Differential Equations

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

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 $y=e$,

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

Given, $ \frac{d y}{d x}+12 y=\cos \left(\frac{\pi}{12} x\right) $

This is a linear differential equation.

Where P = 12 and Q = $\cos \left(\frac{\pi}{12} x\right)$

$ \text { I.F. }=e^{\int P d x}=e^{\int 12 d x}=e^{12 x} $

Solution of differencial equation is

$y \times I.F = \int {Q \times \left( {I.F} \right)} dx$

$y \times \text { I.F. }=\int e^{12 x} \cdot \cos \left(\frac{\pi}{12} x\right) d x $

$ \Rightarrow y e^{12 x}=\frac{e^{12 x}}{12^2+\left(\frac{\pi}{12}\right)^2}\left(12 \cos \frac{\pi}{12} x+\frac{\pi}{12} \sin \frac{\pi}{12} x\right)+C $

[As $\int {{e^{ax}}.\cos \left( {bx} \right)dx} = {{{e^{ax}}} \over {{a^2} + {b^2}}}\left\{ {a\cos \left( {bx} \right) + b\sin \left( {bx} \right)} \right\} + C$]

$ \Rightarrow y=\frac{12}{(12)^4+\pi^2}\left\{(12)^2 \cos \left(\frac{\pi x}{12}\right)+\pi \sin \left(\frac{\pi x}{12}\right)\right\} +\frac{C}{e^{12 x}} $

$ \begin{aligned} & \text { Given }, y(0)=0 \\\\ & \Rightarrow 0=\frac{12 \cdot\left(12^2\right)}{12^4+\pi^2}+C \\\\ & \Rightarrow C=\frac{-12^3}{12^4+\pi^2} \\\\ & \therefore y=\frac{12}{12^4+\pi^2}\left[12^2 \cos \left(\frac{\pi x}{12}\right)+\pi \sin \left(\frac{\pi x}{12}\right)-12^2 \cdot e^{-12 x}\right] \end{aligned} $

$y(x)$ is not a periodic function as $e^{-12 x}$ is a non-periodic function.

$ \therefore $ Option (D) is a wrong statement.

$ \begin{aligned} & \text { Now } \frac{d y}{d x}=\frac{12}{12^4+\pi^2}[{-12 \pi \sin \left(\frac{\pi x}{12}\right)+\frac{\pi^2}{12} \cos \left(\frac{\pi x}{12}\right)}+12^3 \mathrm{e}^{-12 x}] \end{aligned} $

Values of ${-12 \pi \sin \left(\frac{\pi x}{12}\right)+\frac{\pi^2}{12} \cos \left(\frac{\pi x}{12}\right)}$ varies between $\left(-\sqrt{144 \pi^2+\frac{\pi^4}{144}},-12 \pi \sqrt{1+\frac{\pi^2}{12^4}}\right)$

So, $\mathrm{f}(\mathrm{x})$ is neither increasing nor decreasing.

$ \therefore $ Option (A) and (B) are wrong statements.

For some $\beta \in R, y=\beta$ intersects $y=f(x)$ at infinitely many points.

So option $\mathrm{C}$ is correct.

Given, $x^k+y^k=c^2$, since two constant $c_1$ and $c_2$

$ \begin{aligned} & \text { So, } & k & =2 \\ & & x^2+y^2 & =c^2 \\ \Rightarrow & & 2 x+2 y \frac{d y}{d x} & =0 \\ \Rightarrow & & x+y \frac{d y}{d x} & =0 \\ \Rightarrow & & \frac{d y}{d x}+\frac{x}{y} & =0 \end{aligned} $

Given that $y=a x^2+b \log x$

$ \Rightarrow \quad \frac{y}{x^2}=a+b \log x $

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

$ \begin{array}{ll} & \frac{1}{x^2} y^{\prime}-\frac{2}{x^3} y=\frac{b}{x} \\ \Rightarrow & \frac{1}{x} y^{\prime}-\frac{2}{x^2} y=b \end{array} $

