Differential Equations

We have,

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

On integrating both sides, we get

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

As, $y(1)=2$

$ \begin{aligned} 1 & =\frac{1}{2} \log \left|\frac{1+2}{1-2}\right|+C \\\\ \Rightarrow C & =1-\frac{1}{2} \log 3 \end{aligned} $

Now, substitute $x=2$ as $y(2)=\alpha$

$ 2=\frac{1}{2} \log \left|\frac{1+2 \alpha}{1-2 \alpha}\right|+1-\frac{1}{2} \log 3 $

$ \begin{aligned} 1+\frac{1}{2} \log 3 & =\frac{1}{2} \log \left|\frac{1+2 \alpha}{1-2 \alpha}\right| \\\\ 2+\log 3 & =\log \left|\frac{1+2 \alpha}{1-2 \alpha}\right| \end{aligned} $

$ \begin{aligned} &\Rightarrow \frac{1+2 \alpha}{1-2 \alpha} \mid =3 e^2 \\\\ &\Rightarrow \frac{1+2 \alpha}{1-2 \alpha}= \pm \frac{3 e^2}{1} \\\\ & \Rightarrow\frac{1}{2 \alpha} =\frac{ \pm 3 e^2+1}{ \pm 3 e^2-1} \\\\ &\Rightarrow 2 \alpha=\frac{ \pm 3 e^2-1}{ \pm 3 e^2+1} \end{aligned} $

$ \begin{array}{ll} \Rightarrow \alpha=\frac{1}{2}\left(\frac{ \pm 3 e^2-1}{1 \pm 3 e^2}\right) \\\\ \therefore \alpha=\frac{3 e^2-1}{2\left(1+3 e^2\right)} \text { or } \frac{3 e^2+1}{2\left(3 e^2-1\right)} \end{array} $

Given that

$ x^2 f(x)-x=4 \int_0^x t f(t) d t $

On differentiating both sides with respect to $x$, we get

$ \begin{array}{rlrl} & 2 x f(x)+x^2 f^{\prime}(x)-1 =4 x f(x) \\\\ &\Rightarrow x^2 f^{\prime}(x)-2 x f(x)-1 =0 \\\\ &\Rightarrow x^2 \frac{d y}{d x}-2 x y =1 ~~~~~~~(Let, y=f(x) ]\\\\ &\frac{d y}{d x}-\frac{2}{x} y =\frac{1}{x^2} \end{array} $

On comparing above equation with

$\frac{d y}{d x}+P y=Q$, where $P=-\frac{2}{x}, Q=\frac{1}{x^2}$

Now, IF $=e^{\int(-2 / x) d x}=e^{-2 \log x}=1 / x^2$

Solution is $y\left(\frac{1}{x^2}\right)=\int\left(\frac{1}{x^2}\right) \frac{1}{x^2} d x$

$ \begin{array}{ll} &\Rightarrow \frac{y}{x^2}=\int \frac{1}{x^4} d x \Rightarrow \frac{y}{x^2}=-\frac{1}{3 x^3}+C \\\\ &\Rightarrow y =-\frac{1}{3 x}+C x^2 \end{array} $

Given, $f(1)=\frac{2}{3}$.

So, $\frac{2}{3}=-\frac{1}{3}+C \Rightarrow C=1$

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

$ \therefore 18 f(3)=18\left\{9-\frac{1}{9}\right\}=18 \times \frac{80}{9}=160 $

We have, $\left(1+\log _e x\right) \frac{d x}{d y}-x \log _e x=e^{y}, x>0$

Put $x \log _e x=t$

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

Now, IF $=e^{\int-d y}=e^{-y}$

$\begin{aligned} & \therefore \text { General solution, } t\left(e^{-y}\right)=\int\left(e^y \cdot e^{-y}\right) d y+c \\\\ & \Rightarrow t e^{-y}=\int d y+c \\\\ & \Rightarrow t e^{-y}=y+c \\\\ & \Rightarrow \left(x \log _e x\right) e^{-y}=y+c ..........(i)\end{aligned}$

Equation (i), passes through the point $(1,0)$

$ \begin{aligned} & \therefore 0=0+c \\\\ & \Rightarrow c=0 \\\\ & \therefore \left(x \log _e x\right) e^{-y}=y \\\\ & \Rightarrow x \log _e x=y e^y \end{aligned} $

Which passes through $(\alpha, 2)$

$ \begin{aligned} & \therefore \alpha \log _e \alpha=2 e^2 \\\\ & \Rightarrow \log _e \alpha^\alpha=2 e^2 \\\\ & \Rightarrow \alpha^\alpha=e^{2 e^2} \end{aligned} $

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

So, the general solution is

$ \begin{aligned} & \Rightarrow t \times x=\int x \cdot \frac{1}{x^2} d x \\\\ & \Rightarrow y^2 \cdot x=\ln x+c \\\\ & \text { Also,y(2) }=\sqrt{\log _e 2} \\\\ & \log _e^2 2=\log _e 2+c \Rightarrow c=\log _e 2 \\\\ & \Rightarrow y^2 x=\ln x+\ln 2 \\\\ & \Rightarrow y^2 x=\ln 2 x \\\\ & 2 x=\exp \left(x^1 y^2\right) \end{aligned} $

By compare it with given solution we get,

$ \begin{aligned} & \alpha=2,\beta=1, \gamma=2 \\\\ & \Rightarrow \alpha+\beta-\gamma=2+1-2=1 \end{aligned} $

$ \begin{aligned} & \frac{d y}{d x}+\frac{x+a}{y-2}=0 \\\\ & \frac{d y}{d x}=\frac{x+a}{2-y} \\\\ & (2-y) d y=(x+a) d x \\\\ & 2 y \frac{-y}{2}=\frac{x^2}{2}+\mathrm{ax}+\mathrm{c} \\\\ & \mathrm{a}+\mathrm{c}=-\frac{1}{2} \text { as } \mathrm{y}(1)=0 \\\\ & \mathrm{X}^2+\mathrm{y}^2+2 \mathrm{ax}-4 \mathrm{y}-1-2 \mathrm{a}=0 \\\\ & \pi \mathrm{r}^2=4 \pi \\\\ & \mathrm{r}^2=4 \\\\ & 4=\sqrt{a^2+4+1+2 a} \\\\ & (\mathrm{a}+1)^2=0 \end{aligned} $

$ P, Q=(0,2 \pm \sqrt{3}) $

Equation of normal at $\mathrm{P}, \mathrm{Q}$ are $\mathrm{y}-2=\sqrt{3}(\mathrm{x}-1)$

