Mathematics M.C.Q (1 Marks)

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

==> $\frac{1}{{{2^x}}} - \frac{1}{{{2^y}}} = \frac{{{c_1}}}{{\log 2}} = c$

==> $\log (y{\tan ^{ - 1}}x) + \log c = 0$ ==>$y {\tan ^{ - 1}}x = c$.

==> $\log ({e^y} + 1) + \log \sin x = \log K$ ==> $({e^y} + 1)\sin x = K$.

After integration, we get $\log x - \frac{{{x^2}}}{2} = \log y + \log c$

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

==> $\frac{{c{x^2}}}{{{y^2}}} = {e^x}^2$ ==> $c{x^2} = {y^2}{e^{{x^2}}}$.

Put $x + y = t$ ==> $\frac{{dy}}{{dx}} = \frac{{dt}}{{dx}} - 1$

$\therefore \frac{{dy}}{{dx}} = \frac{{1 - t}}{{2t - 3}}$  ==> $\frac{{dt}}{{dx}} - 1 = \frac{{1 - t}}{{2t - 3}}$ ==> $\frac{{dt}}{{dx}} = \frac{{t - 2}}{{2t - 3}}$

==> $\frac{{2t - 3}}{{t - 2}}dt = dx$. Integrating both sides, we get
$\int_{}^{} {\frac{{2t - 4}}{{t - 2}}dt} - \int_{}^{} {\frac{{3 - 4}}{{t - 2}}dt} = \int_{}^{} 1 dx$

==> $2t + \log (t - 2) = x + c$

==> $2(x + y) + \log (x + y - 2) = x + c$

==> $2y + x + \log (x + y - 2) = c$.

Integrating both sides,$\int_{}^{} {dy} = \int_{}^{} {{{\sin }^{ - 1}}a\,dx} $

$y = x{\sin ^{ - 1}}a + c$ and $y(0) = 0 + c = 1$, $\therefore c = 1$

$\therefore y = x{\sin ^{ - 1}}a + 1$ ==> $a = \sin \frac{{y - 1}}{x}$.

Put $x + y = v$. Differentiate $1 + \frac{{dy}}{{dx}} = \frac{{dv}}{{dx}}$

Put these values in $(i),$ $\cos v\,\left( {\frac{{dv}}{{dx}} - 1} \right) = 1$

==> $\cos v\,\frac{{dv}}{{dx}} = 1 + \cos v$ ==> $\frac{{\cos v}}{{1 + \cos v}}dv = dx$

==>$\left[ {\frac{{2{{\cos }^2}(v/2) - 1}}{{2{{\cos }^2}(v/2)}}} \right]\,dv = dx$ ==> $\left[ {1 - \frac{1}{2}{{\sec }^2}(v/2)} \right]\,dv = dx$

Integrate, $v - \tan (v/2) = x + c$

$x + y - \tan \left( {\frac{{x + y}}{2}} \right) = x + c$ ==> $y = \tan \left( {\frac{{x + y}}{2}} \right) + c$.

Integrating both sides, $\int {\frac{{dy}}{{y + 1}} = \int {\frac{{dx}}{{x - 1}}} } $

==> $\log (y + 1) = \log (x - 1) + \log c$ ==> $(y + 1) = (x - 1)c$

At $x = 1$ ==> $y = - 1$ where as $y(1) = 2$

Hence, the solution is not possible.

==> $\frac{{dy}}{{dx}} = \sin \left( {\frac{{x - y}}{2}} \right) - \sin \left( {\frac{{x + y}}{2}} \right)$

==> $\frac{{dy}}{{dx}} = - 2\sin \,\left( {\frac{y}{2}} \right)\,.\cos \,\left( {\frac{x}{2}} \right)$

==> ${\rm{cos}}{\rm{ec}}\left( {\frac{y}{2}} \right).dy = - 2\cos \left( {\frac{x}{2}} \right)\,dx$

Integrating both sides,

$\int {{\rm{cosec}}\left( {\frac{y}{{\rm{2}}}} \right)dy = - \int {2\cos \left( {\frac{x}{2}} \right)dx + c} } $.

==> $\frac{{\log \,\tan \frac{y}{4}}}{{1/2}} = - \frac{{2\sin \left( {x/2} \right)}}{{1/2}} + c$

==> $\log (\tan \frac{y}{4}) = c - 2\sin (x/2)$.

