Differential Equations

$ \left(x^2-4\right) y^{\prime}-2 x y+2 x\left(4-x^2\right)^2=0, x>2 $

rearranging terms.

$ \begin{aligned} & \left(x^2-4\right) \frac{d y}{d x}-2 x y=-2 x\left(x^2-4\right)^2 \\ & \frac{d y}{d x}-\frac{2 x}{x^2-4} y=-2 x\left(x^2-4\right) \end{aligned} $

this is a linear differential equation of the form $\frac{d y}{d x}+P y=Q$

$ P=-\frac{2 x}{x^2-4}, Q=-2 x\left(x^2-4\right) $

finding integrating factor (I.F.).

$ \begin{aligned} \text { I.F. } & =e^{\int P d x}=e^{\int-\frac{2 x}{x^2-4} d x} \\ \text { I.F. } & =e^{-\ln \left(x^2-4\right)}=\frac{1}{x^2-4} \end{aligned} $

the solution is $y \cdot(I . F)=.\int Q \cdot(I . F) d x+$.

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

substitute point $(3,15)$

$ 15=\left(3^2-4\right)\left(C-3^2\right) $

$ \begin{aligned} & 15=5(C-9) \Rightarrow 3=C-9 \Rightarrow C=12 \\ & f(x)=\left(x^2-4\right)\left(12-x^2\right) \\ & f(x)=-x^4+16 x^2-48 \end{aligned} $

finding critical points for local maximum.

$ \begin{aligned} & f^{\prime}(x)=-4 x^3+32 x \\ & f^{\prime}(x)=0 \Rightarrow-4 x\left(x^2-8\right)=0 \\ & \text { since } x>2, x^2=8 \Rightarrow x=\sqrt{8}=2 \sqrt{2} \end{aligned} $

$f^{\prime}\left((2 \sqrt{2})^{-}\right)=$positive, $f^{\prime}\left((2 \sqrt{2})^{+}\right)=$negative

so sign of $f^{\prime}(x)$ changing positive to negative about $x=2 \sqrt{2}$ so $x=2 \sqrt{2}$ is point of local maxima

calculating local maximum value.

$ \begin{aligned} & f(2 \sqrt{2})=\left((2 \sqrt{2})^2-4\right)\left(12-(2 \sqrt{2})^2\right) \\ & =(8-4)(12-8) \\ & =4 \times 4=16 \end{aligned} $

the local maximum value is 16 .

Let $y=f(x)$

$ y^2=25+\int_0^x\left(f(t)^2+f^{\prime}(t)^2\right) d t $

Differentiating w.r.t. $x$

$ \begin{aligned} & 2 y \frac{d y}{d x}=y^2+\left(\frac{d y}{d x}\right)^2 \\ & \Rightarrow\left(\frac{d y}{d x}-y\right)^2=0 \Rightarrow \frac{d y}{y}=d x \\ & \Rightarrow \ln |y|=x+c \end{aligned} $

Such that

$ \begin{aligned} & f(0)^2=25+\int_0^0\left(f^2(t)+f^{\prime}(t)^2\right) d t \\ & \Rightarrow f(0)=5 \\ & \Rightarrow \ln 5=c \end{aligned} $

$ \begin{aligned} & \Rightarrow \ln |y|=x+\ln 5 \\ & \Rightarrow \quad y=5 e^x \\ & f(\ln k)=5 e^{\ln k}=5 k \\ & \text { Mean of }\{5,10, \ldots 625 \times 5\} \\ & =\frac{(5+10+\ldots 5 \times 625)}{625}=\frac{5(625)(626)}{625 \times 2} \\ & =5 \times 313 \\ & =1565 \end{aligned} $

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

$\begin{aligned} &\begin{aligned} & \text { I.F. }=e^{2 \tan x} \\ & y \cdot e^{2 \tan x}=\int e^{2 \tan x}(2+3 \tan x) \sec ^2 x d x \end{aligned}\\ &\text { Put } \tan x=u\\ &\begin{aligned} & \sec ^2 x d x=d u \\ & y \cdot e^{2 u}=\int e^{2 u}(2+3 u) d u \\ & y \cdot e^{2 u} \Rightarrow \frac{2 e^{2 u}}{2}+3 \int e^{2 u} \cdot u d u \\ & y \cdot e^{2 u}=e^{2 u}+3\left[\frac{u e^{2 u}}{2}-\int \frac{e^{2 u}}{2}\right] \\ & y e^{2 u}=e^{2 u}+3\left[\frac{u e^{2 u}}{2}-\frac{e^{2 u}}{4}\right]+C \\ & y e^{2 \tan x}=e^{2 \tan x}+3\left[\frac{\tan x e^{2 \tan x}}{2}-\frac{e^{2 \tan x}}{4}\right]+C \end{aligned} \end{aligned}$

$\begin{aligned} & F(0)=\frac{5}{4} \\ & \frac{5}{4}=1-\frac{3}{4}+C \\ & \frac{5}{4}-\frac{1}{4}=C \\ & 1=C \\ & y=1+3\left(\frac{\tan x}{2}-\frac{1}{4}\right)+1 \cdot e^{-2 \tan x} \\ & y\left(\frac{\pi}{4}\right)=1+3\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{e^2} \\ & y\left(\frac{\pi}{4}\right)=\frac{7}{4}+\frac{1}{e^2} \\ & 12\left(y\left(\frac{x}{4}\right)-\frac{1}{e^2}\right)=12\left(\frac{7}{4}+\frac{1}{e^2}-\frac{1}{e^2}\right)=21 \end{aligned}$

$\begin{aligned} & \frac{d y}{d x}+\frac{\left(\sin ^{-1} \frac{x}{2}\right)}{\sqrt{4-x^2}} y=\frac{\left(\sin ^{-3} \frac{x}{2}\right)^3}{\sqrt{4-x^2}} \\ & y e^{\frac{\left(\sin ^{-1} \frac{x}{2}\right)^2}{2}}=\int \frac{\left(\sin ^{-3} \frac{x}{2}\right)^3}{4-x^2} e^{\frac{\left(\sin ^{-1} \frac{x}{2}\right)^2}{2}} d x \\ & y=\left(\sin ^{-1} \frac{x}{2}\right)^2-2+c \cdot e^{\frac{-\left(\sin ^{-1} \frac{x}{2}\right)^2}{2}} \\ & y(2)=\frac{\pi^2}{4}-2 \Rightarrow c=0 \\ & y(0)=-2 \end{aligned}$

$\begin{aligned} & \frac{d y}{d x}+2 y \tan x=\sin x \\ & \text { I.F. }=e^{2 \int \tan x d x}=\sec ^2 x \\ & y \sec ^2 x=\int \frac{\sin x}{\cos ^2 x} d x \\ & =\int \tan x \sec x d x \\ & =\sec x+C \\ & C=-2 \\ & y=\cos x-2 \cos ^2 x \\ & y\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}-1 \\ & y^{\prime}=-\sin x+4 \cos ^2 x \sin x \\ & y^{\prime}\left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}+2 \\ & y^{\prime}\left(\frac{\pi}{4}\right)+y\left(\frac{\pi}{4}\right)=1 \end{aligned}$

$\begin{aligned} &\text { Differentiate both sides }\\ &\begin{aligned} & 4(x+2) f(x)+2(x+2)^2 f^{\prime}(x)-6(x+2)=10(x+2) f(x) \\ & 2(x+2)^2 f^{\prime}(x)-6(x+2) f(x)=6(x+2) \\ & (x+2) \frac{d y}{d x}-3 y=3 \\ & \int \frac{d y}{d x}=3 \int \frac{d x}{x+2} \\ & \ln (y+1)=3 \ln (x+2)+C \\ & (y+1)=C(x+2)^3 \\ & f(0)=\frac{3}{2} \\ & f(2)=19 \end{aligned} \end{aligned}$

I.F. $\mathrm{e}^{-\frac{1}{2} \int \frac{2 \mathrm{x}}{1-\mathrm{x}^2} \mathrm{dx}}=\mathrm{e}^{-\frac{1}{2} \ln \left(1-\mathrm{x}^2\right)}=\sqrt{1-\mathrm{x}^2}$

$y \times \sqrt{1-x^2}=\int\left(x^6+4 x\right) d x=\frac{x^7}{7}+2 x^2+c$

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

$y=\frac{\frac{x^7}{7}+2 x^2}{\sqrt{1-x^2}}$

$\begin{aligned} &\text { Now, } 6 \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{\frac{x^7}{7}+2 x^2}{\sqrt{1-x^2}} d x=6 \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{2 x^2}{\sqrt{1-x^2}} d x\\ &=24 \int_0^{\frac{1}{2}} \frac{\mathrm{x}^2}{\sqrt{1-\mathrm{x}^2}} \mathrm{dx} \end{aligned}$