$ \begin{aligned} & \mathrm{y}-2=-\sqrt{3}(\mathrm{x}-1) \\\\ & \mathrm{R}=\left(1-\frac{2}{\sqrt{3}}, 0\right) \\\\ & \mathrm{S}=\left(1+\frac{2}{\sqrt{3}}, 0\right) \\\\ & \mathrm{RS}=\frac{4}{\sqrt{3}}=4 \frac{\sqrt{3}}{3} \end{aligned} $

$\frac{d y}{d x}+y \tan x=x \sec x$

$\therefore $ I.F $=e^{\int \tan x d x}=\sec x$

$\Rightarrow y \sec x=\int x \sec ^{2} x d x$

$\Rightarrow y \sec x=x \tan x-\ln |\sec x|+c$

Given, $ y(0)=1 $

$\Rightarrow 1=c$

$ \therefore $ $ y \sec x=x \tan x-\ln |\sec x|+1$

$\Rightarrow y=x \sin x-\cos x \ln |\sec x|+cosx$

Now, at $x=\pi / 6$, we have

$ \begin{aligned} & y=\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _e \frac{2}{\sqrt{3}}+\frac{\sqrt{3}}{2} \\\\ & =\frac{\pi}{12}-\frac{\sqrt{3}}{2}\left[\log _e \frac{2}{\sqrt{3}}-\log _e e\right]\\\\ &=\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _e\left(\frac{2}{e \sqrt{3}}\right) \end{aligned} $

$\left(3 y^{2}-5 x^{2}\right) y \cdot d x+2 x\left(x^{2}-y^{2}\right) d y=0$

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

Put $\mathrm{y}=\mathrm{mx}$

$ \Rightarrow m+x \cdot \frac{d m}{d x}=\frac{m\left(5-3 m^{2}\right)}{2\left(1-m^{2}\right)} $

$ \begin{aligned} & x \cdot \frac{d m}{d x}=\frac{\left(5-3 m^{2}\right) m-2 m\left(1-m^{2}\right)}{2\left(1-m^{2}\right)} \\\\ & \Rightarrow \frac{\mathrm{dx}}{\mathrm{x}}=\frac{2\left(\mathrm{~m}^{2}-1\right)}{\mathrm{m}\left(\mathrm{m}^{2}-3\right)} \mathrm{dm} \\\\ & \Rightarrow \frac{d x}{x}=\left(\frac{2}{m}-\frac{\frac{4}{3}}{m}+\frac{\frac{4 m}{3}}{\mathrm{~m}^{2}-3}\right) d m \end{aligned} $

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

$\Rightarrow \ln |\mathrm{x}|=\frac{2}{3} \ln |\mathrm{m}|+\frac{2}{3} \ln \left|\mathrm{m}^{2}-3\right|+\mathrm{C}$

Or, $\ln |\mathrm{x}|=\frac{2}{3} \ln \left|\frac{\mathrm{y}}{\mathrm{x}}\right|+\frac{2}{3} \ln \left|\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}-3\right|+\mathrm{C}$

Put $(\mathrm{x}=1, \mathrm{y}=1)$ : we get $\mathrm{c}=-\frac{2}{3} \ln (2)$

$\Rightarrow \ln |\mathrm{x}|=\frac{2}{3} \ln \left|\frac{\mathrm{y}}{\mathrm{x}}\right|+\frac{2}{3} \ln \left|\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}-3\right|-\frac{2}{3} \ln (2)$

$\Rightarrow\left(\frac{\mathrm{y}}{\mathrm{x}}\right)\left[\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}-3\right]=2 .\left(\mathrm{x}^{3 / 2}\right)$ ........(1)

Put $\mathrm{x}=2$ in equation (1), we get

$\Rightarrow \mathrm{y}\left(\mathrm{y}^{2}-12\right)=4 \times 2 \times 2 \times 2 \sqrt{2}$

$\Rightarrow \mathrm{y}^{3}-12 \mathrm{y}=32 \sqrt{2}$

$\Rightarrow\left|\mathrm{y}^{3}(2)-12 \mathrm{y}(2)\right|=32 \sqrt{2}$

Differentiating both sides we get,

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

So, $x=3, y=2$

$\Rightarrow c=\frac{16}{3}$

Now, $x=8$

$8 f(8)=\frac{27}{3}-3+\frac{16}{3}=\frac{34}{3}$

$12 f(8)=\frac{34}{3} \times \frac{12}{8}=17$

$y = vx$

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

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

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

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

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

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

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

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

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

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

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

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

$\therefore$ Solution is

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

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

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

$x = 1$

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

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

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

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

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

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

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

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

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

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

Putting $x = e$ in (i),

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

Given,

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

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

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

Integrating both sides, we get

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

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

Given $y(1) = e$

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

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

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

$ \Rightarrow 0 - 1 = 1 + C$

$ \Rightarrow C = - 2$

$\therefore$ Equation (1) becomes,

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

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

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

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

Now,

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

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

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

$ = 0$

$\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}$

$ \begin{aligned} & \text { I.F. }=e^{\int \alpha d t}=e^{\alpha t} \\\\ & \Rightarrow y \cdot e^{\alpha t}=\gamma \int e^{(\alpha-\beta) t} d t=\gamma \frac{e^{(\alpha-\beta) t}}{(\alpha-\beta)}+C \\\\ & \Rightarrow y=\frac{\gamma}{(\alpha-\beta)} e^{-\beta t}+C e^{-\alpha t} \\\\ & \lim _{x \rightarrow \infty} y(t)=\lim _{x \rightarrow \infty}\left[\frac{\gamma}{(\alpha-\beta)} e^{-\beta t}+C e^{-\alpha t}\right]=0 \end{aligned} $

$ \begin{aligned} & \frac{d y}{d x}-\frac{y}{x}=y^3\left(1+\log _e x\right) \\\\ & \frac{1}{y^3} \frac{d y}{d x}-\frac{1}{x y^2}=1+\log _e x \\\\ & \text { Let }-\frac{1}{y^2}=t \Rightarrow \frac{2}{y^3} \frac{d y}{d x}=\frac{d t}{d x} \\\\ & \therefore \frac{d t}{d x}+\frac{2 t}{x}=2\left(1+\log _e x\right) \\\\ & \text { I.F. }=e^{\int \frac{2}{x} d x}=x^2 \\\\ & \frac{-x^2}{y^2}=\frac{2}{3}\left(\left(1+\log _e x\right) x^3-\frac{x^3}{3}\right)+C \\\\ & y(1)=3 \\\\ & \frac{y^2}{9}=\frac{C}{5-2 x^3\left(2+\log _e x^3\right)} \end{aligned} $

Given,

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

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

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

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

Let ${y^2} = t$

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

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

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

This is linear Differential Equation.