 ${v^2}\left( {\frac{{dv}}{{dx}} - 1} \right) = {a^2}$

==> $\frac{{dv}}{{dx}} = \frac{{{a^2}}}{{{v^2}}} + 1 = \frac{{{a^2} + {v^2}}}{{{v^2}}}$ ==> $\frac{{{v^2}}}{{{a^2} + {v^2}}}dv = dx$

==> $\left( {1 - \frac{{{a^2}}}{{{a^2} + {v^2}}}} \right)dv = dx$ ==> $v - a{\tan ^{ - 1}}\frac{v}{a} = x + c$

==> $y = a{\tan ^{ - 1}}\left( {\frac{{x + y}}{a}} \right)+ c.$

==> $y(1 - ay) = \left( {a + x} \right)\,.\frac{{dy}}{{dx}}$ ==> $\frac{{dx}}{{(a + x)}} = \frac{{dy}}{{y(1 - ay)}}$

Integrating both sides, $\int {\frac{{dx}}{{(a + x)}} = } \int {\frac{{dy}}{{y(1 - ay)}}} $

==> $\int {\frac{{dx}}{{a + x}} = \int {\left[ {\frac{1}{y} + \frac{a}{{(1 - ay)}}} \right]\,dx} } $

$\log (a + x) = \log y + \frac{{a\log (1 - ay)}}{{ - a}}$

==> $\log (a + x) = \log y - \log (1 - ay) + \log c$

==> $\log (x + a)(1 - ay) = \log cy$ ==> $(x + a)(1 - ay) = cy$.

==> ${e^{ - by}}dy = {e^{ax}}dx$ ==> $\frac{{{e^{ - by}}}}{{ - b}} = \frac{{{e^{ax}}}}{a} + c$.

Separating the variables and integrating

$\int {(\sin y + y\cos y)dy = \int {(x\log {x^2} + x)dx} } $

==> $ - \cos y + y\sin y + \cos y$

$ = \frac{{{x^2}}}{2}\log {x^2} - \int {\frac{{{x^2}}}{2}.\frac{1}{{{x^2}}}.2xdx + \int {x\,dx + c} } $

==> $y\sin y = \frac{{{x^2}}}{2}2\log x - \int {x\,dx + \int {xdx + c} } $

==> $y\sin y = {x^2}\log x + c$.

$y = \int {\log (x + 1)dx} = x.\log (x + 1) - \int {\frac{x}{{x + 1}}dx} $

$ = x.\log (x + 1) - \int {\left( {1 - \frac{1}{{x + 1}}} \right)} \,dx$

$ = x.\log (x + 1) - x + \log (x + 1) + c$

$ = (x + 1)\log (x + 1) - x + c$

$x = 0$ at $y = 3$

$3 = (1)\log (1) - 0 + c$ ==> $3 = 0 + c$ ==> $c = 3$

 $y = (x + 1)\log |x + 1| - x + 3$.

$\frac{{dy}}{{dx}}(\tan y) = 2\sin x\cos y$ ==> $\frac{{\sin y}}{{{{\cos }^2}y}}dy = 2\sin xdx$

==> $\int {\frac{{\sin y}}{{{{\cos }^2}y}}} dy = 2\int {\sin xdx} $ ==> $\frac{1}{{\cos y}} = - 2\cos x + c$

$\therefore$  $\sec y + 2\cos x = c$.

Integrating, we get $\log |x| = \frac{1}{2}\log (1 + {y^2}) + \log c$

or $|x| = c\sqrt {(1 + {y^2})} $

But it passes through $(1, 0)$, so we get $c = 1$

$\therefore $Solution is ${x^2} = {y^2} + 1$or${x^2} - {y^2} = 1$.

==> $\int_{}^{} {dy} = \int_{}^{} {\left( {x + \frac{1}{{{x^2}}}} \right)} {\rm{ }}dx$ ==> $y = \frac{{{x^2}}}{2} - \frac{1}{x} + c$

Since it passes through $(3, 9)$, therefore

$9 = \frac{9}{2} - \frac{1}{3} + c$ ==> $c = \frac{{29}}{6}$

$y = \frac{{{x^2}}}{2} - \frac{1}{x} + \frac{{29}}{6}$ ==> $6xy = 3{x^3} + 29x - 6$.

