5}{8}=\frac{-3}{8}+C \Rightarrow C=1 \\ & \therefore y=-3+7+\mathrm{e}^{2 \mathrm{x}} \\ & y=\mathrm{e}^{2 \mathrm{x}}+4 \\ & \therefore \mathrm{k}=8 \end{aligned}$

$\begin{aligned} & \frac{d x}{d y}+\frac{x}{1+y^2}=\frac{2 e^{\tan ^{-1} y}}{1+y^2} \\ & \text { I.F. }=e^{\tan ^{-1} y} \\ & x^{\tan ^{-1} y}=\int \frac{2\left(e^{\tan ^{-1} y}\right)^2 d y}{1+y^2} \\ & \text { Put } \tan ^{-1} y=t, \frac{d y}{1+y^2}=d t \\ & x e^{\tan ^{-1} y}=\int 2 e^{2 t} d t \\ & x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+c \\ & x=e^{\tan ^{-1} y}+c e^{-\tan ^{-1} y} \\ & \because y=0, x=1 \\ & 1=1+c \Rightarrow c=0 \\ & y=\frac{1}{\sqrt{3}}, x=e^{\pi / 6} \end{aligned}$

$\begin{aligned} & y^2 d x+\left(x-\frac{1}{y}\right) d y=0 \\ & y^2 d x=\left(\frac{1}{y}-x\right) d y \\ & \Rightarrow y^2 \frac{d x}{d y}=\frac{1}{y}-x \\ & \Rightarrow \frac{d x}{d y}+\frac{x}{y^2}=\frac{1}{y^3} \\ & \text { I.F. }=e^{\int \frac{1}{y^2} d y}=e^{\frac{-1}{y}}\end{aligned}$

$\therefore$ Solution is

$ x e^{\frac{-1}{y}}=\int e^{-\frac{1}{y}} \times \frac{1}{y^3} d y+C $

Let $\frac{-1}{y}=t$

$ \Rightarrow \frac{1}{y^2} d y=d t $

$\begin{aligned} & \Rightarrow x e^{-\frac{1}{y}}=-\int e^t t d t+C \\ & \Rightarrow x e^{-\frac{1}{y}}=-e^t(t-1)+C \\ & \Rightarrow x e^{-\frac{1}{y}}=-e^{\frac{-1}{y}}\left(\frac{-1}{y}-1\right)+C\end{aligned}$

$\begin{aligned} & f(x+y)=f(x) f^{\prime}(y)+f^{\prime}(x) f(x) \\ & \text { Put }=x=y=0 \\ & f(0)=f(0) f^{\prime}(0)+f^{\prime}(0) f(0) \\ & f^{\prime}(0)=\frac{1}{2} \\ & \text { Put } y=0 \\ & f(x)=f(x) f^{\prime}(0)+f^{\prime}(x) f(0) \\ & f(x)=\frac{1}{2} f(x)+f^{\prime}(x) \\ & f^{\prime}(x)=\frac{f(x)}{2}\end{aligned}$

$ \frac{d y}{d x}=\frac{y}{2} \Rightarrow \int \frac{d y}{y}=\int \frac{d x}{2} $

$ \Rightarrow \ln y=\frac{x}{2}+c $

$\begin{aligned} & \because \mathrm{f}(0)=1 \Rightarrow \mathrm{C}=0 \\ & \ell \mathrm{ny}=\frac{\pi}{2} \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x} / 2} \\ & \ln \mathrm{f}(\mathrm{n})=\frac{\mathrm{n}}{2} \\ & \sum_{\mathrm{n}=1}^{100} \ell \mathrm{f}(\mathrm{n})=\frac{1}{2} \sum_{\mathrm{n}=1}^{100} \mathrm{n}=\frac{5050}{2} \\ & =2525\end{aligned}$

$ \begin{aligned} & \mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{y}) \\ & \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{e}^{\lambda \mathrm{x}} \mathrm{f}^{\prime}(0)=4 \mathrm{a} \\ & \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\lambda \mathrm{e}^{\lambda \mathrm{x}} \Rightarrow \lambda=4 \mathrm{a} \end{aligned} $