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

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

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

When, $x=1, y=1$

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

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

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

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

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

$x^{3} d y+x y d x-d x=0$

$\Rightarrow \frac{d y}{d x}=\frac{1-x y}{x^{3}}$

$\Rightarrow \frac{d y}{d x}+\frac{y}{x^{2}}=\frac{1}{x^{3}}$

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

$\therefore \quad y e^{-\frac{1}{x}}=\int \frac{e^{-\frac{1}{x}}}{x^{3}} d x$

For RHS put $-\frac{1}{x}=t \Rightarrow \frac{d x}{x^{2}}=d t$

$\therefore y e^{-\frac{1}{x}}=-\int t e^{t} d t$

$\Rightarrow y e^{-\frac{1}{x}}=-\left[t e^{t}-e^{t}\right]+c$

$\Rightarrow y e^{-\frac{1}{x}}=\frac{e^{-\frac{1}{x}}}{x}+e^{-\frac{1}{x}}+c$

$ \downarrow y\left(\frac{1}{2}\right)=3-e $

$\Rightarrow(3-e) e^{-2}=2 e^{-2}+e^{-2}+c$

$\Rightarrow \quad c=-\frac{1}{e}$

For $y(1)$ put $x=1, c=-e^{-1}$ in equation (i) we get $y e^{-1}=e^{-1}+e^{-1}-e^{-1}$

$\Rightarrow y=1$

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

Let $x-1=X, y-1=Y$

$ \frac{d Y}{d X}=\frac{X+Y}{X-Y} $

Let $Y=t X \Rightarrow \frac{d Y}{d X}=t+X \frac{d t}{d X}$

$t+X \frac{d t}{d X}=\frac{1+t}{1-t}$

$X \frac{d t}{d X}=\frac{1+t}{1-t}-t=\frac{1+t^{2}}{1-t}$

$\int \frac{1-t}{1+t^{2}} d t=\int \frac{d X}{X}$

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

Curve passes through $(2,1)$

$0-0=0+c \Rightarrow c=0$

If $(k+1,2)$ also satisfies the curve

$ \begin{aligned} &\tan ^{-1}\left(\frac{1}{k}\right)-\frac{1}{2} \ln \left(\frac{1+k^{2}}{k^{2}}\right)=\ln k \\\\ &2 \tan ^{-1}\left(\frac{1}{k}\right)=\ln \left(1+k^{2}\right) \end{aligned} $

$\frac{d y}{d x}+\left(\frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6}\right) y=\frac{(x+3)}{x+1}, x>-1$,

Integrating factor I.F. $=e^{\int \frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6} d x}$

$ \begin{aligned} & \text { Let } \frac{2 x^{2}+11 x+13}{(x+1)(x+2)(x+3)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3} \\\\ & A=2, B=1, C=-1 \\\\ & \text { I.F. }=e^{(2 \ln |x+1|+\ln |x+2|-\ln |x+3|)} \\\\ & =\frac{(x+1)^{2}(x+2)}{x+3} \end{aligned} $

Solution of differential equation

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

Curve passes through $(0,1)$

$ \begin{gathered} 1 \times \frac{1 \times 2}{3}=0+c \Rightarrow c=\frac{2}{3} \\\\ \text { So, } y(1)=\frac{\frac{1}{3}+\frac{3}{2}+2+\frac{2}{3}}{\frac{\left(2^{2} \times 3\right)}{4}}=\frac{3}{2} \end{gathered} $

D.E. $(1 + {e^{2x}})\left( {{{dy} \over {dx}} + y} \right) = 1$

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

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

$\therefore$ Solution

${e^x}y(x) = \int {{{{e^x}} \over {1 + {e^{2x}}}}dx} $

$ \Rightarrow {e^x}y(x) = {\tan ^{ - 1}}({e^x}) + C$

$\because$ It passes through $\left( {0,{\pi \over 2}} \right),\,C = {\pi \over 2} - {\pi \over 4} = {\pi \over 4}$

$\therefore$ $\mathop {\lim }\limits_{x \to \infty } {e^x}y(x) = \mathop {\lim }\limits_{x \to \infty } {\tan ^{ - 1}}({e^x}) + {\pi \over 4}$

$ = {{3\pi } \over 4}$

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

Integrating factor I.F. $ = {e^{\int {{1 \over {{x^2} - 1}}dx} }} = {e^{{1 \over 2}\ln \left| {{{x - 1} \over {x + 1}}} \right|}}$

$ = \sqrt {{{x - 1} \over {x + 1}}} $

Solution of differential equation

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

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

Curve passes through $\left( {2,\sqrt {{1 \over 3}} } \right)$

${1 \over {\sqrt 3 }} \times {1 \over {\sqrt 3 }} = 2 - 2\ln 3 + C$

$C = 2\ln 3 - {5 \over 3}$

$y(8) \times {{\sqrt 7 } \over 3} = 8 - 2\ln 9 + 2\ln 3 - {5 \over 3}$

$\sqrt 7 \,.\,y(8) = 19 - 6\ln 3$

Family of circles passing through the points (0, 2) and (0, $-$2)

${x^2} + (y - 2)(y + 2) + \lambda x = 0,\,\lambda \in R$

${x^2} + {y^2} + \lambda x - 4 = 0$ ...... (1)

Differentiate w.r.t x

$2x + 2y{{dy} \over {dx}} + \lambda = 0$ ....... (2)

Using (1) and (2), eliminate $\lambda$

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

$2xy{{dy} \over {dx}} + {x^2} - {y^2} + 4 = 0$

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

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

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

Let ${y \over x} = v$

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

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

OR $\ln \left( {v + \sqrt {1 + {v^2}} } \right) = \ln x + C$

at $x = 1,\,y = 0$

$ \Rightarrow C = 0$

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

At $x = 2$,

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

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

OR $y = {3 \over 2}$

$({\sin ^2}2x){{dy} \over {dx}} + (8{\sin ^2}2x + 2\sin 4x)y$

$ = 2{e^{ - 4x}}(2\sin 2x + \cos 2x)$

${{dy} \over {dx}} + (8 + 4\cot 2x)y = 2{e^{ - 4x}}\left( {{{2\sin 2x + \cos 2x} \over {{{\sin }^2}2x}}} \right)$