Integrating, we get $\frac{{{y^2}}}{2} + \frac{{{x^2}}}{2} = ax + c$

or ${x^2} + {y^2} - 2ax + k = 0$,

which represents a set of circles having centre on $x$-axis.

This passes through $\left( {2,\frac{7}{2}} \right)$,  $\frac{7}{2} = 2 + \frac{1}{2} + c$

==> $c = 1$

Thus the equation of the curve is $y = x + \frac{1}{x} + 1$ or$xy = {x^2} + x + 1$.

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

==> $\int_{}^{} {\frac{1}{{y - 1}}} dy = \int_{}^{} {\left( {\frac{1}{x} - \frac{1}{{x + 1}}} \right){\rm{ }}} dx + c$ ==> $\frac{{(y - 1)(x + 1)}}{x} = k$

Putting $x = 1$,$y = 0$, we get $k = - 2$

Hence the equation is $(y - 1)(x + 1) + 2x = 0$.

Integrating, we get ${y^2} = x + c$

This passes through $(4,3)$,  $9 = 4 + c$ ==> $c = 5$

Hence the equation of the curve is ${y^2} = x + 5$.

Putting $t = 0,\;x = 0$, we get $\log 1 = c$ ==> $c = 0$

$\therefore t = \log (x + 1)$. For $x = 99,\;t = {\log _e}100 = 2{\log _e}10$.

On integrating, $\log x = \log y + \log c$

==> $\log \frac{x}{y} = \log c$ ==> $x = cy$

It is a straight line passing through origin.

${\left( {\frac{{dy}}{{dx}}} \right)_{(1,\,0)}} = \frac{{1 + 0}}{{1 - 0}} = 1$.

==> $\frac{{dy}}{{y(1 - ay)}} = \frac{{dx}}{{x + a}}$

On integrating both sides, we get

==> $\log y - \log (1 - ay) = \log (x + a) + \log c$

==> $\frac{y}{{(1 - ay)}} = c(x + a)$or $c(x + a)(1 - ay) = y$.

Integrating both sides, $\int {\frac{{dy}}{y}} = \int {\frac{{ - x}}{{\sqrt {x + a} }}dx} $

$\log y = - \int_{}^{} {\frac{{x + a - a}}{{\sqrt {x + a} }}} dx$$ = - \int_{}^{} {\sqrt {x + a} } dx + \int_{}^{} {\frac{a}{{\sqrt {x + a} }}} dx$

==> $\log y = - \frac{2}{3}{(x + a)^{3/2}} + 2a\sqrt {x + a} + \log A$

$y = A{e^{ - 2/3{{(x + a)}^{3/2}} + 2a\sqrt {x + a} }}$$ = A{e^{\left[ {(\sqrt {x + a} \left( { - \frac{2}{3}(x + a) + 2a} \right)} \right]}}$

$ = A{e^{\left[ {\sqrt {x + a} \left( {\frac{{ - 2x - 2a + 6a}}{3}} \right)} \right]}}$$ = A{e^{[ - 2/3\sqrt {x + a} (x - 2a)]}}$

or $y = A{e^{[2/3\sqrt {x + a} (2a - x)]}}$.

Putting $y = vx$so that $\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}$, we get

$v + x\frac{{dv}}{{dx}} = v - {\cos ^2}v$ or $\frac{{dv}}{{{{\cos }^2}v}} = - \frac{{dx}}{x}$

On integrating, we get $\tan v = - \log x + \log c$

==> $\tan \left( {\frac{y}{x}} \right) = - \log x + \log C$

This passes through $\left( {1,\frac{\pi }{4}} \right)$, therefore$1 = \log c$

==>$\tan \left( {\frac{y}{x}} \right) = - \log x + \log e$==>$y = x{\tan ^{ - 1}}\left[ {\log \left( {\frac{e}{x}} \right)} \right]$.

It is given that $y\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} = \sqrt {{x^2} + {y^2}} $

(Radius vector $ = r = \sqrt {{x^2} + {y^2}} $)

==> ${y^2} + {y^2}{\left( {\frac{{dy}}{{dx}}} \right)^2} = {x^2} + {y^2}$ ==> ${y^2}{\left( {\frac{{dy}}{{dx}}} \right)^2} = {x^2}$

==> $ydy \pm xdx = 0$ ==> ${y^2} \pm {x^2} = k$.