So, $\mathrm{f}(\mathrm{x})=\mathrm{e}^{4 \mathrm{ax}}$

$ \begin{aligned} & \mathrm{f}^{\prime \prime}(\mathrm{x})-3 \mathrm{af}^{\prime}(\mathrm{x})-\mathrm{f}(\mathrm{x})=0 \\ & \Rightarrow \lambda^2-3 \mathrm{a} \lambda-1=0 \\ & \Rightarrow 16 \mathrm{a}^2-12 \mathrm{a}^2-1=0 \Rightarrow 4 \mathrm{a}^2=1 \Rightarrow \mathrm{a}=\frac{1}{2} \end{aligned} $

$ \begin{aligned} &\text { Equation of circle is }\\ &\begin{aligned} & (x-\alpha)^2+(y-\beta)^2=a^2 \\ & \Rightarrow \quad(x-\alpha)+2(y-\beta) \times y_1=0 \\ & \Rightarrow \quad x-\alpha+(y-\beta) \times y_1=0 \\ & \Rightarrow \quad(y-\beta)^2\left(1+y_1^2\right)=a^2 \\ & \Rightarrow \quad(y-\beta)^2=\frac{a^2}{1+y_1^2} \\ & \Rightarrow \quad y-\beta=\frac{a}{\sqrt{1+y_1^2}} \\ & \Rightarrow \quad y-\frac{a}{\sqrt{1+y_1^2}}=\beta \\ & \Rightarrow \quad y_1=\left(\frac{-a \times \frac{1}{2 \sqrt{1+y_1^2}} \times 2 y_1 y_2}{\left(1+y_1^2\right)}\right)=0 \\ & \Rightarrow \quad y_1+\frac{a y_1 y_2}{\left(1+y_1^2\right)^3}=0 \\ & \Rightarrow \quad \frac{a^2 y_1^2 y_2^2}{\left(1+y_1^2\right)^3}=y_1^2 \\ & \Rightarrow \quad a^2 y_2^2=\left(1+y_1^2\right)^3 \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } \begin{aligned} & \frac{d x}{d y}+x \cdot \frac{1}{1+y^2}=\frac{\tan ^{-1} y}{1+y^2} \\ & \quad \text { IF }=e^{\int \frac{d y}{1+y^2}} \\ &=e^{\tan ^{-1} y} \\ & x e^{\tan ^{-1} y}=\int \frac{e^{\tan ^{-1} y} \cdot \tan ^{-1} y}{1+y^2} d y \\ & \Rightarrow x e^{\tan ^{-1} y}=\int e^\theta \cdot \theta d \theta \quad[\text { put } y=\tan \theta] \\ & \Rightarrow x e^{\tan ^{-1} y}=\theta \int \theta d \theta-\int\left(\frac{d \theta}{d \theta} \int e^\theta d \theta\right) d \theta \\ & \Rightarrow \theta e^\theta-e^\theta+C \\ & \Rightarrow x e^{\tan ^{-1} y}=\tan ^{-1} y e^{\tan ^{-1} y}-e^{\tan ^{-1} y}+C \\ & \Rightarrow x=\tan ^{-1} y-1+c e^{-\tan ^{-1} y} \end{aligned} \end{aligned} $

So, $f(y)=\tan ^{-1} y-1$

$ \begin{aligned} & \text { } f^{\prime}(x)-\frac{4 \sin 2 x}{1+\cos ^2 x}-\frac{f(x) \sin 2 x}{1+\cos ^2 x}=0 \\ & f^{\prime}(x)-\frac{f(x) \sin 2 x}{1+\cos ^2 x}=\frac{4 \sin 2 x}{1+\cos ^2 x} \\ & \text { Let } y=f(x) \\ & \Rightarrow \quad \frac{d y}{d x}-\frac{y \sin 2 x}{1+\cos ^2 x}=\frac{4 \sin 2 x}{1+\cos ^2 x} \end{aligned} $