Integrating factor

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

$ = {e^{8x + 2\ln \sin 2x}}$

Solution of differential equation

$y.\,{e^{8x + 2\ln \sin 2x}}$

$ = \int {2{e^{(4x + 2\ln \sin 2x)}}{{(2\sin 2x + \cos 2x)} \over {{{\sin }^2}2x}}dx} $

$ = 2\int {{e^{4x}}(2\sin 2x + \cos 2x)dx} $

$y.\,{e^{8x + 2\ln \sin 2x}} = {e^{4x}}\sin 2x + c$

$y\left( {{\pi \over 4}} \right) = {e^{ - \pi }}$

${e^{ - \pi }}\,.\,{e^{2\pi }} = {e^\pi } + c \Rightarrow c = 0$

$y\left( {{\pi \over 6}} \right) = {{{e^{{{2\pi } \over 3}}}{{\sqrt 3 } \over 2}} \over {{e^{\left( {{{4\pi } \over 3} + 2\ln {{\sqrt 3 } \over 2}} \right)}}}}$

$ = {e^{{{ - 2\pi } \over 3}}}\,.\,{2 \over {\sqrt 3 }}$

${{dy} \over {dx}} = x + y$

Let $x + y = t$

$1 + {{dy} \over {dx}} = {{dt} \over {dx}}$

${{dt} \over {dx}} - 1 = t \Rightarrow \int {{{dt} \over {t + 1}} = \int {dx} } $

$\ln |t + 1| = x + C'$

$|t + 1| = C{e^x}$

$|x + y + 1| = C{e^x}$

For ${y_1}(x),\,{y_1}(0) = 0 \Rightarrow C = 1$

For ${y_2}(x),\,{y_2}(0) = 1 \Rightarrow C = 2$

${y_1}(x)$ is given by $|x + y + 1| = {e^x}$

${y_2}(x)$ is given by $|x + y + 1| = 2{e^x}$

At point of intersection

${e^x} = 2{e^x}$

No solution

So, there is no point of intersection of ${y_1}(x)$ and ${y_2}(x)$.

${{dy} \over {dx}} + {{xy} \over {{x^2} - 1}} = {{{x^4} + 2x} \over {\sqrt {1 - {x^2}} }}$

which is first order linear differential equation.

Integrating factor $(I.F.) = {e^{\int {{x \over {{x^2} - 1}}dx} }}$

$ = {e^{{1 \over 2}\ln |{x^2} - 1|}} = \sqrt {|{x^2} - 1|} $

$ = \sqrt {1 - {x^2}} $

$\because$ $x \in ( - 1,1)$

Solution of differential equation

$y\sqrt {1 - {x^2}} = \int {({x^4} + 2x)dx = {{{x^5}} \over 5} + {x^2} + c} $

Curve is passing through origin, $c = 0$

$y = {{{x^5} + 5{x^2}} \over {5\sqrt {1 - {x^2}} }}$

$\int\limits_{{{ - \sqrt 3 } \over 2}}^{{{\sqrt 3 } \over 2}} {{{{x^5} + 5{x^2}} \over {5\sqrt {1 - {x^2}} }}dx = 0 + 2\int\limits_0^{{{\sqrt 3 } \over 2}} {{{{x^2}} \over {\sqrt {1 - {x^2}} }}dx} } $

put $x = \sin \theta $

$dx = \cos \theta \,d\theta $

$I = 2\int\limits_0^{{\pi \over 3}} {{{{{\sin }^2}\theta \,.\,\cos \theta d\theta } \over {\cos \theta }}} $

$ = \int\limits_0^{{\pi \over 3}} {(1 - \cos 2\theta )d\theta } $

$ = \left. {\left( {\theta - {{\sin 2\theta } \over 2}} \right)} \right|_0^{{\pi \over 3}}$

$ = {\pi \over 3} - {{\sqrt 3 } \over 4}$

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

which is a first order linear differential equation.

Integrating factor (I. F.) $ = {e^{\int {2\tan x\,dx} }}$

$ = {e^{2\ln |\sec x|}} = {\sec ^2}x$

Solution of differential equation can be written as

$y\,.\,{\sec ^2}x = \int {\sin x\,.\,{{\sec }^2}x\,dx = \int {\sec \,x\,.\,\tan x\,dx} } $

$y~{\sec ^2}x = \sec x + C$

$y\left( {{\pi \over 3}} \right) = 0,0 = \sec {\pi \over 3} + C \Rightarrow \,\,\,\,C = - 2$

$y = {{\sec x - 2} \over {{{\sec }^2}x}} = \cos x - 2{\cos ^2}x$

$ = {1 \over 8} - 2{\left( {\cos x - {1 \over 4}} \right)^2}$

${y_{\max }} = {1 \over 8}$

${{dy} \over {dx}} \propto {{ - y} \over x}$

${{dy} \over {dx}} = {{ - ky} \over x} \Rightarrow \int {{{dy} \over y} = - K\int {{{dx} \over x}} } $

$\ln |y| = - K\ln |x| + C$

If the above equation satisfy (1, 2) and (8, 1)

$\ln 2 = - K \times 0 + C \Rightarrow C = \ln 2$

$\ln 1 = - K\ln 8 + \ln 2 \Rightarrow K = {1 \over 3}$

So, at $x = {1 \over 8}$

$\ln |y| = - {1 \over 3}\ln \left( {{1 \over 8}} \right) + \ln 2 = 2\ln 2$

$|y| = 4$

$\frac{d y}{d x}=\frac{2 e^{2 x}-6 e^{-x}+9}{2+9 e^{-2 x}}=e^{2 x}-\frac{6 e^{-x}}{2+9 e^{-2 x}}$

$ \begin{aligned} &\int d y=\int e^{2 x} d x-3 \int \underbrace{1+\left(\frac{3 e^{-x}}{\sqrt{2}}\right)^{2}}_{\text {put } e^{-x}=t} d x \\\\ &=\frac{e^{2 x}}{2}+3 \int \frac{d t}{1+\left(\frac{3 t}{\sqrt{2}}\right)^{2}} \\\\ &=\frac{e^{2 x}}{2}+\sqrt{2} \tan ^{-1} \frac{3 t}{\sqrt{2}}+C \end{aligned} $

$y=\frac{e^{2 x}}{2}+\sqrt{2} \tan ^{-1}\left(\frac{3 e^{-x}}{\sqrt{2}}\right)+C$

It is given that the curve passes through

$ \left(0, \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}\right) $

$ \begin{aligned} & \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}=\frac{1}{2}+\sqrt{2} \tan ^{-1}\left(\frac{3}{\sqrt{2}}\right)+C \end{aligned} $