Again differentiating w.r.t. $x$,

$ \begin{aligned} & \frac{1}{x} y^{\prime \prime}-\frac{y^{\prime}}{x^2}+\frac{4 y}{x^3} \cdot \frac{2}{x^2} y^{\prime} =0 \\ \Rightarrow & -\frac{1}{x} y^{\prime \prime}+\frac{3 y^{\prime}}{x^2} =\frac{4 y}{x^3} \\ \Rightarrow & -x y^{\prime \prime}+3 y^{\prime} =\frac{4 y}{x} \end{aligned} $

On comparing with

$ \begin{aligned} & \quad f(x) y^{\prime \prime}+g(x) y^{\prime}=\frac{4 y}{x} \\ & \therefore \quad f(x)=-x \\ & \text { and } g(x)=3 \end{aligned} $

an $m=1$ degree of differential equation

$ \therefore \quad f(1)+g(1)=-1+3=2 $

Order of differential equation i.e.

$ f(m)+g(m)=l $

Given,

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

On putting

$ \begin{array}{rlrl} 2 x+3 y-4 =z & \\ 2 \frac{d x}{d y}+3 =\frac{d z}{d y} \\ \Rightarrow \quad \frac{d x}{d y} =\frac{1}{2} \frac{d z}{d y}-\frac{3}{2} \end{array} $

Now, $\frac{1}{2} \frac{d z}{d y}-\frac{3}{2}=z$

$ \begin{array}{lc} \Rightarrow & \frac{d z}{d y}=2 z+3 \\ \Rightarrow & \frac{d z}{2 z+3}=d y \\ \Rightarrow & \frac{1}{2} \log (2 z+3)+c_1=y \end{array} $

$ \begin{array}{lr} \Rightarrow & \log (4 x+6 y-8+3)+2 c_1=2 y \\ \Rightarrow & 6 y-3 \log (4 x+6 y-5)=6 c_1 \\ \Rightarrow & 6 y-3 \log (4 x+6 y-5)=c \end{array} $

Given, differential equation

$ \begin{aligned} \left(\frac{d^4 y}{d x^4}+\frac{d^2 y}{d x^2}\right)^{3 / 2} & =5 \frac{d^3 y}{d x^3} \\ \Rightarrow \quad\left(\frac{d^4 y}{d x^4}+\frac{d^2 y}{d x^2}\right)^3 & =25\left(\frac{d^3 y}{d x^3}\right)^2 \end{aligned} $

Order $=4$

So, number of arbitrary constants $=4$

Assertion (A) :

$ y^{\prime}+2 x y^{\prime}+\log _e\left(\frac{d y}{d x}\right)=0 $

Here, degree is not defined. So, Assertion is false but Reason (R) is true.

Given, $S$ be the family of curves given by general solution of differential equation,

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

Curve passes through $\left(\pi^2, 1\right)$

$ \begin{aligned} & & -e & =\sin \pi+c \\ \Rightarrow & & c & =-e \\\therefore & & S:-e^{1 / y} & =\sin \sqrt{x}-e \\ \Rightarrow& & \sin \sqrt{x} & +e^{1 / y}=e \end{aligned} $

Statement I: If the centre of circle lies on $Y$-axis and radius $k$, then equation of circle is $(x)^2+(y-h)^2=k^2$, where $(0, h)$ is the centre of circle.

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

Now, substitute value of $y-h$ in equation of circle,

$ \begin{gathered} x^2+\frac{x^2}{\left(\frac{d y}{d x}\right)^2}=k^2 \\ \Rightarrow \quad x^2\left(\frac{d y}{d x}\right)^2+x^2=k^2\left(\frac{d y}{d x}\right)^2 \\ \Rightarrow\left(x^2-k^2\right)\left(\frac{d y}{d x}\right)^2+x^2=0 \end{gathered} $

∴ Statement I is true.

Statement II: If the centre of circle lies on $X$-axis and circle passes through origin is $(x-h)^2+y^2=h^2$, where centre of circle is $(h, 0)$ and radius is $h$ unit. On differentiate both sides w.r.t. $x$, we get

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

Now, substitute the value of $h$ in the equation of circle

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

∴ Statement II is also true.