$ \begin{array}{ll} \Rightarrow & y \times \mathrm{IF}=\int(Q \times \mathrm{IF}) d x \\ \Rightarrow & \mathrm{IF}=e^{-\int \frac{\sin 2 x d x}{1+\cos ^2 x}} \\ \Rightarrow & \mathrm{IF}=1+\cos ^2 x \\ \Rightarrow & y \times\left(1+\cos ^2 x\right)=\int \frac{4 \sin 2 x}{1+\cos ^2 x} \times 1+\cos ^2 x d x \end{array} $

$ \begin{array}{cc} \Rightarrow & y\left(1+\cos ^2 x\right)=4 \int \sin 2 x d x \\ \Rightarrow & y\left(1+\cos ^2 x\right)=-2 \cos 2 x+C \\ & y(0)=0 \\ & 0(1+1)=-2+C \Rightarrow C=2 \\ \Rightarrow & y\left(1+\cos ^2 x\right)=2(1-\cos 2 x) \\ \Rightarrow & y(x)=\frac{4 \sin ^2 x}{1+\cos ^2 x}=f(x) \\ \Rightarrow & f\left(\frac{\pi}{3}\right)=\frac{4 \sin ^2 60^{\circ}}{1+\cos ^2 60^{\circ}} \\ & =\frac{4 \times \frac{3}{4}}{1+\frac{1}{4}} \\& = \frac{12}{5} \end{array} $

$ \begin{aligned} & \text { } \frac{x^2}{a^2}+\frac{y^2}{4}=1 \\ & \frac{2 x}{a^2}+\frac{2 y \cdot y^{\prime}}{4}=0 \end{aligned} $

(differentiating both sides)

$ \begin{aligned} & \frac{x}{a^2}=\frac{-y y^{\prime}}{4} \\ \Rightarrow \quad & \frac{1}{a^2}=-\frac{y y^{\prime}}{4 x} \end{aligned} $

So, differential equation of such ellipse.

$ \begin{array}{rlrl} & x^2\left(\frac{-y y^{\prime}}{4 x}\right)+\frac{y^2}{4}=1 \\ \Rightarrow & \frac{-x y y^{\prime}}{4}+\frac{y^2}{4}=1 \\ \Rightarrow & \frac{y^2-4}{4}=\frac{x y y^{\prime}}{4} \\ \Rightarrow & x y \frac{d y}{d x}=y^2-4 \end{array} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \frac{d y}{d x}+(\sec x \cdot \operatorname{cosec} x) y=\cos ^2 x \\ & \text { IF }=\mathrm{e}^{\int \sec x \operatorname{cosec} x d x} \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =e^{\int \frac{2}{\sin 2 x} d x}=e^{2 \int \operatorname{cosec} 2 x d x} \\ & =e^{2 \frac{\ln |\operatorname{cosec} 2 x-\cot 2 x|}{2}} \\ & =(\operatorname{cosec} 2 x-\cot 2 x) \\ & =\frac{1-\cos 2 x}{\sin 2 x}=\frac{2 \sin ^2 x}{2 \sin x \cos x} \\ & =\frac{\sin x}{\cos x} \end{aligned}\\ &\text { Solution }\\ &\begin{aligned} y(\mathrm{IF}) & =\int \cos ^2 x(\mathrm{IF}) d x+k \\ \Rightarrow y \tan x & =\int \sin x \cos x d x+k \\ \Rightarrow y \tan x & =\frac{\sin ^2 x}{2}+k \\ \Rightarrow 2 y \tan x & =\sin ^2 x+C \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } y=A e^x+B \sin x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \frac{d y}{d x}=A e^x+B \cos x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \frac{d^2 y}{d x^2}=A e^x-B \sin x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