==> $\frac{{dy}}{{dx}} = (1 + x)(1 + {y^2})$
==> $\frac{{dy}}{{1 + {y^2}}} = (1 + x)dx$
Integrating both sides, $\int {\frac{{dy}}{{1 + {y^2}}} = \int {(1 + x)dx} } $
==> ${\tan ^{ - 1}}y = x + \frac{{{x^2}}}{2} + c$
Put $y(0) = 0$, ?$0 = 0 + 0 + c \Rightarrow c = 0$
$\therefore$ ${\tan ^{ - 1}}y = x + \frac{{{x^2}}}{2}$

==> $y = \tan \left( {x + \frac{{{x^2}}}{2}} \right)$.

$\Rightarrow \frac{d y}{e^{y}}=e^{x} d x$

$\Rightarrow \mathrm{e}^{-\mathrm{y}} \mathrm{dy}=\mathrm{e}^{\mathrm{x}} \mathrm{d} \mathrm{x}$

Intergrating both sides, we get:

$\int e^{-y} d y=\int e^{x} d x$

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

$\Rightarrow \mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{y}}=-\mathrm{k}$

$\Rightarrow e^{x}+e^{-y}=c \quad(c=-k)$

Hence, the correct answer is $\mathrm{C}.$

Now put $y = vx$and $\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}$

Therefore, $v + x\frac{{dv}}{{dx}} = \frac{{{x^2} + v{x^2} + {v^2}{x^2}}}{{{x^2}}} = 1 + v + {v^2}$

==> $x\frac{{dv}}{{dx}} = 1 + {v^2}$

==> $\frac{{dv}}{{1 + {v^2}}} = \frac{{dx}}{x}$

Now integrating both sides, we get ${\tan ^{ - 1}}v = \log x + c$

==> ${\tan ^{ - 1}}\left( {\frac{y}{x}} \right) = \log x + c$ 

 $\{\because\,\,y=vx\Rightarrow v=y/x\}$

Put $y = vx$ and $\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}$

So, we get $x\frac{{dv}}{{dx}} = \frac{{1 + {v^2}}}{{2v}}$

==> $\frac{{2vdv}}{{1 + {v^2}}} = \frac{{dx}}{x}$

On integrating, we get ${x^2} + {y^2} = p{x^3}$.

Now solve it by putting $y = vx$ and $\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}$.

It is a homogeneous equation so putting $y = vx$

and $\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}},$ we get

$v + x\frac{{dv}}{{dx}} = \frac{{x + vx}}{{x - vx}} = \frac{{1 + v}}{{1 - v}}$

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

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

==> ${\log _e}x = {\tan ^{ - 1}}v - \frac{1}{2}\log (1 + {v^2}) + {\log _e}c$

Substituting $v = \frac{y}{x},$we get

${\log _e}x = {\tan ^{ - 1}}\frac{y}{x} - \frac{1}{2}\log \left[ {1 + {{\left( {\frac{y}{x}} \right)}^2}} \right] + {\log _e}c$

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

So, now put $y = vx$and $\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}},$we get

$v + x\frac{{dv}}{{dx}} = - \frac{{3{x^2}v + {x^2}{v^2}}}{{{x^2} + {x^2}v}}$==>$x\frac{{dv}}{{dx}} = \frac{{ - 2v(v + 2)}}{{v + 1}}$

==> $\frac{1}{x}dx = - \frac{{v + 1}}{{2v(v + 2)}}dv = - \left[ {\frac{1}{{2(v + 2)}} + \frac{1}{{2v(v + 2)}}} \right]\,dv$

==> $ - \frac{2}{x}dx = \left[ {\frac{1}{{v + 2}} + \frac{1}{{2v}} - \frac{1}{{2v(v + 2)}}} \right]$

On integrating, we get

$ - 2{\log _e}x = \frac{1}{2}\log (v + 2) + \frac{1}{2}\log v + \log c$

==> $v(v + 2){x^4} = {c^2}$ ==> $\frac{y}{x}\left( {\frac{y}{x} + 2} \right){x^4} = {c^2}$,$\left( \because v=\frac{y}{x} \right)$

Hence required solution is $({y^2} + 2xy){x^2} = {c^2}$.

$y = vx$ and $\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}$.