$\begin{aligned} & \text { Put } x=\sin \theta \\ & d x=\cos \theta d \theta \\ & =24 \int_0^{\frac{\pi}{6}} \frac{\sin ^2 \theta}{\cos \theta} \cos \theta d \theta \\ & =24 \int_0^{\frac{\pi}{6}}\left(\frac{1-\cos 2 \theta}{2}\right) d \theta=12\left[\theta-\frac{\sin 2 \theta}{2}\right]_0^{\frac{\pi}{6}} \\ & =12\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right) \\ & =2 \pi-3 \sqrt{3} \\ & \alpha^2=(3 \sqrt{3})^2=27 \end{aligned}$

$\begin{aligned} & f^{\prime}(x)=3 f(x)+\alpha \\ & \Rightarrow \frac{d y}{3 y+\alpha}=d x \\ & \Rightarrow \frac{1}{3} \ln (3 y+\alpha)=x+C \\ & y(0)=1 \Rightarrow C=\frac{1}{3} \ln (3+\alpha) \\ & \frac{1}{3} \ln \left(\frac{3 y+\alpha}{3+\alpha}\right)=x \\ & \Rightarrow y=\frac{1}{3}\left((3+\alpha) e^{3 x}-\alpha\right)=f(x) \\ & \lim _{x \rightarrow-\infty} f(x)=7 \Rightarrow \alpha=-21 \\ & \Rightarrow f(x)=7-6 e^{3 x} \\ & 9 f(-\ln 3)=61 \end{aligned}$

$\begin{aligned} & \alpha|x|=|y| e^{x y-\beta} \\ & \frac{x d y-y d x}{y^2}+\frac{x y(x d y+y d x)}{y^2}=0 \\ & -d\left(\frac{x}{y}\right)+\frac{x}{y} d(x y)=0 \end{aligned}$

$\begin{aligned} & \int d(x y)=\int \frac{d\left(\frac{x}{y}\right)}{\frac{x}{y}} \\ & x y=\ln \left|\frac{x}{y}\right|+\ln c \\ & x y=\ln \left(\left|\frac{x}{y}\right| \cdot c\right) \\ & \because y(1)=2 \\ & 2=\ln \left|\frac{1}{2}\right| c \Rightarrow c=2 e^2 \\ & \therefore \quad \operatorname{solution} x y=\ln \left(\left|\frac{x}{y}\right| \cdot 2 e^2\right) \\ & e^{x y}=\frac{|x|}{|y|} \cdot 2 e^2 \\ & 2|x|=|y| e^{x y-2} \\ & \Rightarrow \alpha=2, \beta=2, \alpha+\beta=4 \end{aligned}$

Given the differential equation

$\left(e^y + 1\right) \cos x \, dx + e^y \sin x \, dy = 0,$

we aim to find the value of $ e^{y\left(\frac{\pi}{6}\right)} $ given that the solution $ y(x) $ passes through the point $\left(\frac{\pi}{2}, 0\right)$.

First, we recognize that the differential equation can be rearranged as:

$d\left(e^y \sin x\right) + \cos x \, dx = 0.$

Integrating this expression, we obtain:

$ e^y \sin x + \sin x = C,$

where $ C $ is a constant. Given that the solution passes through the point $\left(\frac{\pi}{2}, 0\right)$, we substitute these values into the equation to find $ C $:

$ e^0 \sin\left(\frac{\pi}{2}\right) + \sin\left(\frac{\pi}{2}\right) = C \Rightarrow 1 + 1 = C \Rightarrow C = 2.$

Thus, the equation simplifies to:

$e^y \sin x + \sin x = 2.$

We now need to determine the value of $ e^{y\left(\frac{\pi}{6}\right)} $. Substituting $ x = \frac{\pi}{6} $ into the equation, we get:

$ \left(e^{y\left(\frac{\pi}{6}\right)} + 1\right) \cdot \sin\left(\frac{\pi}{6}\right) = 2.$

Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$, the equation becomes:

$ \left(e^{y\left(\frac{\pi}{6}\right)} + 1\right) \cdot \frac{1}{2} = 2 \Rightarrow e^{y\left(\frac{\pi}{6}\right)} + 1 = 4 \Rightarrow e^{y\left(\frac{\pi}{6}\right)} = 3.$

Therefore, the value of $ e^{y\left(\frac{\pi}{6}\right)} $ is $ 3 $.

$\frac{d y}{d x}+\frac{2 x}{\left(1+x^2\right)^2} y=x e^{\frac{1}{1+x^2}} ; y(0)=0$

I.F. of linear differential equation,

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

Area between curve $y e^{\left(\frac{-1}{1+x^2}\right)}=\frac{x^2}{2}$ and $y-x=4$

$\begin{aligned} & \Rightarrow 2(x+4)=x^2 \Rightarrow x^2-2 x-8=0 \\ & \Rightarrow(x-4)(x+2)=0 \\ & \int_{-2}^4\left[(x+4)-\frac{x^2}{2}\right] d x=\frac{x^2}{2}+4 x-\left.\frac{x^3}{6}\right|_{-2} ^4 \\ & =\left(8+16-\frac{64}{6}\right)-\left(2-8+\frac{8}{6}\right) \\ & =30-12=18 \end{aligned}$

$\begin{aligned} & \frac{d y}{d x}=(x+y+z)^2 \\ & \text { Put } x+y+z=t \\ & \Rightarrow 1+\frac{d y}{d x}=\frac{d t}{d x} \\ & \text { Given DE } \Rightarrow \frac{d t}{d x}-1=t^2 \\ & \Rightarrow \frac{d t}{1+t^2}=d x \Rightarrow \tan ^{-1} t=x+c \\ & \Rightarrow x+y+z=\tan (x+c) \\ & \Rightarrow y(x)=\tan (x+c)-x-2 \\ & \because y(0)=-2 \Rightarrow-2=\tan c-0-2 \\ & \qquad \Rightarrow c=0 \\ & \Rightarrow y(x)=\tan x-x-2 \\ & \frac{d y}{d x}=\sec ^2 x-1 \geq 0 \end{aligned}$

$\Rightarrow y(x)$ is increasing if $x \in\left(0, \frac{\pi}{3}\right)$

$\begin{aligned} & \Rightarrow \alpha=y\left(\frac{\pi}{3}\right), \beta=y(0) \\ & \Rightarrow \alpha=-\frac{\pi}{3}-2+\sqrt{3} \text { and } \beta=-2 \end{aligned}$

Now, $(3 \alpha+\pi)^2+\beta^2=(6+3 \sqrt{3})^2+(-2)^2$

$=67-36 \sqrt{3}=y+\delta \sqrt{3}$.

$\Rightarrow \gamma+\delta=31$

$\begin{aligned} & \frac{d y}{d x}-y=1+4 \sin x \\ & \text { Integrating factor }=e^{-\int d x}=e^{-x} \end{aligned}$

Solution is $y e^{-x}=\int(1+4 \sin x) e^{-x} d x$

$\begin{aligned} & =-e^{-x}+2 \cdot e^{-x}(-\sin x-\cos x)+C \\ y(\pi) & =1 \Rightarrow C=0 \end{aligned}$

Hence $y(x)=-1-2(\sin x+\cos x)$

$y\left(\frac{\pi}{2}\right)+10=7$

$\frac{d x}{d y}-\frac{x}{y}=\frac{1-y^2}{y}$

Integrating factor $=\mathrm{e}^{\int-\frac{1}{y} d y}=\frac{1}{y}$

$\begin{aligned} & x \cdot \frac{1}{y}=\int \frac{1-y^2}{y^2} d y \\\\ & \frac{x}{y}=\frac{-1}{y}-y+c \\\\ & x=-1-y^2+c y\end{aligned}$

$\begin{aligned} & x(1)=1 \\\\ & 1=-1-1+c \Rightarrow c=3 \\\\ & x=-1-y^2+3 y \\\\ & 5 x(2)=5(-1-4+6) \\\\ & =5\end{aligned}$

