Differential Equations

We have,

$ \begin{aligned} & (3 y-7 x+7) d x+(7 y-3 x+3) d y=0 \\ & \Rightarrow(7 x-3 y-7) d x+(3 x-7 y-3) d y=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Given equation is non-homogeneous and $a_1 b_2-a_2 b_1=-40 \neq 0$

On solving $7 x-3 y-7=0$ and $3 x-7 y-3=0$

we get $x=1, y=0$

Now, substitude $x=1+u$ and $y=0+v=v$

$\therefore d x=d u$ and $d y=d v \ldots$ in Eq. (i), we get $(7 u-3 v) d u+(3 u-7 v) d v=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

This homogeneous equation in $u$ and $v$. Put, $u=t v \Rightarrow d u=t d v+v d t$ in Eq. (ii), we get

$ \begin{aligned} & 7 t v-3 v)(t d v+v d t)+(3 t v-7 v) d v=0 \\ \Rightarrow & (7 t-3)(t d v+v d t)+(3 t-7) d v=0 \\ \Rightarrow & \left(7 t^2-7\right) d v+v(7 t-3) d t=0 \\ \Rightarrow & \int \frac{d v}{v}+\int \frac{7 t-3}{7 t^2-7} d t=0 \\ \Rightarrow & \log |v|+\int \frac{t-3 / 7}{t^2-1} d t=C_1 \\ \Rightarrow & \log |v|+\left[\int \frac{2}{7(t-1)} d t+\int \frac{\frac{5}{7}}{(t+1)} d t\right] =\log C_1 \end{aligned} $

$ \begin{aligned} & \begin{array}{r} \log |v|+\frac{2}{7} \log |t-1|+\frac{5}{7} \log |t+1| =\log C_1 \end{array} \\ & \begin{array}{r} \Rightarrow \log |v|+\frac{2}{7} \log \left|\frac{u}{v}-1\right|+\frac{5}{7} \log \left|\frac{u}{v}+1\right| =\log C_1 \end{array} \\ & \begin{array}{r} \Rightarrow \log |v|+\frac{2}{7} \log \left(\frac{u-v}{v}\right) +\frac{5}{7} \log \left(\frac{u+v}{v}\right)=\log C_1 \end{array} \\ & \begin{array}{r} \Rightarrow \log |v|+\log \left(\frac{u-v}{v}\right)^{\frac{2}{7}}+\log \left(\frac{u+v}{v}\right)^{\frac{5}{7}} =\log C_1 \end{array} \end{aligned} $

$ \begin{aligned} & \Rightarrow \log \left[|v| \cdot \frac{(u-v)^{2 / 7}}{(v)^{2 / 7}} \cdot \frac{(u+v)^{5 / 7}}{(v)^{5 / 7}}\right]=\log C_1 \\ & \Rightarrow \log \left[(u-v)^{2 / 7}(u+v)^{5 / 7}\right]=\log C_1 \\ & \Rightarrow \log \left[(x-y-1)^{2 / 7}(x+y-1)^{5 / 7}\right] =\log C_1 \\ & \Rightarrow \quad(x-y-1)^{2 / 7}(x+y-1)^{5 / 7}=C_1 \\ & \Rightarrow \quad(x-y-1)^2(x+y-1)^5=C \end{aligned} $

We have,

$ \begin{aligned} & (3 y-7 x+7) d x+(7 y-3 x+3) d y=0 \\ & \Rightarrow(7 x-3 y-7) d x+(3 x-7 y-3) d y=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Given equation is non-homogeneous and $a_1 b_2-a_2 b_1=-40 \neq 0$

On solving $7 x-3 y-7=0$ and $3 x-7 y-3=0$

we get $x=1, y=0$

Now, substitude $x=1+u$ and $y=0+v=v$

$\therefore d x=d u$ and $d y=d v \ldots$ in Eq. (i), we get $(7 u-3 v) d u+(3 u-7 v) d v=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

This homogeneous equation in $u$ and $v$.