From Eqs. (i) and (iii),

$ \frac{1}{2}\left(y+\frac{d^2 y}{d x^2}\right)=A e^x $

From Eqs. (i) and (ii),

$ B=\left(y-\frac{d y}{d x}\right) \frac{1}{\sin x-\cos x} $

∴ From Eq. (iii),

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

$ \begin{aligned} & \Rightarrow \frac{1}{2} \frac{d^2 y}{d x^2}=\frac{y}{2}-\frac{\sin x}{\sin x-\cos x}\left(y-\frac{d y}{d x}\right) \\ & \begin{array}{r} \Rightarrow(\sin x-\cos x) \frac{d^2 y}{d x^2}=(\sin x-\cos x) y \\ \quad-2 \sin x \cdot y+2 \sin x \frac{d y}{d x} \end{array} \\ & \Rightarrow(\sin x-\cos x) \frac{d^2 y}{d x^2}-2 \sin x \frac{d y}{d x} +(2 \sin x-\sin x+\cos x) y=0 \\ & \Rightarrow(\sin x-\cos x) \frac{d^2 y}{d x^2}-2 \sin x \frac{d y}{d x} +(\sin x+\cos x) y=0 \\ & \begin{array}{r} \therefore f(x)+g(x)+h(x) \\ =\sin x-\cos x-2 \sin x+\sin x+\cos x \\ \end{array} \end{aligned} $

$ =0 $

Let centre $(h, k)$ lies on $y=2 x$

$ \begin{array}{ll} \Rightarrow & k=2 h \\ \therefore & \frac{(x-h)^2}{a^2}-\frac{(y-2 h)^2}{b^2}=1 \\ \Rightarrow & \frac{(x-h)^2}{a^2}-\frac{(y-2 h)^2}{2 a^2}=1 \\ & {\left[\because e=\sqrt{3} \Rightarrow b^2=2 a^2\right]} \end{array} $

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

$ \begin{aligned} & 4(x-h)-2(y-2 h) \frac{d y}{d x}=0 \\ \Rightarrow \quad & (y-2 h) y_1=2(x-h) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Again, differentiating,

$ (y-2 h) y_2=2-y_1^2 $

From Eq. (i), $h=\frac{2 x-y y_1}{2\left(1-y_1\right)}$

∴ Differentiating equation is

$ \begin{aligned} & \left(y-2\left(\frac{2 x-y y_1}{2\left(1-y_1\right)}\right)\right) y_2=2-y_1^2 \\ \Rightarrow \quad & \left(y-y y_1-2 x+y y_1\right) y_2=2+y_1^3 -2 y_1-y_1^2 \\ \Rightarrow \quad & (y-2 x) y_2+y_1^2+2 y_1=y_1^3+2 \end{aligned} $

Given, $\left(x^3-y^3\right) d x=\left(x^2 y-x y^2\right) d y$

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

Homogeneous differential equation

Put $\quad y=v x$

$ \begin{aligned} & \frac{d y}{d x}=v+x \frac{d v}{d x} \\ \Rightarrow & v+\frac{x d v}{d x}=\frac{1+v+v^2}{v} \\ \Rightarrow & x \frac{d v}{d x}=\frac{1+v}{v} \\ \Rightarrow & \frac{v}{v+1} d v=\frac{d x}{x} \\ \Rightarrow & \frac{v+1-1}{v+1} d v=\frac{d x}{x} \end{aligned} $

On taking integration both sides, we get

$ \begin{aligned} & \int \frac{v+1}{v+1} d v-\int \frac{1}{v+1} d v=\int \frac{d x}{x} \\ & v-\ln |v+1|=\ln x+\ln c \\ \Rightarrow & v=\ln |x(v+1) c| \\ \Rightarrow & \frac{y}{x}=\log [(y+x) c] \\ \Rightarrow & y=x \log (c|x+y|) \end{aligned} $

Given, $t^2 d x+\left(x^2-t x+t^2\right) d t=0$

$ \Rightarrow \quad \frac{d t}{d x}=-\frac{t^2}{x^2-t x+t^2} $

On putting $t=V x$

$ \begin{aligned} & \Rightarrow \quad \frac{d t}{d x}=V+x \frac{d V}{d x} \\ & \therefore \quad\left(V+x \frac{d V}{d x}\right)=-\frac{V^2 x^2}{x^2-V x^2+V^2 x^2} \\ & \Rightarrow \quad V+x \frac{d V}{d x}=-\frac{V^2}{1-V+V^2} \\ & \Rightarrow \quad x \frac{d V}{d x}=-\frac{V^2}{1-V+V^2}-V \\ & \Rightarrow \quad x \frac{d V}{d x}=\frac{-V^2-V+V^2-V^3}{1-V+V^2} \\ & \Rightarrow \quad x \frac{d V}{d x}=-\frac{V+V^3}{1-V+V^2} \\ & \Rightarrow \quad \frac{1-V+V^2}{V+V^3} d V=-\frac{d x}{x} \end{aligned} $

Which can be solve by variables separable method.