$\begin{aligned} & (\mathrm{t}+1) \mathrm{dx}=\left(2 \mathrm{x}+(\mathrm{t}+1)^4\right) \mathrm{dt} \\\\ & \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{2 \mathrm{x}+(\mathrm{t}+1)^4}{\mathrm{t}+1} \\\\ & \frac{\mathrm{dx}}{\mathrm{dt}}-\frac{2 \mathrm{x}}{\mathrm{t}+1}=(\mathrm{t}+1)^3\end{aligned}$

$I.F=e^{-\int \frac{2}{t+1} d t}=e^{-2 \ln (t+1)}=\frac{1}{(t+1)^2}$

$\begin{aligned} & \frac{x}{(t+1)^2}=\int \frac{1}{(t+1)^2}(t+1)^3 d t+C\end{aligned}$

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

Now $x(0)=2$

$ \begin{aligned} & \Rightarrow C=2 \\\\ & \therefore x=\left(\frac{t^2}{2}+t+2\right)(t+1)^2 \\\\ & x(1)=\left(\frac{1}{2}+1+2\right)(1+1)^2 \\\\ & =\frac{7}{2} \times 4=14 \end{aligned} $

$\begin{aligned} & \sec ^2 x \frac{d x}{d y}+e^{2 y} \tan ^2 x+\tan x=0 \\ & \left(\text { Put } \tan x=t \Rightarrow \sec ^2 x \frac{d x}{d y}=\frac{d t}{d y}\right) \\ & \frac{d t}{d y}+e^{2 y} \times t^2+t=0 \\ & \frac{d t}{d y}+t=-t^2 \cdot e^{2 y} \\ & \frac{1}{t^2} \frac{d t}{d y}+\frac{1}{t}=-e^{2 y} \\ & \left(\text { Put } \frac{1}{t}=u \frac{-1}{t^2} \frac{d t}{d y}=\frac{d u}{d y}\right) \\ & \frac{-d u}{d y}+u=-e^{2 y} \\ & \frac{d u}{d y}-u=e^{2 y} \\ & \text { I.F. }=e^{-\int d y}=e^{-y} \\ & u e^{-y}=\int e^{-y} \times e^{2 y} d y \\ & \frac{1}{\tan x} \times e^{-y}=e^y+c \\ & x=\frac{\pi}{4}, y=0, c=0 \\ & \begin{aligned} & \mathrm{x}=\frac{\pi}{6}, \quad \mathrm{y}=\alpha \\ & \sqrt{3} \mathrm{e}^{-\alpha}=\mathrm{e}^\alpha+0 \\ & \mathrm{e}^{2 \alpha}=\sqrt{3} \\ & \mathrm{e}^{8 \alpha}=9 \end{aligned} \end{aligned}$

$\mathrm{A}=\frac{1}{2}\left(\frac{-\mathrm{y}}{\mathrm{Y}^{\prime}(\mathrm{x})}+\mathrm{x}\right)(\mathrm{y}-\mathrm{xY} / \mathrm{x})=\frac{-\mathrm{y}^2}{2 \mathrm{Y}^{\prime}(\mathrm{x})}+1$

$\Rightarrow\left(-y+x Y^{\prime}(x)\right)\left(y-x Y^{\prime}(x)\right)=-y^2+2 Y^{\prime}(x)$

$\begin{aligned} -y^2+x y Y^{\prime}(x)+x y Y^{\prime}(x) & -x^2\left[Y^{\prime}(x)\right]^2 \\ = & -y^2+2 Y^{\prime}(x) \end{aligned}$

$\begin{aligned} & 2 x y-x^2 Y^{\prime}(x)=2 \\ & \frac{d y}{d x}=\frac{2 x y-2}{x^2} \\ & \frac{d y}{d x}-\frac{2}{x} y=\frac{-2}{x^2} \\ & \text { I.F. }=e^{-2 \ln x}=\frac{1}{x^2} \\ & y \cdot \frac{1}{x^2}=\frac{2}{3} x^{-3}+c \\ & \text { Put } x=1, y=1 \\ & 1=\frac{2}{3}+c \Rightarrow c=\frac{1}{3} \\ & Y=\frac{2}{3} \cdot \frac{1}{X}+\frac{1}{3} X^2 \\ \Rightarrow \quad & 12 Y(2)=\frac{5}{3} \times 12=20 \end{aligned}$

$\begin{aligned} & \frac{d y}{d x}-\frac{x y}{1-x^2}=\frac{\left(x^3+2\right) \sqrt{3\left(1-x^2\right)}}{1-x^2} \\ & \mathrm{IF}=e^{-\int \frac{x}{1-x^2} d x}=e^{+\frac{1}{2} \ln \left(1-x^2\right)}=\sqrt{1-x^2} \\ & y \sqrt{1-x^2}=\sqrt{3} \int\left(x^3+2\right) d x \\ & y \sqrt{1-x^2}=\sqrt{3}\left(\frac{x^4}{4}+2 x\right)+c \\ & \Rightarrow y(0)=0 \quad \therefore c=0\\ & y\left(\frac{1}{2}\right)=\frac{65}{32}=\frac{m}{n} \\ & m+n=97 \end{aligned}$

$\begin{aligned} & \int\left(\frac{1}{x}+\frac{\ln x}{x}\right) d x+\int \frac{d y}{1+y^2}=0 \\ & \ln x+\frac{(\ln x)^2}{2}+\tan ^{-1} y=C \end{aligned}$

Put $x=y=1$

$\begin{aligned} & \therefore C=\frac{\pi}{4} \\ & \Rightarrow \ln x+\frac{(\ln x)^2}{2}+\tan ^{-1} y=\frac{\pi}{4} \end{aligned}$

Put $x=e$

$\begin{aligned} & \Rightarrow \mathrm{y}=\tan \left(\frac{\pi}{4}-\frac{3}{2}\right)=\frac{1-\tan \frac{3}{2}}{1+\tan \frac{3}{2}} \\ & \therefore \alpha=1, \beta=1 \\ & \Rightarrow \alpha+2 \beta=3 \end{aligned}$

$\begin{aligned} & \frac{d y}{d x}=\frac{x+y-2}{x-y} \\ & \mathrm{x}=\mathrm{X}+\mathrm{h}, \mathrm{y}=\mathrm{Y}+\mathrm{k} \\ & \frac{d Y}{d X}=\frac{X+Y}{X-Y} \\ & \left.\begin{array}{l} \mathrm{h}+\mathrm{k}-2=0 \\ \mathrm{~h}-\mathrm{k}=0 \end{array}\right\} \mathrm{h}=\mathrm{k}=1 \\ & \mathrm{Y}=\mathrm{vX} \\ & v+\frac{d v}{d X}=\frac{1+v}{1-v} \Rightarrow X-\frac{d v}{d X}=\frac{1+v^2}{1-v} \\ & \end{aligned}$

$\begin{aligned} & \frac{1-v}{1+v^2} d v=\frac{d X}{X} \\ & \tan ^{-1} v-\frac{1}{2} \ln \left(1+v^2\right)=\ln |X|+C \end{aligned}$

As curve is passing through $(2,1)$

$\begin{aligned} & \tan ^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{2} \ln \left(1+\left(\frac{y-1}{x-1}\right)^2\right)=\ln |x-1| \\ & \therefore \alpha=1 \text { and } \beta=2 \\ & \Rightarrow 5 \beta+\alpha=11 \end{aligned}$

$\begin{array}{ll} 2 x+3 y-2=t & 4 x+6 y-4=2 t \\ 2+3 \frac{d y}{d x}=\frac{d t}{d x} & 4 x+6 y-7=2 t-3 \end{array}$

$\begin{aligned} & \frac{d y}{d x}=\frac{-(2 x+3 y-2)}{4 x+6 y-7} \\ & \frac{d t}{d x}=\frac{-3 t+4 t-6}{2 t-3}=\frac{t-6}{2 t-3} \\ & \int \frac{2 t-3}{t-6} d t=\int d x \\ & \int\left(\frac{2 t-12}{t-6}+\frac{9}{t-6}\right) \cdot d t=x \\ & 2 t+9 \ln (t-6)=x+c \\ & 2(2 x+3 y-2)+9 \ln (2 x+3 y-8)=x+c \\ & x=0, y=3 \\ & c=14 \\ & 4 x+6 y-4+9 \ln (2 x+3 y-8)=x+14 \\ & x+2 y+3 \ln (2 x+3 y-8)=6 \\ & \alpha=1, \beta=2, \gamma=8 \\ & \alpha+2 \beta+3 \gamma=1+4+24=29 \end{aligned}$

We can solve the given differential equation using an integrating factor.