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

Now if $\left(\alpha, \frac{1}{2} e^{2 \alpha}\right)$ satisfies the curve, then

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

$\tan ^{-1}\left(\frac{3}{\sqrt{2}}\right)-\tan ^{-1}\left(\frac{3 e^{-\alpha}}{\sqrt{2}}\right)=\frac{\pi}{2 \sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{\pi}{4}$

$ \frac{\frac{3}{\sqrt{2}}-\frac{3 e^{-\alpha}}{\sqrt{2}}}{1+\frac{9}{2} e^{-\alpha}}=1 $

$ \frac{3}{\sqrt{2}} e^{\alpha}-\frac{3}{\sqrt{2}}=e^{\alpha}+\frac{9}{2} $

$ e^{\alpha}=\frac{\frac{9}{2}+\frac{3}{\sqrt{2}}}{\frac{3}{\sqrt{2}}-1}=\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right) $

$\left(x-y^{2}\right) d x+y\left(5 x+y^{2}\right) d y=0$

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

Let $y^{2}=t$ $ \frac{1}{2} \cdot \frac{d t}{d x}=\frac{t-x}{5 x+t} $

Now substitute, $t=v x$

$ \begin{aligned} & \frac{d t}{d x}=v+x \frac{d v}{d x} \\\\ & \frac{1}{2}\left\{v+x \frac{d v}{d x}\right\}=\frac{v-1}{5+v} \\\\ & x \frac{d v}{d x}=\frac{2 v-2}{5+v}-v=\frac{-3 v-v^{2}-2}{5+v} \\\\ & \int \frac{5+v}{v^{2}+3 v+2} d v=\int-\frac{d x}{x} \\\\ & \int \frac{4}{v+1} d v-\int \frac{3}{v+2} d v=-\int \frac{d x}{x} \\\\ & 4 \ln |v+1|-3 \ln |v+2|=-\ln x+\ln C \\\\ & \left|\frac{(v+1)^{4}}{(v+2)^{3}}\right|=\frac{c}{x} \\\\ & \left| \frac{\left(\frac{y^{2}}{x}+1\right)^{4}}{\left(\frac{y^{2}}{x}+2\right)^{3}}\right|=\frac{c}{x} \\\\ & \left|\left(y^{2}+x\right)^{4}\right|=C\left|\left(y^{2}+2 x\right)^{3}\right| \end{aligned} $

Given,

${{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}}$

$ \Rightarrow bxdy + cydy + ady = axdx - bydx + adx$

$ \Rightarrow bxdy + bydx + (cy + a)dy = (ax + a)dx$

$ \Rightarrow b(xdy + ydx) + (cy + a)dy = (ax + a)dx$

$ \Rightarrow bd(xy) + (cy + a)dy = (ax + a)dx$

Integrating both sides, we get

$ \Rightarrow b\int {d(xy) + \int {(cy + a)dy = \int {(ax + a)dx} } } $

$ \Rightarrow b\,.\,xy + c\,.\,{{{y^2}} \over 2} + ay = {{a{x^2}} \over 2} + ax + k$

$ \Rightarrow {{a{x^2}} \over 2} - {{c{y^2}} \over 2} - bxy + ax - ay + k = 0$

For equation of circle,

Coefficient of x² = Coefficient of y²

$\therefore$ ${a \over 2} = - {c \over 2}$

$ \Rightarrow a = - c$

And coefficient of $xy = 0$

$\therefore$ $ - b = 0$

$ \Rightarrow b = 0$

$\therefore$ Circle equation becomes,

${{a{x^2}} \over 2} + {{a{y^2}} \over 2} + ax - ay + k = 0$

$ \Rightarrow {x^2} + {y^2} + 2x - 2y + {{2k} \over a} = 0$

$\therefore$ Center $ = ( - g, - f) = ( - 1,1) = (\alpha ,\beta )$

$\therefore$ $\alpha = - 1$ and $\beta = 1$

$\therefore$ $\alpha + 2\beta = - 1 + 2 \times (1) = 1$

Given,

$(1 + {e^{2x}}){{dy} \over {dx}} + 2(1 + {y^2}){e^x} = 0$

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

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

$ \Rightarrow {\tan ^{ - 1}}(y) = \int {{{ - 2{e^x}dx} \over {(1 + {e^{2x}})}}} $

Now,

Let ${e^x} = t$

$ \Rightarrow {e^x}dx = dt$

$ \Rightarrow {\tan ^{ - 1}}(y) = \int {{{ - 2dt} \over {1 + {t^2}}}} $

$ \Rightarrow {\tan ^{ - 1}}(y) = - 2{\tan ^{ - 1}}(t) + C$

$ \Rightarrow {\tan ^{ - 1}}(y) = - 2{\tan ^{ - 1}}({e^x}) + C$ [Putting value of t]

Given, y(0) = 0 means when x = 0 the y = 0

$\therefore$ ${\tan ^{ - 1}}(0) = - 2{\tan ^{ - 1}}({e^o}) + C$

$ \Rightarrow 0 = - 2 \times {\pi \over 4} + C$

$ \Rightarrow C = {\pi \over 2}$

$\therefore$ ${\tan ^{ - 1}}(y) = 2{\tan ^{ - 1}}({e^x}) + {\pi \over 2}$

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

Differentiating both sides, we get

$y' = {\sec ^2}\left( { - 2{{\tan }^{ - 1}}({e^x}) + {\pi \over 2}} \right) - 2\,.\,{{{e^x}} \over {1 + {e^{2x}}}}$

$ = {\sec ^2}\left( { - 2{{\tan }^{ - 1}}({e^x}) + {\pi \over 2}} \right) \times {{ - 2{e^x}} \over {1 + {e^{2x}}}}$

$\therefore$ $y'(0) = {\sec ^2}\left( { - 2{{\tan }^{ - 1}}({e^o}) + {\pi \over 2}} \right) \times {{ - 2{e^o}} \over {1 + {e^o}}}$

$ = {\sec ^2}\left( { - 2 \times {\pi \over 4} + {\pi \over 2}} \right) \times {{ - 2} \over 2}$

$ = {\sec ^2}(0)x - 1$

$ = 1 \times 1$

$ = - 1$

And

$y\left( {\log _e^{\sqrt 3 }} \right) = \tan \left( { - 2{{\tan }^{ - 1}}\left( {{e^{\log _e^{\sqrt 3 }}}} \right) + {\pi \over 2}} \right)$

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

$ = \tan \left( { - 2 \times {\pi \over 3} + {\pi \over 2}} \right)$

$ = \tan \left( { - {\pi \over 6}} \right)$

$ = - \tan \left( {{\pi \over 6}} \right)$

$ = - {1 \over {\sqrt 3 }}$

$\therefore$ $6\left( {y'(0) + {{\left( {y(\log _e^{\sqrt 3 })} \right)}^2}} \right)$

$ = 6\left( { - 1 + {{\left( {{{ - 1} \over {\sqrt 3 }}} \right)}^2}} \right)$

$ = 6\left( { - 1 + {1 \over 3}} \right) = 6 \times {{ - 2} \over 3} = - 4$

Given,

$x{{dy} \over {dx}} - y = \sqrt {{y^2} + 16x} $

$ \Rightarrow x{{dy} \over {dx}} = y + \sqrt {{y^2} + 16x} $

$ \Rightarrow {{dy} \over {dx}} = {y \over x} + \sqrt {{{\left( {{y \over x}} \right)}^2} + 16} $

This is a homogenous different equation.