==> $\frac{{dy}}{{dx}} = - \frac{{x + y}}{x}$

It is homogenous equation, hence put $y = vx$ and $\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}},$ we get $v + x\frac{{dv}}{{dx}} = - \frac{{x + vx}}{x} = - \frac{{1 + v}}{1}$

==> $x\frac{{dv}}{{dx}} = - 1 - 2v$==> $\int_{}^{} {\frac{{dv}}{{1 + 2v}}} = - \int_{}^{} {\frac{{dx}}{x}} $

==> $\frac{1}{2}\log (1 + 2v) = - \log x + \log c$ ==> $\log \left( {1 + 2\frac{y}{x}} \right) = 2\log \frac{c}{x}$ 

==> $\frac{{x + 2y}}{x} = {\left( {\frac{c}{x}} \right)^2}$ ==> ${x^2} + 2xy = c$.

Put $y = vx$and $\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}$

$\therefore \frac{1}{v} + v + x\frac{{dv}}{{dx}} = 2$ ==> $v + x.\frac{{dv}}{{dx}} = \frac{{2v - 1}}{v}$

==> $\frac{v}{{{{(v - 1)}^2}}}dv = - \frac{{dx}}{x}$ ==> $\frac{{v - 1 + 1}}{{{{(v - 1)}^2}}}dv = - \frac{{dx}}{x}$

$\left[ {\frac{1}{{(v - 1)}} + \frac{1}{{{{(v - 1)}^2}}}} \right]dv = - \frac{{dx}}{x}$

Integrating both sides, $\int_{}^{} {\frac{{dv}}{{v - 1}}} + \int_{}^{} {\frac{{dv}}{{{{(v - 1)}^2}}}} = - \int_{}^{} {\frac{{dx}}{x}} $

==> $\log (v - 1) - \frac{1}{{v - 1}} = - \log x + c$ ==> $\log (y - x) = \frac{x}{{y - x}} + c$.

$\therefore v + x\frac{{dv}}{{dx}} = \frac{{(x)(vx)}}{{{x^2} + {v^2}{x^2}}}$

==> $v + x.\frac{{dv}}{{dx}} = \frac{v}{{1 + {v^2}}}$ ==> $x\frac{{dv}}{{dx}} = \frac{{ - {v^3}}}{{1 + {v^2}}}$

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

Integrating both sides, $\int_{}^{} {\frac{{dv}}{{{v^3}}}} + \int_{}^{} {\frac{{dv}}{v}} = - \int_{}^{} {\frac{{dx}}{x}} $

==> $ - \frac{1}{{2{v^2}}} + \log v = - \log x - \log c$

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

==> $cy = {e^{{x^2}/2{y^2}}}$ ==> ${c^2}{y^2} = {e^{{x^2}/{y^2}}}$

$\therefore {y^2}a = {e^{{x^2}/{y^2}}}$, where ${c^2} = a$.

Put $y = vx$ ==> $\frac{{dy}}{{dx}} = v + x.\frac{{dv}}{{dx}}$

$\therefore v + x.\frac{{dv}}{{dx}} = v(\log v + 1)$

$v + x\frac{{dv}}{{dx}} = v\log v + v$ ==> $x\frac{{dv}}{{dx}} = v\log v$==> $\frac{{dv}}{{v\log v}} = \frac{{dx}}{x}$

Integrating both sides, $\int_{}^{} {\frac{{dv}}{{v\log v}}} = \int_{}^{} {\frac{{dx}}{x}} $

$\log \log v = \log x + \log c$$ \Rightarrow $$\log v = xc$ ==> $\log (y/x) = \,x\,c$.

$\therefore v + x\frac{{dv}}{{dx}} = \frac{{v - 1}}{{v + 1}}$ ==> $x\frac{{dv}}{{dx}} = \frac{{v - 1}}{{v + 1}} - v$

==> $x\frac{{dv}}{{dx}} = - \frac{{{v^2} + 1}}{{v + 1}}$ ==> $\int_{}^{} {\frac{{dx}}{x}} = - \int_{}^{} {\frac{{v + 1}}{{{v^2} + 1}}} {\rm{ }}dv$

==> $ - {\log _e}x = \frac{1}{2}\int_{}^{} {\frac{{2v}}{{{v^2} + 1}}} dv + \int_{}^{} {\frac{1}{{{v^2} + 1}}} dv$

==> $ - {\log _e}x = \frac{1}{2}\log ({v^2} + 1) + {\tan ^{ - 1}}v + c$

==> $ - 2{\log _e}x = \log \left( {\frac{{{x^2} + {y^2}}}{{{x^2}}}} \right) + 2{\tan ^{ - 1}}\left( {\frac{y}{x}} \right) + c$

==> ${\log _e}({x^2} + {y^2}) + 2{\tan ^{ - 1}}\frac{y}{x} + c = 0$.