Equation of parabola with its axis parallel to the $Y$-axis.

$ \begin{aligned} & y=a x^2+b x+c, a \neq 0 \\ \Rightarrow \quad & \frac{d y}{d x}=2 a x+b \Rightarrow \frac{d^2 y}{d x^2}=2 a \\ \Rightarrow \quad & \frac{d^3 y}{d x^3}=0 \end{aligned} $

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

$ \begin{aligned} & x \neq(2 n+1) \frac{\pi}{2}, n \in Z \\ \Rightarrow \quad & \frac{d y}{d x}=\frac{y \sin x}{\cos x}-\frac{1}{\cos x} \\ & =y \tan x-\sec x \\ \Rightarrow \quad & \frac{d y}{d x}-y \tan x=-\sec x \end{aligned} $

This is in the form $\frac{d y}{d x}+P(x) y=Q(x)$,

where, $P(x)=-\tan x$ and $Q(x)=-\sec x$

$ \begin{aligned} \mathrm{IF} & =e^{\int P(x) d x}=e^{\int-\tan x d x} \\ & =e^{\ln |\cos x|}=|\cos x| \end{aligned} $

So, the general solution is

$ \begin{aligned} y \cdot \text { IF } & =\int A(x) \cdot \text { IF } d x+C \\ y \cdot \cos x & =\int-\sec x \cdot \cos x d x+C \\ & =-\int d x+C=-x+C \\ & =y=\frac{-x+C}{\cos x} \end{aligned} $

Given, $f(0)=1$

$ \Rightarrow \quad y=1 \text { and } x=0 $

So, $1=\frac{-0+C}{\cos (0)}$

$ \Rightarrow 1=C $

So, $y=\frac{-x+1}{\cos x}=(1-x) \sec x$

Given, differential equation

$ \begin{aligned} & 2 d x+d y=(6 x y+4 x-3 y) d x \\ & \Rightarrow \quad d y=(6 x y+4 x-3 y-2) d x \\ & \Rightarrow \quad \frac{d y}{d x}=6 x y+4 x-3 y-2 \\ & \Rightarrow \quad \frac{d y}{d x}=6 x y+4 x-2-3 y \\ & \Rightarrow \quad 2 x(3 y+2)-1(3 y+2) \\ & \Rightarrow \quad(3 y+2)(2 x-1) \\ & \Rightarrow \quad \frac{d y}{3 y+2}=d x(2 x-1) \end{aligned} $

Integrating to both sides w.r.t $x$, we get

$ \begin{aligned} & \int \frac{d y}{3 y+2}=\int(2 x-1) d x \\ & \Rightarrow \frac{1}{3} \log |3 y+2|=\frac{2 x^2}{2}-x+C_1 \\ & \Rightarrow \frac{1}{3} \log |3 y+2|=x^2-x+C_1 \\ & \Rightarrow \quad \log |3 y+2|=3 x^2-3 x+3 C_1 \\ & \quad=3 x^2-3 x+C \end{aligned} $

(where $C=3 C_1$ )

So, $\log |3 y+2|=3 x^2-3 x+C$

$\because y=A t^2+\frac{B}{t}$

$\Rightarrow y^{\prime}=2 A t-B t^{-2}$ and $y^{\prime \prime}=2 A+2 B t^{-3}$

Substituting into the differential equation

$ \begin{aligned} & f(t)\left(2 A+2 B t^{-3}\right)+g(t)\left(2 A t-B t^{-2}\right) +h(t)\left(A t^2+B t^{-1}\right)=0 \\ & \Rightarrow A\left[2 f(t)+2 t g(t)+t^2 h(t)\right] +B\left[2 t^{-3} f(t)-t^{-2} g(t)+t^{-1} h(t)\right]=0 \end{aligned} $

the coefficients must be zero.