Let ${y \over x} = v$

$ \Rightarrow y = vx$

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

$\therefore$ $v + x{{dv} \over {dx}} =v+ \sqrt {{v^2} + 16} $

$ \Rightarrow $ $ x{{dv} \over {dx}} = \sqrt {{v^2} + 16} $

$ \Rightarrow {{dv} \over {\sqrt {{v^2} + 16} }} = {{dx} \over x}$

Integrating both sides, we get

$\int {{{dv} \over {\sqrt {{v^2} + 16} }} = \int {{{dx} \over x}} } $

$ \Rightarrow \ln \left| {v + \sqrt {{v^2} + 16} } \right| = \ln x + \ln c$

$ \Rightarrow v + \sqrt {{v^2} + 16} = cx$

Now putting, $v = {y \over x}$, we get

${y \over x} + \sqrt {{{{y^2}} \over {{x^2}}} + 16} = cx$

$ \Rightarrow {y \over x} + \sqrt {{{{y^2} + 16{x^2}} \over {{x^2}}}} = cx$

$ \Rightarrow y + \sqrt {{y^2} + 16{x^2}} = c{x^2}$ ...... (1)

Given, $y(1) = 3$

$\therefore$ When x = 1 then y = 3.

Putting in equation (1) we get,

$3 + \sqrt {9 + 16} = c.\,1$

$ \Rightarrow c = 8$

$\therefore$ Solution of equation,

$y + \sqrt {{y^2} + 16{x^2}} = 8{x^2}$

Now, y(2) means when x = 2 then y = ?

$\therefore$ $y + \sqrt {{y^2} + 16 \times 4} = 8 \times 4$

$ \Rightarrow y = 15$

Given differential equation

$2y{e^{{x \over {{y^2}}}}}dx + \left( {{y^2} - 4x{e^{{x \over {{y^2}}}}}} \right)dy = 0,\,x(1) = 0$

$ \Rightarrow {e^{{x \over {{y^2}}}}}[2ydx - 4xdy] = - {y^2}dy$

$ \Rightarrow {e^{{x \over {{y^2}}}}}\left[ {{{2{y^2}dx - 4xydy} \over {{y^4}}}} \right] = {{ - 1} \over y}dy$

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

$ \Rightarrow 2{e^{{x \over {{y^2}}}}} = - \ln y + c$ ...... (i)

Now, using x(1) = 0, c = 2

So, for x(e), Put y = e in (i)

$2{e^{{x \over {{e^2}}}}} = - 1 + 2$

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

${{dy} \over {dx}} = 2\tan x(\cos x - y)$

$ \Rightarrow {{dy} \over {dx}} + 2\tan xy = 2\sin x$

$I.F. = {e^{\int {2\tan xdx} }} = {\sec ^2}x$

$\therefore$ Solution of D.E. will be

$y(x){\sec ^2}x = \int {2\sin x{{\sec }^2}xdx} $

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

$\because$ Curve passes through $\left( {{\pi \over 4},0} \right)$

$\therefore$ $c = - 2\sqrt 2 $

$\therefore$ $y = 2\cos x - 2\sqrt 2 {\cos ^2}x$

$\therefore$ $\int_0^{\pi /2} {ydx = \int_0^{\pi /2} {(2\cos x - 2\sqrt 2 {{\cos }^2}x)\,dx} } $

$ = 2 - 2\sqrt 2 \,.\,{\pi \over 4} = 2 - {\pi \over {\sqrt 2 }}$

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

Putting y = tx

$\left( {{1 \over {\sqrt {1 - {t^2}} }} + {e^t}} \right)\left( {t + x{{dt} \over {dx}}} \right) = 1 + \left( {{1 \over {\sqrt {1 - {t^2}} }} + {e^t}} \right)t$

$ \Rightarrow x\left( {{1 \over {\sqrt {1 - {t^2}} }} + {e^t}} \right){{dt} \over {dx}} = 1$

$ \Rightarrow {\sin ^{ - 1}}t + {e^t} = \ln x + C$

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

at x = 1, y = 0

So, $0 + {e^0} = 0 + C \Rightarrow C = 1$

at $(2\alpha ,\alpha )$

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

$ \Rightarrow {\pi \over 6} + {e^{{1 \over 2}}} - 1 = \ln (2\alpha )$

$ \Rightarrow \alpha = {1 \over 2}{e^{\left( {{\pi \over 6} + {e^{{1 \over 2}}} - 1} \right)}}$

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

$IF = {e^{\int {{{3{x^2} - 1} \over {x - {x^3}}}dx} }} = {e^{ - \ln |{x^3} - x|}} = {e^{ - \ln ({x^3} - x)}} = {1 \over {{x^3} - x}}$

Solution of D.E. can be given by

$y.\,{1 \over {{x^3} - x}} = \int {{{4{x^3}} \over {x(1 - {x^2})}}.\,{1 \over {x({x^2} - 1)}}dx} $

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

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

at x = 2, y = $-$2

${{ - 2} \over 6} = {2 \over 3} + c \Rightarrow c = - 1$

at $x = 3 \Rightarrow {y \over {24}} = {2 \over 8} - 1 \Rightarrow y = - 18$

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

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

$I.F. = {e^{\int {{1 \over {1 + {y^2}}}dy} }} = {e^{{{\tan }^{ - 1}}y}}$

$\therefore$ Solution

$x.\,{e^{{{\tan }^{ - 1}}y}} = \int {{{{e^{{{\tan }^{ - 1}}y}}{{\tan }^{ - 1}}y} \over {1 + {y^2}}}dy} $