The integrating factor is given by :

$ \mu(x) = e^{\int \frac{4x}{x^2 - 1} dx} = e^{2\ln(x^2 - 1)} = (x^2 - 1)^2 $

Multiplying both sides of the differential equation by $\mu(x)$, we get :

$(x^2-1)^2 \frac{d y}{d x} + 4x y = \frac{x+2}{\left(x^{2}-1\right)^{\frac{1}{2}}}$

We can rewrite the left-hand side using the product rule:

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

Integrating both sides with respect to $x$, we get:

$(x^2-1)^2 y = \int \frac{x+2}{\left(x^{2}-1\right)^{\frac{1}{2}}} dx = \sqrt{x^2-1}+2 \ln\left|x+\sqrt{x^2-1}\right|+C$

where $C$ is the constant of integration. Using the initial condition $y(2) = \frac{2}{9} \log_e(2+\sqrt{3})$, we can solve for $C$:

At $x=2$,

$9 \cdot \frac{2}{9} \ln (2+\sqrt{3})=2 \ln (2+\sqrt{3})+\sqrt{3}+C$

$C=-\sqrt{3}$

At $x=\sqrt{2}$

$y(\sqrt{2})=2 \ln (1+\sqrt{2})+1-\sqrt{3}$

$\beta=1, \alpha=2, \gamma=3$

$ \Rightarrow \alpha \beta \gamma=6 $

Let the equation of tangent to the curve at $(x, y)$.

Whose slope is $\frac{d y}{d x}$ is



$ \begin{aligned} &\mathrm{Y}-y=\frac{d y}{d x}(\mathrm{X}-x)\\\\ &\begin{aligned} & \text { Putting } Y=0 \Rightarrow X=x-\frac{y}{\left(\frac{d y}{d x}\right)} \mathrm{t} \\\\ & \Rightarrow \alpha=x-\frac{y}{\left(\frac{d y}{d x}\right)} \end{aligned} \end{aligned} $

and putting $\mathrm{X}=0 \Rightarrow \mathrm{Y}=y-x \frac{d y}{d x}$

$ \Rightarrow \beta=y-x \frac{d y}{d x} $

$\because \mathrm{P}$ divides $\mathrm{AB}$ in $1: k$

$ x=\frac{k \alpha+0}{k+1} \text { and } y=\frac{k \times 0+\beta}{k+1} $

$ \begin{aligned} & \Rightarrow x(k+1)=k\left(x-\frac{y}{\frac{d y}{d x}}\right) \\\\ & \Rightarrow x k+x=x k-\frac{y k}{\frac{d y}{d x}} \\\\ &\Rightarrow x \frac{d y}{d x}=-y k \end{aligned} $

$ \begin{aligned} & \text { or } \int \frac{d y}{y}=-k \times \int \frac{1}{x} d x \\\\ & \Rightarrow \log y=-k \log x+\log \mathrm{C} \\\\ & \text { or } \log y \times x^k=\log \mathrm{C} \\\\ & \Rightarrow y x^k=\mathrm{C} \\\\ & \text { putting } x=1, y=1 \Rightarrow c=1 \\\\ & \text { so } y x^k=1 \end{aligned} $

Putting $x=\frac{1}{10}, y=100 \Rightarrow 100 \times\left(\frac{1}{100}\right)^k=1 \Rightarrow k=2$

so $y x^2=1$ or $y=\frac{1}{x^2}$

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

$ \Rightarrow \frac{d y}{d x}=\log _e\left(k x+\frac{k}{2}\right)=\log _e(2 x+1) $

On integrating

$ \begin{aligned} & y=\int 1 \cdot \log _e(2 x+1) d x \\\\ & =x \log _e(2 x+1)-\int \frac{1 \times 2}{2 x+1} \times x d x \\\\ & =x \log _e(2 x+1)-\int 1-\frac{1}{2 x+1} d x \\\\ & y=x \log _e(2 x+1)-x+\frac{1}{2} \log _e(2 x+1)+c \end{aligned} $

Put $x=0, y=k=2 \Rightarrow c=2$ putting $x=1$

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

$ \begin{aligned} & (\cos y)(\ln (\cos y))^2 d x=(1+3 x \ln \cos y) \sin y d y \\\\ & \Rightarrow \frac{d x}{d y}=\frac{(1+3 x \ln \cos y) \sin y}{(\ln \cos y)^2 \cos y} \\\\ & =\tan y\left[\frac{1}{(\ln \cos y)^2}+\frac{3 x}{\ln \cos y}\right] \\\\ & \Rightarrow \frac{d x}{d y}-\left(\frac{3 \tan y}{\ln \cos y}\right) x=\frac{\tan y}{(\ln \cos y)^2} \end{aligned} $

Which is a linear differential equation.

$ \text { I.F. }=e^{-\int \frac{3 \tan y}{\ln \cos y} d y}=(\ln \cos y)^3 $

So, the solution is :

$ \begin{aligned} & x \times(\ln \cos y)^3=\int\left((\ln \cos y)^3 \times \frac{\tan y}{(\ln \cos y)^2}\right) d y \\\\ & x \times(\ln \cos y)^3=\frac{-(\ln \cos y)^2}{2}+C \end{aligned} $

$ \text { At } y=\frac{\pi}{3} \text {, } $

$ \begin{aligned} & \frac{1}{2 \ln 2} \times\left(\ln \left(\frac{1}{2}\right)\right)^3=-\frac{\left(\ln \left(\frac{1}{2}\right)\right)^2}{2}+C \\\\ & \Rightarrow C=0 \end{aligned} $

$ \begin{aligned} & \text { So, } x \times \ln ^3 \cos y=\frac{-\ln ^2 \cos y}{2} \\\\ & \text { At } y=\frac{\pi}{6}, x \times\left(\ln \left(\frac{\sqrt{3}}{2}\right)\right)^3=-\frac{1}{2}\left(\ln \left(\frac{\sqrt{3}}{2}\right)\right)^2 \\\\ & \Rightarrow x=-\frac{1}{2 \ln \left(\frac{\sqrt{3}}{2}\right)} \end{aligned} $

$ \begin{aligned} & =-\frac{1}{2[\ln \sqrt{3}-\ln 2]}=\frac{-1}{2\left[\frac{1}{2} \ln 3-\ln 2\right]} \\\\ & =\frac{-1}{2\left[\frac{\ln 3-\ln 4}{2}\right]}=\frac{1}{\ln 4-\ln 3} \\\\ & \Rightarrow m=4, n=3 \\\\ & \Rightarrow m n=12 \end{aligned} $

The given differential equation is,

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

It is a linear differential equation.

$ \therefore \text { I.F. }=e^{\int \frac{1}{2 x \log x} d x} $

Put $\log x=t \Rightarrow \frac{1}{x} d x=d t$

$ \therefore \text { I.F. }=e^{\int \frac{1}{2 t} d t}=e^{\log (t)^{\frac{1}{2}}}=\sqrt{t}=\sqrt{\log x} $

So, required solution is,

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

Now, this curve passes through $\left(e, \frac{4}{3}\right)$ and $\left(e^4, \alpha\right)$

$ \begin{aligned} & \therefore \frac{4}{3} \sqrt{\log e}=\frac{2}{3}(\log e)^{3 / 2}+C \\\\ & \Rightarrow C=\frac{4}{3}-\frac{2}{3}=\frac{2}{3} \end{aligned} $

Also, $\alpha \sqrt{\log e^4}=\frac{2}{3}\left(\log e^4\right)^{3 / 2}+\frac{2}{3}$

$ \begin{aligned} & \Rightarrow 2 \alpha=\frac{2}{3} \times(4)^{3 / 2}+\frac{2}{3}=\frac{16}{3}+\frac{2}{3}=\frac{18}{3} \\\\ & \Rightarrow \alpha=3 \end{aligned} $

Given, differential equation

$ \begin{aligned} & (x \cos x) d y+(x y \sin x+y \cos x-1) d x=0,0 < x < \frac{\pi}{2} \\\\ & \Rightarrow \frac{d y}{d x}+\frac{x y \sin x+y \cos x-1}{x \cos x}=0 \\\\ & \Rightarrow \frac{d y}{d x}+\left(\frac{x y \sin x+y \cos x}{x \cos x}\right)=\frac{1}{x \cos x} \\\\ & \Rightarrow \frac{d y}{d x}+\left(\tan x+\frac{1}{x}\right) y=\frac{1}{x \cos x} \end{aligned} $