$v + x\,\frac{{dv}}{{dx}} = \frac{{x - vx}}{{x + vx}}$

==> $v + x\,\frac{{dv}}{{dx}} = \frac{{1 - v}}{{1 + v}}$ ==> $\frac{{1 + v}}{{2 - {{(1 + v)}^2}}}dv = \frac{{dx}}{x}$

Integrating both sides, $\int {\frac{{1 + v}}{{2 - {{(1 + v)}^2}}}} \,dv = \int {\frac{{dx}}{x}} $

Put ${(1 + v)^2} = t \Rightarrow 2(1 + v)dv = dt$

==> $\frac{1}{2}\int_{}^{} {\frac{{dt}}{{2 - t}}} = \int_{}^{} {\frac{{dx}}{x}} $ ==> $ - \frac{1}{2}\log (2 - t) = \log xc$

==> $ - \frac{1}{2}\log [2 - {(1 + v)^2}] = \log xc$

==> $ - \frac{1}{2}\log [ - {v^2} - 2v + 1] = \log xc$

==> $\log \frac{1}{{\sqrt {1 - 2v - {v^2}} }} = \log xc$

==> ${x^2}{c^2}(1 - 2v - {v^2}) = 1$ ==> ${y^2} + 2xy - {x^2} = {c_1}$.

$\frac{{2v}}{{1 - {v^2}}}.dv = \frac{{dx}}{x}$

Integrating both sides, $ - \log (1 - {v^2}) = \log x + \log c$

$ - \log \left( {1 - \frac{{{y^2}}}{{{x^2}}}} \right) = \log x + \log c$…..$(i)$

This passes through $(2,\,1)$

$ - \log \,\left( {1 - \frac{1}{4}} \right) = \log 2 + \log c$ ==> $c = \frac{2}{3}$

From equation $(i),$ $\log \left( {\frac{{{x^2}}}{{{x^2} - {y^2}}}} \right) = \log xc$

$2\,({x^2} - {y^2}) = 3x$.

 The given differential equation becomes

$v + x\frac{{dv}}{{dx}} = v + \frac{{\phi \,(v)}}{{\phi '\,(v)}}$ ==> $\frac{{\phi '(v)}}{{\phi (v)}}dv = \frac{{dx}}{x}$

==> $\log \phi (v) = \log x + \log k$ ==> $\phi (v) = kx$ ==> $\phi \,\left( {\frac{y}{x}} \right) = kx$.

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

Put $v = x/y$ ==> $x = vy$ ==> $\frac{{dx}}{{dy}} = v + y\frac{{dv}}{{dy}}$

$v + y\frac{{dv}}{{dy}} + {v^2} - v + 1 = 0$ ==> $\frac{{dv}}{{{v^2} + 1}} + \frac{{dy}}{y} = 0$

==> $\int {\frac{{dv}}{{{v^2} + 1}} + \int {\frac{{dy}}{y} = 0} } $ ==> ${\tan ^{ - 1}}(v) + \log y + C = 0$

==> ${\tan ^{ - 1}}(x/y) + \log y + c = 0$.

Now integrating factor $(I.F.)$$ = {e^{\int {Pdx} }} = {e^{\int {\tan dx} }}$

$ = {e^{\log \sec x}} = \sec x$

Thus solution is given by, $y.{e^{\int {Pdx} }} = \int Q .\,{e^{\int {Pdx} }}dx + c$

==> $y.\sec x = \int {{x^m}} .\cos x.\sec xdx + c$ ==> $y\sec x = \frac{{{x^{m + 1}}}}{{m + 1}} + c$

==> $(m + 1)y = {x^{m + 1}}\cos x + c(m + 1)\cos x$.

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

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

This is equation of the form $\frac{{dx}}{{dy}} + Px = Q$

So, $I.F.$ $ = {e^{\int {P\,dy} }} = {e^{\int {\frac{1}{{1 + {y^2}}}.dy} }} = {e^{{{\tan }^{ - 1}}y}}$.