$y^2-5 a x-5 a^{\frac{3}{2}}=0$

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

$ \begin{array}{rlrl} & (2 y) \frac{d y}{d x}-5 a =0 \\ a & =\left(\frac{2}{5} y\right) \frac{d y}{d x} \\ \Rightarrow \quad y^2-5\left(\frac{2}{5} y \frac{d y}{d x}\right) x & =5\left(\frac{2}{5} y \frac{d y}{d x}\right)^{\frac{3}{2}} \\ \Rightarrow \quad & y^2-(2 x y) \frac{d y}{d x} =(2 y) \frac{d y}{d x}\left(\left(\frac{2}{5} y\right) \frac{d y}{d x}\right)^{\frac{1}{2}} \\ \Rightarrow \quad & y-(2 x) \frac{d y}{d x} =\left(\frac{2 d y}{d x}\right)\left(\left(\frac{2}{5} y\right) \frac{d y}{d x}\right)^{\frac{1}{2}} \end{array} $

On squaring both sides, we get

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

$ \left(\left(\frac{2}{5} y\right) \frac{d y}{d x}\right) $

$ \Rightarrow y^2+4 x^2\left(\frac{d y}{d x}\right)^2-2 x y\left(\frac{d y}{d x}\right)=\frac{8}{5} y\left(\frac{d y}{d x}\right)^3 $

Order of differential equation is 1 and degree is 3

$ \begin{array}{ll} \therefore & m=1 \text { and } n=3 \\ & m-n=1-3=-2 \end{array} $

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

On variable separable,

$ \begin{aligned} & \int \frac{\log y}{4} d y=\int \sin x d x-\int \frac{2 x d x}{1+x^2} \\ \Rightarrow & \frac{(\log y)^2}{2}=-\cos x-\log \left(1+x^2\right)+c_1 \\ \Rightarrow & (\log y)^2+2 \cos x+2 \log \left(1+x^2\right)=2 c_1 \\ \Rightarrow & (\log y)^2+2 \cos x+\log \left(1+x^2\right)^2=c \end{aligned} $

$ \begin{aligned} &\text { Let the equation of the ellipse be }\\ &\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \end{aligned} $

Differentiating with resp. to $x$,

$ \Rightarrow \quad \frac{2 x}{a^2}+\frac{2 y y^{\prime}}{b^2}=0 \Rightarrow \frac{y y^{\prime}}{x}=\frac{-b^2}{a^2} $

Again, differentiating w.r. to $x$,

$ \begin{aligned} & \Rightarrow \quad \frac{x\left(x y^{\prime \prime}+\left(y^{\prime}\right)^2\right)-\left(y y^{\prime}\right)}{x^2}=0 \\ & \Rightarrow \quad x\left[y y^{\prime}+\left(y^{\prime}\right)^2\right]-y y^{\prime}=0 \\ & \Rightarrow \quad x y y^{\prime}+x\left(y^{\prime}\right)^2-y y^{\prime}=0 \end{aligned} $

$\left(\frac{d^2 y}{d x^2}\right)^{4 / 3}+x\left(\frac{d y}{d x}\right)^2-y \cos \left(\frac{d y}{d x}\right)=0$

The degree of given differential equation is not defined as it is not a polynomial differential equation.

$ \begin{aligned} & \text { } \begin{aligned} \frac{d y}{d x} & =\frac{2 x-3 y+5}{6 x-9 y+7}=\frac{1(2 x-3 y)+5}{3(2 x-3 y)+7} \\Let\,\,\,\,\,\,\,\,\, u & =2 x-3 y \\ \frac{d u}{d x} & =2-3 \frac{d y}{d x} \\ \Rightarrow \quad \frac{d y}{d x} & =\left(2-\frac{d u}{d x}\right) \frac{1}{3} \end{aligned} \end{aligned} $

$ \begin{aligned} & \frac{1}{3}\left(2-\frac{d u}{d x}\right)=\frac{u+5}{3 u+7} \\ \Rightarrow \quad 2-\frac{d u}{d x} & =\frac{3 u+15}{3 u+7} \\ \frac{d u}{d x} & =2-\left(\frac{3 u+15}{3 u+7}\right) \end{aligned} $