Which is linear differential equation in the form of

$ \begin{aligned} & \frac{d y}{d x}+P y=Q \\\\ & \therefore \mathrm{IF}=e^{\int\left(\tan x+\frac{1}{x}\right) d x}=e^{(\log \sec x+\log x)}=e^{\log (x \sec x)}=x \sec x \end{aligned} $

$\therefore$ The general solution of the given differential equation

$ \begin{array}{rlrl} y \cdot \mathrm{IF} =\int(Q \times \mathrm{IF}) d x+c \\\\ \Rightarrow y(x \sec x) =\int\left(\frac{1}{x \cos x} x \sec x\right) d x+c \\\\ \Rightarrow x y \sec x =\int \sec ^2 x d x+c \\\\ \Rightarrow x y \sec x =\tan x+c ........(i) \end{array} $

Since, $\frac{\pi}{3} y\left(\frac{\pi}{3}\right)=\sqrt{3} \Rightarrow y\left(\frac{\pi}{3}\right)=\frac{3 \sqrt{3}}{\pi}$

$ \begin{aligned} & \therefore \frac{\pi}{3}\left(\frac{3 \sqrt{3}}{\pi}\right) \sec \left(\frac{\pi}{3}\right)=\tan \left(\frac{\pi}{3}\right)+c \\\\ & \Rightarrow \sqrt{3}(2)=\sqrt{3}+c \\\\ & \begin{aligned} \Rightarrow & c=\sqrt{3} \end{aligned} \end{aligned} $

On putting the value of $c$ in Eq. (i), we get

$ \begin{aligned} &x y \sec x =\tan x+\sqrt{3} \\\\ &\Rightarrow y =\frac{1}{x}(\sin x+\sqrt{3} \cos x) \\\\ & y^{\prime} =\frac{1}{x}(\cos x-\sqrt{3} \sin x)-\frac{1}{x^2}(\sin x+\sqrt{3} \cos x) \end{aligned} $

$\begin{aligned} & \text { and } y^{\prime \prime}=\frac{1}{x}(-\sin x-\sqrt{3} \cos x)-\frac{1}{x^2}(\cos x-\sqrt{3} \sin x) -\frac{1}{x^2}(\cos x-\sqrt{3} \sin x) \\ & -\frac{2}{x^3}(\cos x-\sqrt{3} \sin x) \\\\ & \therefore\left|\frac{\pi}{6} y^{\prime \prime}\left(\frac{\pi}{6}\right)+2 y^{\prime}\left(\frac{\pi}{6}\right)\right|=|-2|=2 \\\\ & \end{aligned}$

${{dy} \over {dx}} + y\left( {{{4x} \over {\sin (2{x^2})\ln (\tan {x^2})}}} \right) = {{4\sqrt 2 x\sin \left( {{x^2} - {\pi \over 4}} \right)} \over {\sin (2{x^2})\ln (\tan {x^2})}}$

$I.F = {e^{\int {{{4x} \over {\sin (2{x^2})\ln (\tan {x^2})}}dx} }}$

$ = {e^{\ln |\ln (\tan {x^2})}} = \ln (\tan {x^2})$

$\therefore$ $\int {d(y.\ln (\tan {x^2})) = \int {{{4\sqrt 2 x\sin \left( {{x^2} - {\pi \over 4}} \right)} \over {\sin (2{x^2})}}dx} } $

$ \Rightarrow y\ln (\tan {x^2}) = \ln \left| {{{\sec {x^2} + \tan {x^2}} \over {{\mathop{\rm cosec}\nolimits} \,{x^2} - \cot {x^2}}}} \right| + C$

$\ln \left( {{1 \over {\sqrt 3 }}} \right) = \ln \left( {{{{3 \over {\sqrt 3 }}} \over {2 - \sqrt 3 }}} \right) + C$

$e = \ln \left( {{1 \over {\sqrt 3 }}} \right) - \ln \left( {{{\sqrt 3 } \over {2 - \sqrt 3 }}} \right)$

For $y\left( {\sqrt {{\pi \over 3}} } \right)$

$y\ln \left( {\sqrt 3 } \right) = \ln \left| {{{2 + \sqrt 3 } \over {{2 \over {\sqrt 3 }} + {1 \over {\sqrt 3 }}}}} \right| + \ln \left( {{1 \over {\sqrt 3 }}} \right) - \ln \left( {{{\sqrt 3 } \over {2\sqrt 3 }}} \right)$

$ = \ln \left( {2 + \sqrt 3 } \right) + \ln \left( {{1 \over {\sqrt 3 }}} \right) + \ln \left( {{1 \over {\sqrt 3 }}} \right) - \ln \left( {{{\sqrt 3 } \over {2 - \sqrt 3 }}} \right)$

$ \Rightarrow y\ln \sqrt 3 = \ln \left( {{1 \over {\sqrt 3 }}} \right)$

$ \Rightarrow {y \over 2}\ln 3 = - {1 \over 2}\ln 3$

$ \Rightarrow y = 1$

$\therefore$ $\left| {y\left( {\sqrt {{\pi \over 3}} } \right)} \right| = 1$.

IF $ = {e^{-x}}$

$y\,.\,{e^{-x}} = - 2{e^{ - x}} + {{{e^{ - 2x}}} \over 2} + C$

$ \Rightarrow y = - 2 + {e^{ - x}} + C{e^x}$

$\mathop {\lim }\limits_{x \to \infty } \,y(x)$ is finite so $C = 0$

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

$ \Rightarrow {\left. {{{dy} \over {dx}} = - {e^{ - x}} \Rightarrow {{dy} \over {dx}}} \right|_{x = 0}} = - 1$

Equation of tangent

$y + 1 = - 1(x - 0)$

or $y + x = - 1$

So $a = - 1,\,b = - 1$

$ \Rightarrow a - 4b = 3$

$\int\limits_3^x {f(x)dx = {{\left( {{{f(x)} \over x}} \right)}^3}} $

${x^3}\,.\,\int\limits_3^x {f(x)dx = {f^3}(x)} $

Differentiate w.r.t. x

${x^3}f(x) + 3{x^2}\,.\,{{{f^3}(x)} \over {{x^3}}} = 3{f^2}(x)f'(x)$

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

$3xy{{dy} \over {dx}} = {x^4} + 3{y^2}$

Let ${y^2} = t$

${3 \over 2}{{dt} \over {dx}} = {x^3} + {{3t} \over x}$

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

$I.F. = \,.\,{e^{\int { - {2 \over x}dx} }} = {1 \over {{x^2}}}$

Solution of differential equation

$t\,.\,{1 \over {{x^2}}} = \int {{2 \over 3}x\,dx} $

${{{y^2}} \over {{x^2}}} = {{{x^2}} \over 3} + C$

${y^2} = {{{x^4}} \over 3} + C{x^2}$

Curve passes through $(3,3) \Rightarrow C = - 2$

${y^2} = {{{x^4}} \over 3} - 2{x^2}$

Which passes through $\left( {\alpha ,6\sqrt {10} } \right)$

${{{\alpha ^4} - 6{\alpha ^2}} \over 3} = 360$

${\alpha ^4} - 6{\alpha ^2} - 1080 = 0$

$\alpha = 6$

${{dy} \over {dx}} = {y \over x}{{(4{y^2} + 2{x^2})} \over {(3{y^2} + {x^2})}}$

Put $y = vx$

$ \Rightarrow {{dy} \over {dx}} = v + x{{dv} \over {dx}}$

$ \Rightarrow v + x{{dv} \over {dx}} = {{v(4{v^2} + 2)} \over {(3{v^2} + 1)}}$

$ \Rightarrow x{{dv} \over {dx}} = v\left( {{{(4{v^2} + 2 - 3{v^2} - 1)} \over {3{v^2} + 1}}} \right)$

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

$ \Rightarrow \ln |{v^3} + v| = \ln x + c$

$ \Rightarrow \ln \left| {{{\left( {{y \over x}} \right)}^3} + \left( {{y \over x}} \right)} \right| = \ln x + C$

$ \downarrow \,y(1) = 1$

$ \Rightarrow C = \ln 2$

$\therefore$ for $y(2)$

$\ln \left( {{{{y^3}} \over 8} + {y \over 2}} \right) = 2\ln 2 \Rightarrow {{{y^3}} \over 8} + {y \over 2} = 4$

$ \Rightarrow [y(2)] = 2$

$ \Rightarrow n = 3$

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

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

$ \begin{aligned} & =e^{2 \int\left(\frac{x-1}{x-1}+\frac{1}{x-1}\right) d x} \\\\ & =e^{2 x+2 \ln (x-1)} \\\\ & =\mathrm{e}^{2 x}(\mathrm{x}-1)^{2} \end{aligned} $