Let ${e^{{{\tan }^{ - 1}}y}} = t$

${{{e^{{{\tan }^{ - 1}}y}}} \over {1 + {y^2}}} = dt$

$ = x{e^{{{\tan }^{ - 1}}y}} = \int {\ln tdt = t\ln t - t + c} $

$\therefore$ $ = x{e^{{{\tan }^{ - 1}}y}} = {e^{{{\tan }^{ - 1}}y}}{\tan ^{ - 1}}y - {e^{{{\tan }^{ - 1}}y}} + c$ ..... (i)

$\because$ It passes through (1, 0) $\Rightarrow$ c = 2

Now put y = tan1, then

$ex = e - e + 2$

$ \Rightarrow x = {2 \over e}$

${{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}}$

$ = bx\,dy + cy\,dy + a\,dy = ax\,dx - by\,dx + a\,dx$

$ = cy\,dy + a\,dy - ax\,dx - a\,dx + b(x\,dy + y\,dx) = 0$

$ = c\int {y\,dy + a\int {x\,dx - a\int {dx + b\int {d(xy) = 0} } } } $

$ = {{c{y^2}} \over 2} + ay - {{a{x^2}} \over 2} - ax + bxy = k$

$ = a{x^2} - c{y^2} + 2ax - 2ay - 2bxy = k$

Above equation is circle

$\Rightarrow$ a = $-$ c and b = 0

$a{x^2} + a{y^2} + 2ax - 2ay = k$

$ \Rightarrow {x^2} + {y^2} + 2x - 2y = \lambda \,\,\,\,\,\,\,\left[ {\lambda = {k \over a}} \right]$

Passes through (2, 5)

$4 + 25 + 4 - 10 = \lambda \Rightarrow \lambda = 23$

Circle $ \equiv {x^2} + {y^2} + 2x - 2y - 23 = 0$

Centre ($-$1, 1) $r = \sqrt {{{( - 1)}^2} + {1^2} + 23} = 5$

Shortest distance of $(11,6) = \sqrt {{{12}^2} + {5^2}} - 5$

$ = 13 - 5$

$ = 8$

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

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

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

$ = {{{{\log }_e}({2^y} - 1)} \over {{{\log }_e}2}} = - {{{{\log }_e}({2^x} - 1)} \over {{{\log }_e}2}} + {{{{\log }_e}c} \over {{{\log }_e}2}}$

$ = |({2^y} - 1)({2^x} - 1)| = c$

$\because$ $y(1) = 1$

$\therefore$ $c = 1$

$ = |({2^y} - 1)({2^x} - 1)| = 1$

For $x = 2$

$|({2^y} - 1)3| = 1$

${2^y} - 1 = {1 \over 3} \Rightarrow 2y = {4 \over 3}$

Taking log to base 2.

$\therefore$ $y = 2 - {\log _2}3$

$x{{dy} \over {dx}} + 2y = x{e^x},\,\,y(1) = 0$

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

$y\,.\,{x^2} = \int {{x^2}{e^x}dx} $

$y{x^2} = {x^2}{e^x} - \int {2x{e^x}dx} $

$ = {x^2}{e^x} - 2(x{e^x} - {e^x}) + c$

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

$y{x^2} = ({x^2} - 2x + 2){e^x} + c$

$0 = e + c \Rightarrow c = - e$

$y(x)\,.\,{x^2} - {e^x} = {(x - 1)^2}{e^x} - e$

$z(x) = {(x - 1)^2}{e^x} - e$

For local maximum $z'(x) = 0$

$\therefore$ $2(x - 1){e^x} + {(x - 1)^2}{e^x} = 0$

$\therefore$ $x = - 1$

And local maximum value $ = z( - 1)$

$ = {4 \over e} - e$

$\because$ ${{dy} \over {dx}} + {e^x}({x^2} - 2)y = ({x^2} - 2x)({x^2} - 2){e^{2x}}$

Here, $I.F. = {e^{\int {{e^x}({x^2} - 2)dx} }}$

$ = {e^{({x^2} - 2x){e^x}}}$

$\therefore$ Solution of the differential equation is

$y\,.\,{e^{({x^2} - 2x){e^x}}} = \int {({x^2} - 2x)({x^2} - 2){e^{2x}}\,.\,{e^{({x^2} - 2x){e^x}}}dx} $

$ = \int {({x^2} - 2x){e^x}\,.\,({x^2} - 2){e^x}\,.\,{e^{({x^2} - 2x){e^x}}}dx} $

Let $({x^2} - 2x){e^x} = t$

$\therefore$ $({x^2} - 2){e^x}dx = dt$

$y\,.\,{e^{({x^2} - 2x){e^x}}} = \int {t\,.\,{e^t}dt} $

$y\,.\,{e^{({x^2} - 2x){e^x}}} = ({x^2} - 2x - 1){e^{({x^2} - 2x){e^x}}} + c$

$\therefore$ $y(0) = 0$

$\therefore$ $c = 1$

$\therefore$ $y = ({x^2} - 2x - 1) + {e^{(2x - {x^2}){e^x}}}$

$\therefore$ $y(2) = - 1 + 1 = 0$

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

$ \Rightarrow 2x(xdy - ydx) + 3{y^3}dx = 0$

$ \Rightarrow 2\left( {{{xdy - ydx} \over {{y^2}}}} \right) + 3{{dx} \over x} = 0$

$ \Rightarrow - {{2x} \over y} + 3\ln x = C$

$\because$ $y(e) = {e \over 3} \Rightarrow - 6 + 3 = C \Rightarrow C = - 3$

Now, at $x = 1$, $ - {2 \over y} + 0 = - 3$

$y = {2 \over 3}$

$ \int\left(\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g(x)\left(e^x+1-x e^x\right)}{\left(e^x+1\right)^2}\right) d x=\frac{x g(x)}{e^x+1}+c $

On differentiating both sides w.r.t. $\mathrm{x}$, we get

$ \begin{aligned} &\left(\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g(x)\left(e^x+1-x e^x\right.}{\left(e^x+1\right)^2}\right) \\\\ &=\frac{\left(e^x+1\right)\left(g(x)+x g^{\prime}(x)\right)-e^x \cdot x \cdot g(x)}{\left(e^x+1\right)^2} \\\\ &\left(e^x+1\right) x(\cos x-\sin x)+g(x)\left(e^x+1-x e^x\right) \\\\ &=\left(e^x+1\right)\left(g(x)+x g^{\prime}(x)\right)-e^x \cdot x \cdot g(x) \\\\ &\Rightarrow g g^{\prime}(x)=\cos x-\sin x \\\\ &\Rightarrow g(x)=\sin x+\cos x+C \end{aligned} $