So, $\quad 2 f(t)+2 t g(t)+t^2 h(t)=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and $2 t^{-3} f(t)-t^{-2} g(t)+t^{-1} h(t)=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Multiply Eq. (ii) by $t^3$

$ 2 f(t)-t g(t)+t^2 h(t)=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

Subtracting Eqs. (i) and (ii), we get

$ \begin{gathered} {\left[2 f(t)+2 t g(t)+t^2 h(t)\right]} \\ -\left[2 f(t)-t g(t)+t^2 h(t)\right]=0 \Rightarrow 3 t g(t)=0 \end{gathered} $

$\because t \neq 0$, so put $g(t)=0$ in Eq. (i)

$ 2 f(t)+t^2 h(t)=0 $

$\because \quad g(t)=0$, so $g(t) \cdot f(t)=0$

Hence, $2 f(t)+t^2 h(t)=g(t) \cdot f(t)$

$\because(2 x-y)^2 d y-2(2 x-y)^2 d x-2 d x=0$

put $v=2 x-y \Rightarrow d v=2 d x-d y$

$\Rightarrow d y=2 d x-d v$

So, $v^2(2 d x-d v)-2 v^2 d x-2 d x=0$

$\Rightarrow \quad 2 v^2 d x-v^2 d v-2 v^2 d x-2 d x=0$

$\Rightarrow-v^2 d v-2 d x=0 \Rightarrow v^2 d v+2 d x=0$

$\Rightarrow \int v^2 \cdot d v=-2 \int d x \Rightarrow \frac{v^3}{3}=-2 x+c$

$\Rightarrow \quad \frac{(2 x-y)^3}{3}=-2 x+c$

$\Rightarrow \quad(2 x-y)^3=-6 x+3 c$

$\Rightarrow \quad(2 x-y)^3+6 x=C$, where $C=3 x$

$x \log x d y=(x \log x-y) d x$

$ \begin{aligned} & \Rightarrow \quad x \log x \frac{d y}{d x}=x \log x-y \\ & \Rightarrow \quad \frac{d y}{d x}+\frac{y}{x \log x}=1 \end{aligned} $

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

$ \therefore \mathrm{IF}=e^{\int \frac{1}{x \log x} d x}=e^{\log |\log x|}=\log x $

Multiply IF both sides

$ \begin{array}{ll} & \log x \cdot \frac{d y}{d x}+\frac{y}{x}=\log x \\ \Rightarrow & \frac{d}{d x}(y \log x)=\log x \\ \Rightarrow & y \log x=\int \log x \cdot d x \\ \Rightarrow & y \log x=x \log x-x+C \\ \Rightarrow & \log x(y-x)+x=C \end{array} $

$ \begin{aligned} &\text { We have, }\left(x \sin \frac{y}{x}\right) d y\\ &\begin{aligned} & =\left(y \sin \frac{y}{x}-x\right) d x \\ \Rightarrow \quad \frac{d y}{d x} & =\frac{\frac{y}{x} \cdot \sin \frac{y}{x}-1}{\sin \frac{y}{x}} \end{aligned} \end{aligned} $

$ \begin{aligned} &=\frac{y}{x}-\frac{1}{\sin \frac{y}{x}}\\ &\text { Let } y=v x\\ &\frac{d y}{d x}=v+x \frac{d v}{d x} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad v+x \frac{d v}{d x}=v-\frac{1}{\sin v} \\ & \Rightarrow \quad x \frac{d v}{d x}=-\frac{1}{\sin v} \\ & \Rightarrow \quad \int-\sin v d v=\int \frac{d x}{x} \\ & \Rightarrow \quad \cos v=\ln x+C \\ & \Rightarrow \quad \cos \left(\frac{y}{x}\right)=\ln x+C \\ & \Rightarrow \quad \cos \frac{y}{x}=\log |x|+C \end{aligned} $