$\Rightarrow \int d\left(y \cdot e^{2 x}(x-1)^{2}\right)=\int e^{2 x} d x$

$\Rightarrow \quad y \cdot e^{2 x}(x-1)^{2}=\frac{e^{2 x}}{2}+c$

$\downarrow y(2)=\frac{1+e^{4}}{2 e^{4}}$

$\frac{1+e^{4}}{2 e^{4}} \cdot e^{4}=\frac{e^{4}}{2}+c$

$\Rightarrow \quad c=\frac{e^{4}}{2}\left(\frac{1+e^{4}-e^{4}}{e^{4}}\right)=\frac{1}{2}$

$\Rightarrow \quad y \cdot e^{2 x}(x-1)^{2}=\frac{e^{2 x}+1}{2}$

$ \downarrow y(3)=\frac{e^{\alpha}+1}{\beta e^{\alpha}} $

$\Rightarrow \frac{e^{\alpha}+1}{\beta e^{\alpha}} \cdot e^{6} \cdot 4=\frac{e^{6}+1}{2}$

$\Rightarrow \alpha=6$ and $\beta=8 \Rightarrow \alpha+\beta=14$

$\frac{d y}{d x}+\frac{2 \sqrt{2} y}{1+\cos ^{2} 2 x}=x e^{\tan ^{-1}(\sqrt{2} \cot 2 x)}$

I.F. $=e^{\int \frac{2 \sqrt{2} d x}{1+\cos ^{2} 2 x}}=e^{\sqrt{2} \int \frac{2 \sec ^{2} 2 x}{2+\tan ^{2} 2 x} d x}$ $=e^{\tan ^{-1}\left(\frac{\tan 2 x}{\sqrt{2}}\right)}$

$\Rightarrow y \cdot e^{\tan ^{-1}\left(\frac{\tan 2 x}{\sqrt{2}}\right)}=\int x e^{\tan ^{-1}(\sqrt{2} \cot 2 x)}$

$ \cdot e^{\tan ^{-1}\left(\frac{\tan 2 x}{\sqrt{2}}\right)} d x+c $

$\Rightarrow y \cdot e^{\tan ^{-1}\left(\frac{\tan 2 x}{\sqrt{2}}\right)}=e^{\frac{\pi}{2}} \cdot \frac{x^{2}}{2}+c$

When $x=\frac{\pi}{4}, y=\frac{\pi^{2}}{32}$ gives $c=0$

When $x=\frac{\pi}{3}, y=\frac{\pi^{2}}{18} e^{-\tan ^{-1} \alpha}$

So $\frac{\pi^{2}}{18} e^{-\tan ^{-1} \alpha} \cdot e^{-\tan ^{-1}\left(-\sqrt{\left.\frac{3}{2}\right)}\right.}=e^{\pi / 2} \frac{\pi^{2}}{18}$

$\Rightarrow \quad-\tan ^{-1} \alpha+\tan ^{-1}\left(\sqrt{\frac{3}{2}}\right)=\frac{\pi}{2}$

$\Rightarrow \tan ^{-1}(-\alpha)=\tan ^{-1}\left(\sqrt{\frac{2}{3}}\right)$

$\Rightarrow \alpha=-\sqrt{\frac{2}{3}} \Rightarrow 3 \alpha^{2}=2$

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

$\therefore$ ${{dy} \over {dx}} - {x \over {1 - {x^2}}}y = {{{x^3} + 3} \over {\sqrt {1 - {x^2}} }}$

$\therefore$ $I.F. = {e^{\int { - {x \over {1 - {x^2}}}dx} }} = \sqrt {1 - {x^2}} $

Solution is

$y.\,\sqrt {1 - {x^2}} = \int {\left( {{x^3} + 3} \right)dx} $

$y.\,\sqrt {1 - {x^2}} = {{{x^4}} \over 4} + 3x + c$

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

$\therefore$ $y(x) = {{{x^4} + 12x} \over {4\sqrt {1 - {x^2}} }}$

$\therefore$ $\int_{{{ - 1} \over 2}}^{{1 \over 2}} {\sqrt {1 - {x^2}} y(x)dx = \int_{{{ - 1} \over 2}}^{{1 \over 2}} {\left( {{{{x^4} + 12x} \over 4}} \right)dx = \int_0^{{1 \over 2}} {{{{x^4}} \over 2}dx} } } $

$\therefore$ $k = {1 \over {320}}$

$\therefore$ $ = {k^{ - 1}} = 320$

$(4 + {x^2})dy - 2x({x^2} + 3y + 4)dx = 0$

$ \Rightarrow {{dy} \over {dx}} = \left( {{{6x} \over {{x^2} + 4}}} \right)y + 2x$

$ \Rightarrow {{dy} \over {dx}} - \left( {{{6x} \over {{x^2} + 4}}} \right)y = 2x$

$I.F. = {e^{ - 3\ln ({x^2} + 4)}} = {1 \over {{{({x^2} + 4)}^3}}}$

So ${y \over {{{({x^2} + 4)}^3}}} = \int {{{2x} \over {{{({x^2} + 4)}^3}}}dx + c} $

$ \Rightarrow y = - {1 \over 2}({x^2} + 4) + c{({x^2} + 4)^3}$

When x = 0, y = 0 gives $c = {1 \over {32}}$,

So, for x = 2, y = 12

${{dy} \over {dx}} = {1 \over {1 + \sin 2x}}$

$ \Rightarrow dy = {{{{\sec }^2}xdx} \over {{{(1 + \tan x)}^2}}}$

$ \Rightarrow y = - {1 \over {1 + \tan x}} + c$

When $x = {\pi \over 4}$, $y = {1 \over 2}$ gives c = 1

So $y = {{\tan x} \over {1 + \tan x}} \Rightarrow y = {{\sin x} \over {\sin x + \cos x}}$

Now, $y = \sqrt 2 \sin x \Rightarrow \sin x = 0$

or $\sin x + \cos x = {1 \over {\sqrt 2 }}$

$\sin x = 0$ gives $x = \pi $ only.

and $\sin x + \cos x = {1 \over {\sqrt 2 }} \Rightarrow \sin \left( {x + {\pi \over 4}} \right) = {1 \over 2}$

So $x + {\pi \over 4} = {{5\pi } \over 6}$ or ${{13\pi } \over 6} \Rightarrow x = {{7\pi } \over {12}}$ or ${{23\pi } \over {12}}$

Sum of all solutions $ = \pi + {{7\pi } \over {12}} + {{23\pi } \over {12}} = {{42\pi } \over {12}}$

Hence, $k = 42$.

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

$ \Rightarrow {\left( {{y^{{1 \over 4}}}} \right)^2} - 2x{y^{\left( {{1 \over 4}} \right)}} + 1 = 0$

$ \Rightarrow {y^{{1 \over 4}}} = x + \sqrt {{x^2} - 1} $ or $x - \sqrt {{x^2} - 1} $

So, ${1 \over 4}{1 \over {{y^{{3 \over 4}}}}}{{dy} \over {dx}} = 1 + {x \over {\sqrt {{x^2} - 1} }}$

$ \Rightarrow {1 \over 4}{1 \over {{y^{{3 \over 4}}}}}{{dy} \over {dx}} = {{{y^{{1 \over 4}}}} \over {\sqrt {{x^2} - 1} }}$

$ \Rightarrow {{dy} \over {dx}} = {{4y} \over {\sqrt {{x^2} - 1} }}$ .... (1)

Hence, ${{{d^2}y} \over {d{x^2}}} = 4{{\left( {\sqrt {{x^2} - 1} } \right)y' - {{yx} \over {\sqrt {{x^2} - 1} }}} \over {{x^2} - 1}}$

$ \Rightarrow ({x^2} - 1)y'' = 4{{({x^2} - 1)y' - xy} \over {\sqrt {{x^2} - 1} }}$

$ \Rightarrow ({x^2} - 1)y'' = 4\left( {\sqrt {{x^2} - 1} y' - {{xy} \over {\sqrt {{x^2} - 1} }}} \right)$

$ \Rightarrow ({x^2} - 1)y'' = 4\left( {4y - {{xy'} \over 4}} \right)$ (from I)

$ \Rightarrow ({x^2} - 1)y'' + xy' - 16y = 0$

So, | $\alpha$ $-$ $\beta$ | = 17

$\int {{e^{ - y}}} dy = \int {{e^{\alpha x}}} dx$

$ \Rightarrow {e^{ - y}} = {{{e^{\alpha x}}} \over \alpha } + c$ ..... (i)