$\mathrm{g}(\mathrm{x})$ is increasing in $(0, \pi / 4)$

$g "(x)=-\sin x-\cos x<0$

$\Rightarrow \mathrm{g}^{\prime}(\mathrm{x})$ is decreasing function

let $h(x)=g(x)+g^{\prime}(x)=2 \cos x+C$

$ \Rightarrow \mathrm{h}^{\prime}(\mathrm{x})=\mathrm{g}^{\prime}(\mathrm{x})+\mathrm{g}^{\prime \prime}(\mathrm{x})=-2 \sin \mathrm{x}<0 $

$\Rightarrow \mathrm{h}$ is decreasing

let $\phi(\mathrm{x})=\mathrm{g}(\mathrm{x})-\mathrm{g}^{\prime}(\mathrm{x})=2 \sin \mathrm{x}+\mathrm{C}$

$ \Rightarrow \phi^{\prime}(\mathrm{x})=\mathrm{g}^{\prime}(\mathrm{x})-\mathrm{g}{ }^{\prime \prime}(\mathrm{x})=2 \cos \mathrm{x}>0 $

$\Rightarrow \phi$ is increasing

Hence option D is correct.

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

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

If ${e^{ - \int {{1 \over {x + 1}}x} }} = {e^{ - \log (x + 1)}} = {1 \over {x + 1}}$

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

${y \over {x + 1}} = \int {{e^{3x}}dx} $

${y \over {x + 1}} = {{{e^{3x}}} \over 3} + c$

$\because$ $y(0) = {1 \over 3}$

${1 \over 3} = {1 \over 3} + c$

$\therefore$ $c = 0$

So : $y = {{{e^{3x}}} \over 3}(x + 1)$

$y' = {e^{3x}}(x + 1) + {{{e^{3x}}} \over 3} = {e^{3x}}\left( {x + {4 \over 3}} \right)$

$y'' = 3{e^{3x}}\left( {x + {4 \over 3}} \right) + {e^{3x}} = {e^{3x}}(3x + 5)$

$y' = 0$ at $x = {{ - 4} \over 3}$ & $y'' = {e^{ - 4}}(1) > 0$ at $x = {{ - 4} \over 3}$

$ \Rightarrow x = {{ - 4} \over 3}$ is point of local minima

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

Put $y = vx$ we get

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

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

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

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

As it passes through (1, 1)

$c = {\pi \over 4}$

$ \Rightarrow {\tan ^{ - 1}}\left( {{y \over x}} \right)\ln \left( {{y \over x}} \right) = \ln x + {\pi \over 4}$

Put $y = \sqrt 3 x$ we get

$ \Rightarrow {\pi \over 3} - \ln \sqrt 3 = \ln x + {\pi \over 4}$

$ \Rightarrow \ln x = {\pi \over {12}} - \ln \sqrt 3 = \ln \alpha $

$\therefore$ $\ln \left( {\sqrt 3 \alpha } \right) = \ln \sqrt 3 + \ln \alpha $

$ = \ln \sqrt 3 + {\pi \over {12}} - \ln \sqrt 3 = {\pi \over {12}}$

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

$ \text { If }=e^{\int-\frac{2}{y} d y}=e^{-2 \ln y}=\frac{1}{y^{2}} $

Solution is given by

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

$\Rightarrow x=y^{2}\left(y e^{y}+c\right)$ at, $y=1, x=0$

$\Rightarrow 0=1\left(1 \cdot e^{1}+c\right) \Rightarrow c=-e$ at $y=e$,

$x=e^{2}\left(e . e^{e}-e\right)$

${x^2}dy + \left( {y - {1 \over x}} \right)dx = 0$ ; x > 0, y(1) = 1

${x^2}dy + {{(xy - 1)} \over x}dx = 0$

${x^2}dy = {{(xy - 1)} \over x}dx$

${{dy} \over {dx}} = {{1 - xy} \over {{x^3}}}$

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

${{dy} \over {dx}} = {1 \over {{x^2}}}.y = {1 \over {{x^3}}}$

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

$y{e^{ - {1 \over x}}} = \int {{1 \over {{x^3}}}.{e^{ - {1 \over x}}}} $

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

$1.\,{e^{ - 1}} = {e^{ - 1}}(2) + C$

$C = - {e^{ - 1}} = - {1 \over e}$

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

$y\left( {{1 \over 2}} \right) = 3 - {1 \over e} \times {e^2}$

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

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

$ \Rightarrow \int {{{(1 + {2^y})\ln 2} \over {(y + {2^y})}}dy = \int {dx} } $

$ \Rightarrow \ln \left| {y + {2^y}} \right| = x + c$

x = 0; y = 0 $\Rightarrow$ c = 0

$ \Rightarrow x = \ln \left| {y + {2^y}} \right|$

$\Rightarrow$ at y = 1, x = ln3

$\because$ $3 \in (e,{e^2}) \Rightarrow x \in (1,2)$

Let, $y = tx$

${{dy} \over {dx}} = t + x{{dt} \over {dx}}$

$\therefore$ $tx\left( {t + x{{dt} \over {dx}}} \right) = x\left( {{t^2} + {{\varphi ({t^2})} \over {\varphi '({t^2})}}} \right)$

${t^2} + xt{{dt} \over {dx}} = {t^2} + {{\varphi ({t^2})} \over {\varphi '({t^2})}}$

$\int {{{t\varphi '({t^2})} \over {\varphi ({t^2})}}} dt = \int {{{dx} \over x}} $

Let $\varphi ({t^2}) = p$

$\therefore$ $\varphi '({t^2})2tdt = dp$

$ \Rightarrow \int {{{dy} \over {2p}}} = \int {{{dx} \over x}} $

${1 \over 2}\ln \varphi ({t^2}) = \ln x + \ln c$

$\varphi ({t^2}) = {x^2}k$

$\varphi \left( {{{{y^2}} \over {{x^2}}}} \right) = k{x^2},\varphi (1) = k$

$\varphi \left( {{{{y^2}} \over 4}} \right) = 4\varphi (1)$

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

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

$\int {{{{2^y}} \over {{2^y} - 1}}dy = \int {{2^x}\,dx} } $

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

$ \Rightarrow {\log _2}({2^y} - 1) = {2^x}{\log _2}e + C$

$\because$ $y(0) = 1 \Rightarrow 0 = {\log _2}e + C$

$C = - {\log _2}e$

$ \Rightarrow {\log _2}({2^y} - 1) = ({2^x} - 1){\log _2}e$

put x = 1, ${\log _2}({2^y} - 1) = {\log _2}e$

2^y = e + 1

y = log₂(e + 1) Ans.