$ \begin{aligned} & \frac{d u}{d x}=\frac{6 u+14-3 u-15}{3 u+7} \Rightarrow \frac{d u}{d x}=\frac{3 u-1}{3 u+7} \\ & \int\left(\frac{3 u+7}{3 u-1}\right) d u=\int d x \\ & \Rightarrow \int\left(\frac{3 u-1}{3 u-1}+\frac{8}{3 u-1}\right) d u=x \\ & u+\frac{8}{3} \log |3 u-1|+c=x \\ & 2 x-3 y+\frac{8}{3} \log \\ & |6 x-9 y-1|-x+c=0 \\ & x-3 y+\frac{8}{3} \log |6 x-9 y-1|+c=0 \\ & {[\because u=2 x-3 y]} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } a x^2+b y^2=1 \\ & \Rightarrow \quad 2 a x+2 b y \cdot y^{\prime}=0 \\ & \Rightarrow \quad b y \cdot y^{\prime}=-a x \\ & \Rightarrow \quad \quad \frac{y y^{\prime}}{x}=\frac{-a}{b} \\ & \Rightarrow \quad \frac{x\left(\left(y^{\prime}\right)^2+y \cdot y^{\prime \prime}\right)-y y^{\prime}}{x^2}=0 \\ & \Rightarrow \quad x y \cdot y^{\prime \prime}+x \cdot\left(y^{\prime}\right)^2-y y^{\prime}=0 \end{aligned} \end{aligned} $

$\sqrt{\frac{d^2 y}{d x^2}}=\sqrt[3]{\left[y \frac{d y}{d x}+x \sin \left(\frac{d y}{d x}\right)\right]^2}$

$ \begin{gathered} \Rightarrow\left(\frac{d^2 y}{d x^2}\right)^{\frac{1}{2}}=\left[y \frac{d y}{d x}+x \sin \left(\frac{d y}{d x}\right)\right]^{\frac{2}{3}} \\ \Rightarrow\left(\frac{d^2 y}{d x^2}\right)^{\frac{1}{2} \times 6}=\left[y \frac{d y}{d x}+x \sin \left(\frac{d y}{d x}\right)\right]^{\frac{2}{3} \times 6} \\ \quad\left(\frac{d^2 y}{d x^2}\right)^{\prime}=\left[y \frac{d y}{d x}+x \sin \left(\frac{d y}{d x}\right)\right]^4 \end{gathered} $

Order $=2$

Degree $=$ Not defined

( ∵ The above differential equation is not a polynomial in $\frac{d y}{d x}$ )

$ \begin{aligned} &\text { } \frac{d y}{d x}=\frac{(x-2)(y+1)}{(x+1)(y-2)}\\ &\begin{aligned} & \Rightarrow \quad \int \frac{y-2}{y+1} d y=\int \frac{x-2}{x+1} d x \\ & \Rightarrow \quad \int\left(1-\frac{3}{y+1}\right) d y=\int\left(1-\frac{3}{x+1}\right) d x \\ & \Rightarrow \quad y-3 \ln |y+1|=x-3 \ln |x+1|+c \\ & \Rightarrow \quad x-y+3 \ln \left|\frac{y+1}{x+1}\right|=c \end{aligned} \end{aligned} $

$ \because\left(x-x_1\right)^2+(y-3)^2=r^2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

On differentiating,

$ \begin{array}{lc} & 2\left(x-x_1\right)+2(y-3) \cdot y^{\prime}=0 \\ \Rightarrow & x-x_1=-(y-3) y^{\prime} \end{array} $

$ \begin{aligned} &\text { On putting in Eq. (i), }\\ &\begin{aligned} (y-3)^2\left(y^{\prime}\right)^2+(y-3)^2 & =r^2 \\ 1+\left(\frac{d y}{d x}\right)^2 & =\frac{r^2}{(y-3)^2} \end{aligned} \end{aligned} $