Put (x, y) = (ln2, ln2)

${{ - 1} \over 2} = {{{2^\alpha }} \over \alpha } + C$ ..... (ii)

Put (x, y) $ \equiv $ (0, $-$ln2) in (i)

$ - 2 = {1 \over \alpha } + C$ ..... (iii)

(ii) $-$ (iii)

${{{2^\alpha } - 1} \over \alpha } = {3 \over 2}$

$\Rightarrow$ $\alpha$ = 2 (as $\alpha$ $\in$ N)

$\sec y{{dy} \over {dx}} = 2\sin x\cos y$

${\sec ^2}ydy = 2\sin xdx$

$\tan y = - 2\cos x + c$

$c = 2$

$\tan y = - 2\cos x + 2 \Rightarrow $ at $x = {\pi \over 2}$

$\tan y = 2$

${\sec ^2}y{{dy} \over {dx}} = 2\sin x$

$ \therefore $ $5{{dy} \over {dx}} = 2$

$y' = {{2y} \over {x\ln x}}$

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

$ \Rightarrow \ln |y| = 2\ln |\ln x| + C$

put x = 2, y = (ln2)²

$\Rightarrow$ c = 0

$\Rightarrow$ y = (lnx)²

$\Rightarrow$ f(e) = 1

Let e^y = t

$ \Rightarrow {{dt} \over {dx}} - (2\sin x)t = - \sin x{\cos ^2}x$

$I.F. = {e^{2\cos x}}$

$ \Rightarrow t.{e^{2\cos x}} = \int {{e^{2\cos x}}.( - \sin x{{\cos }^2}x)dx} $

$ \Rightarrow {e^y}.{e^{2\cos x}} = \int {{e^{2z}}.{z^2}dz,z = {e^{2\cos x}}} $

$ \Rightarrow {e^y}.{e^{2\cos x}} = {1 \over 2}.{\cos ^2}x.{e^{2\cos x}} - {1 \over 2}\cos x.{e^{2\cos x}} + {{{e^{2\cos x}}} \over 4} + C$

at $x = {\pi \over 2},y = 0 \Rightarrow C = {3 \over 4}$

$ \Rightarrow {e^y} = {1 \over 2}{\cos ^2}x - {1 \over 2}\cos x + {1 \over 4} + {3 \over 4}.{e^{ - 2\cos x}}$

$ \Rightarrow y = \log \left[ {{{{{\cos }^2}x} \over 2} - {{\cos x} \over 2} + {1 \over 4} + {3 \over 4}{e^{ - 2\cos x}}} \right]$

Put x = 0

$ \Rightarrow y = \log \left[ {{1 \over 4} + {3 \over 4}{e^{ - 2}}} \right] \Rightarrow \alpha = {1 \over 4},\beta = {3 \over 4}$

Let y + 1 = Y and x + 2 = X

dy = dY

dx = dX

$\left( {X{e^{{X \over X}}} + Y} \right)dX = XdY$

$ \Rightarrow {{XdY - YdX} \over {{X^2}}} = {{{e^{{Y \over X}}}} \over X}dX$

$ \Rightarrow {e^{ - {Y \over X}}}d\left( {{Y \over X}} \right) = {{dX} \over X}$

$ \Rightarrow - {e^{ - {Y \over X}}} = \ln |X| + c$

$ \Rightarrow - {e^{ - \left( {{{y + 1} \over {x + 2}}} \right)}} = \ln |x + 2| + c$

$\because$ (1, 1) satisfy this equation

So, $c = - {e^{ - {2 \over 3}}} - \ln 3$

Now, $y = - 1 - (x + 2)\ln \left( {\ln \left( {\left| {{3 \over {x + 2}}} \right|} \right) + {e^{ - {2 \over 3}}}} \right)$

Domain :

$\ln \left| {{3 \over {x + 2}}} \right| > {e^{ - {e^{ - {2 \over 3}}}}}$

$ \Rightarrow {3 \over {\left| {x + 2} \right|}} > {e^{ - {e^{ - {2 \over 3}}}}}$

$ \Rightarrow \left| {x + 2} \right| < 3{e^{{e^{ - {2 \over 3}}}}}$

$ \Rightarrow - 3{e^{{e^{ - {2 \over 3}}}}} - 2 < x < 3{e^{{e^{ - {2 \over 3}}}}} - 2$

So, $\alpha + \beta = - 4$

$ \Rightarrow \left| {\alpha + \beta } \right| = 4$

$\cos \left( {{1 \over 2}{{\cos }^{ - 1}}({e^{ - x}})} \right)dx = \sqrt {{e^{2x}} - 1} dy$

Put cos^$-$1(e^$-$x) $\theta$, $\theta$ $\in$ [0, $\pi$]

$\cos \theta = {e^{ - x}} \Rightarrow 2{\cos ^2}{\theta \over 2} - 1 = {e^{ - x}}$

$\cos {\theta \over 2} = \sqrt {{{{e^{ - x}} + 1} \over 2}} = \sqrt {{{{e^x} + 1} \over {2{c^x}}}} $

$\sqrt {{{{e^x} + 1} \over {2{c^x}}}} dx = \sqrt {{e^{2x}} - 1} dy$

${1 \over {\sqrt 2 }}\int {{{dx} \over {\sqrt {{e^x}} \sqrt {{e^x} - 1} }} = \int {dy} } $

Put ${e^x} = t,{{dt} \over {dx}} = {e^x}$

${1 \over {\sqrt 2 }}\int {{{dx} \over {{e^x}\sqrt {{e^x}} \sqrt {{e^x} - 1} }} = \int {dy} } $

$\int {{{dt} \over {t\sqrt {{t^2} - t} }} = \sqrt 2 y} $

Put $t = {1 \over z},{{dt} \over {dz}} = - {1 \over {{z^2}}}$

$\int {{{ - {{dz} \over {{z^2}}}} \over {{1 \over z}\sqrt {{1 \over {{z^2}}} - {1 \over z}} }} = \sqrt {2y} } $

$ - \int {{{dz} \over {\sqrt {1 - z} }} = \sqrt 2 y} $

${{ - 2{{(1 - z)}^{1/2}}} \over { - 1}} = \sqrt 2 y + c$

$2{\left( {1 - {1 \over t}} \right)^{1/2}} = \sqrt 2 y + c$

$2{(1 - {e^{ - x}})^{1/2}} = \sqrt 2 y + c\buildrel {(0, - 1)} \over \longrightarrow \Rightarrow c = \sqrt 2 $

$2{(1 - {e^{ - x}})^{1/2}} = \sqrt 2 (y + 1)$, passes through ($\alpha$, 0)

$2{(1 - {e^{ - \alpha }})^{1/2}} = \sqrt 2 $

$\sqrt {1 - {e^{ - \alpha }}} = {1 \over {\sqrt 2 }} \Rightarrow 1 - {e^{ - \alpha }} = {1 \over 2}$

${e^{ - \alpha }} = {1 \over 2} \Rightarrow {e^\alpha } = 2$

$xdy - ydx = \sqrt {{x^2} - {y^2}} dx$

dividing both sides by x², we get

${{xdy - ydx} \over {{x^2}}} = {{\sqrt {{x^2} - {y^2}} } \over {{x^2}}}dx$

$ \Rightarrow d\left( {{y \over x}} \right) = {1 \over x}\sqrt {1 - {{\left( {{y \over x}} \right)}^2}} dx$

$ \Rightarrow {{d\left( {{y \over x}} \right)} \over {\sqrt {1 - {{\left( {{y \over x}} \right)}^2}} }} = {{dx} \over x}$

Integrating both side, we get

$ \Rightarrow \int {{{d\left( {{y \over x}} \right)} \over {\sqrt {1 - {{\left( {{y \over x}} \right)}^2}} }} = \int {{{dx} \over x}} } $

${\sin ^{ - 1}}\left( {{y \over x}} \right) = \ln (x) + C$

Given, y(1) = 0 $ \Rightarrow $ at x = 1, y = 0

$ \therefore $ $ \Rightarrow {\sin ^{ - 1}}(0) = \ln (1) + C$

$ \Rightarrow $ C = 0

$ \therefore $ ${\sin ^{ - 1}}\left( {{y \over x}} \right) = \ln (x)$

$ \Rightarrow $ y = x sin(ln(x))

$ \therefore $ Area $ = \int_1^{{e^{\pi {} }}} {x\sin (\ln (x))} dx$

Let, lnx = t

$ \Rightarrow $ x = e^t

$ \Rightarrow $ dx = e^t dt

New lower limit, t = ln(1) = 0

and upper limit t = ln$({e^{\pi {} }})$ = ${\pi {} }$

$ \therefore $ Area = $\int_0^{^{\pi {} }} {{e^t}\sin (t).{e^t}} dt$

$ = \int_0^{^{\pi {} }} {{e^{2t}}\sin t\,} dt$

$ = \left[ {{{{e^{2t}}} \over {({1^2} + {2^2})}}(2\sin t - 1\cos t)} \right]_0^{\pi {} }$

$ = {\left[ {{{{e^{2\pi {} }}} \over 5}(0 - ( - 1)) - {1 \over 5}( - 1)} \right]}$

$ = {{{e^{2\pi {} }}} \over 5} + {1 \over 5}$

$ = \alpha {e^{2\pi {} }} + \beta $

$ \therefore $ $\alpha = {1 \over 5},\beta = {1 \over 5}$

So, $10(\alpha + \beta ) = 4$

Given, ${{dy} \over {dx}}$ = 2(x + 1)

Integrating both sides, we get

$y = {x^2} + 2x + c$

Let the two roots of the quadratic equation $\alpha $ and $\beta $



As parabola intercept the x axis so D > 0

From figure, AB = |$\alpha $ - $\beta $| = ${{\sqrt D } \over {\left| a \right|}}$ = $\sqrt D $

and BC = $ - {D \over {4a}}$ = $ - {D \over 4}$

$ \therefore $ Area of rectangle (ABCD) = AB $ \times $ BC = $\sqrt D \times {D \over 4}$

From property we know,

Area of parabola with the x axis = ${2 \over 3}$(Area of rectangle)

$ \Rightarrow $ ${{4\sqrt 8 } \over 3}$ = ${2 \over 3} \times \sqrt D \times {D \over 4}$

$ \Rightarrow $ $D\sqrt D $ = $8\sqrt 8 $

$ \Rightarrow $ D = 8

$ \therefore $ b² - 4ac = 8

$ \Rightarrow $ 4 - 4c = 8

$ \Rightarrow $ 1 $-$ c = 2 $ \Rightarrow $ c = $-$ 1

Equation of f(x) = x² + 2x $-$ 1

$ \therefore $ f(1) = 1 + 2 $-$ 1 = 2

${y^2} = a\left( {x + {{\sqrt a } \over 2}} \right)$

Differentiating both sides, we get

$2yy' = a$

${y^2} = 2yy'\left( {x + {{\sqrt {2yy'} } \over 2}} \right)$

$y = 2y'\left( {x + {{\sqrt {yy'} } \over {\sqrt 2 }}} \right)$

$y - 2xy' = \sqrt 2 y'\sqrt {yy'} $

${\left( {y - 2x{{dy} \over {dx}}} \right)^2} = 2y{\left( {{{dy} \over {dx}}} \right)^3}$

D = 3 & O = 1

$ \therefore $ D $-$ O = 3 $-$ 1 = 2

e^{sin y} cos y${{dy} \over {dx}}$ + e^{sin y} cos x = cos x

Put e^{sin y} = t

e^{sin y} $\times$ cos y${{dy} \over {dx}}$ = ${{dt} \over {dx}}$

$ \Rightarrow $ ${{dt} \over {dx}}$ + t cos x = cos x

I. F. = ${e^{\int {\cos x\,dx} }} = {e^{\sin x}}$

Solution of differential equation :

$t.{e^{\sin x}} = \int {{e^{\sin x}}.\cos x\,dx} $

${e^{\sin y}}.{e^{\sin x}} = {e^{\sin x}} + c$

at x = 0, y = 0

1 = 1 + c $ \Rightarrow $ c = 0

$ \therefore $ e^{sin x + sin y} = e^{sin x}

$ \Rightarrow $ sin x + sin y = sin x

$ \Rightarrow $ sin y = 0 $ \Rightarrow $ y = 0

$ \Rightarrow y\left( {{\pi \over 6}} \right) = 0,y\left( {{\pi \over 3}} \right) = 0,y\left( {{\pi \over 4}} \right) = 0$

$ \therefore $ $1 + y\left( {{\pi \over 6}} \right) + {{\sqrt 3 } \over 2}y\left( {{\pi \over 3}} \right) + {1 \over {\sqrt 2 }}y\left( {{\pi \over 4}} \right)$

= 1 + 0 + 0 + 0 = 1

Given,

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

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

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

${1 \over y} = z$

$ - {1 \over {{y^2}}}{{dy} \over {dx}} = {{dz} \over {dx}}$

$ \Rightarrow {{dz} \over {dx}} + z\left( {{1 \over x}} \right) = 2$

I. F. $ = {e^{\int {{1 \over x}dx} }} = x$

$ \therefore $ $z(x) = \int {2(x)dx} = {x^2} + c$

$ \Rightarrow {x \over y} = {x^2} + c$

As it passes through P(2, 1)

[Point of intersection of $2x - 3y = 1$ and $3x + 2y = 8$]

$ \therefore $ ${2 \over 1} = 4 + c$

$ \Rightarrow c = - 2$

$ \Rightarrow {x \over y} = {x^2} - 2$

Put x = 1

${1 \over y} = 1 - 2 = - 1$

$ \Rightarrow y(1) = - 1$

$ \Rightarrow |y(1)|\, = 1$

(x + 1)dy = ((x + 1)² + y – 3)dx

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

$ \Rightarrow $ ${{dy} \over {dx}} - {y \over {1 + x}} = \left( {1 + x} \right) - {3 \over {1 + x}}$

I.F = ${e^{ - \int {{{dx} \over {1 + x}}} }}$ = ${1 \over {1 + x}}$

Solution of the differential equation,

$y\left( {{1 \over {1 + x}}} \right)$ = $\int {\left( {\left( {1 + x} \right) - {3 \over {1 + x}}} \right)\left( {{1 \over {1 + x}}} \right)dx} $

$ \Rightarrow $ ${y \over {1 + x}}$ = $\int {{{{x^2} + 2x + 1 - 3} \over {{{\left( {x + 1} \right)}^2}}}dx} $

$ \Rightarrow $ ${y \over {1 + x}}$ = x + ${3 \over {1 + x}}$ + C

As y(2) = 0 $ \Rightarrow $ x = 2, y = 0

$ \therefore $ 0 = 2 + ${3 \over {1 + 2}}$ + C

$ \Rightarrow $ C = -3

So solution is ${y \over {1 + x}}$ = x + ${3 \over {1 + x}}$ - 3

y(3) means x = 3 and find value of y.

${y \over {1 + 3}} = 3 + {3 \over {1 + 3}} - 3$

$ \Rightarrow $ y = 3

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

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

Integrating both sides

$ \begin{aligned} & \Rightarrow \int \frac{d t}{(t-1)^2}=\int \frac{d x}{x} \\ & \Rightarrow \frac{-1}{(t-1)}=\ln x+C \\ & \Rightarrow \frac{-1}{\frac{y}{x}-1}=\ln x+C \\ & \Rightarrow \frac{-x}{y-x}=\ln x+C \end{aligned} $

Given $y(1)=0$

$ \begin{aligned} & 1=C \\ & \Rightarrow \frac{-x}{y-x}=\ln x+1 ..........(i)\end{aligned} $

$\therefore $ Put $x=e$

$\begin{aligned} & \frac{-e}{y-e}=1+1 \\ & \Rightarrow-e=2(y-e) \\ & \Rightarrow e=2(e-y) \\ & \Rightarrow \frac{e}{2}=e-y \\ & \Rightarrow y=e-\frac{e}{2} \Rightarrow \frac{e}{2} \Rightarrow y=\frac{e}{2} \\ & \text { Put } x=e^2 \text { in (i) } \\ & \Rightarrow \frac{-e^2}{y-e^2}=2+1\end{aligned}$

$\begin{aligned} & \Rightarrow-e^2=3\left(y-e^2\right) \\ & \Rightarrow-e^2=3 y-3 e^2 \\ & \Rightarrow 2 e^2=3 y \\ & \Rightarrow y=\frac{2}{3} e^2 \\ & \therefore \frac{2(y(e))^2}{y\left(e^2\right)}=2 \frac{\left(\frac{e}{2}\right)^2}{\frac{2}{3} e^2} \\ & \Rightarrow \frac{\frac{1}{2}}{\frac{2}{3}}=\frac{3}{4}=00.75\end{aligned}$