Put, $u=t v \Rightarrow d u=t d v+v d t$ in Eq. (ii), we get

$ \begin{aligned} & (7 t v-3 v)(t d v+v d t)+(3 t v-7 v) d v=0 \\ & \Rightarrow(7 t-3)(t d v+v d t)+(3 t-7) d v=0 \\ & \Rightarrow \quad\left(7 t^2-7\right) d v+v(7 t-3) d t=0 \\ & \Rightarrow \quad \int \frac{d v}{v}+\int \frac{7 t-3}{7 t^2-7} d t=0 \\ & \Rightarrow \log |v|+\int \frac{t-3 / 7}{t^2-1} d t=C_1 \\ & \Rightarrow \log |v|+\left[\int \frac{2}{7(t-1)} d t+\int \frac{\frac{5}{7}}{(t+1)} d t\right] \left.=\log C_1\right] \end{aligned} $

$ \begin{aligned} & \begin{array}{r} \Rightarrow \log |v|+\frac{2}{7} \log |t-1|+\frac{5}{7} \log |t+1| =\log C_1 \end{array} \\ & \begin{array}{r} \Rightarrow \log |v|+\frac{2}{7} \log \left|\frac{u}{v}-1\right|+\frac{5}{7} \log \left|\frac{u}{v}+1\right| =\log C_1 \end{array} \\ & \begin{array}{r} \Rightarrow \log |v|+\frac{2}{7} \log \left(\frac{u-v}{v}\right) +\frac{5}{7} \log \left(\frac{u+v}{v}\right)=\log C_1 \end{array} \\ & \begin{array}{r} \Rightarrow \log |v|+\log \left(\frac{u-v}{v}\right)^{\frac{2}{7}}+\log \left(\frac{u+v}{v}\right)^{\frac{5}{7}} =\log C_1 \end{array} \end{aligned} $

$ \begin{aligned} & \Rightarrow \log \left[|v| \cdot \frac{(u-v)^{2 / 7}}{(v)^{2 / 7}} \cdot \frac{(u+v)^{5 / 7}}{(v)^{5 / 7}}\right]=\log C_1 \\ & \Rightarrow \log \left[(u-v)^{2 / 7}(u+v)^{5 / 7}\right]=\log C_1 \\ & \Rightarrow \log \left[(x-y-1)^{2 / 7}(x+y-1)^{5 / 7}\right] =\log C_1 \\ & \Rightarrow \quad(x-y-1)^{2 / 7}(x+y-1)^{5 / 7}=C_1 \\ & \Rightarrow \quad(x-y-1)^2(x+y-1)^5=C \end{aligned} $

We have,

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

Put, $y=v x$

$ \begin{array}{lc} \Rightarrow & \frac{d y}{d x}=v+x \frac{d v}{d x} \\ \therefore & v+x \frac{d v}{d x}=\frac{v \cos v+v^2 \sin v}{v \sin v-\cos v} \\ \Rightarrow & x \frac{d v}{d x}=\frac{2 v \cos v}{v \sin v-\cos v} \\ \Rightarrow & \frac{v \sin v-\cos v}{v \cos v} d v=2 \frac{d x}{x} \\ \Rightarrow & \int\left(\tan v-\frac{1}{v}\right) d v=2 \int \frac{d x}{x} \\ \Rightarrow & \log |\sec v|-\log |v|=2 \log |x|+\log \left|C_1\right| \\ \Rightarrow & \log \left|\frac{\sec v}{v}\right|-\log |x|^2=\log \left|C_1\right| \\ \Rightarrow & \log \left|\frac{\sec v}{v x^2}\right|=\log \left|C_1\right| \end{array} $

$ \begin{aligned} & \Rightarrow \quad \frac{\sec v}{v x^2}=C_1 \\ & \Rightarrow \quad \frac{\sec \left(\frac{y}{x}\right)}{\frac{y}{x}\left(x^2\right)}=C_1 \\ & \Rightarrow \quad \frac{\sec \left(\frac{y}{x}\right)}{x y}=C_1 \\ & \Rightarrow \quad \sec \left(\frac{y}{x}\right)=C_1 x y \\ & \Rightarrow \quad \frac{1}{\cos \left(\frac{y}{x}\right)}=\frac{C}{x y}\left(\because \text { where } \frac{1}{C_1}=C\right) \end{aligned} $

We have,

$ \begin{aligned} y & =a e^{4 x}+b e^{-x} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \therefore \quad \frac{d y}{d x} & =4 a e^{4 x}-b e^{-x} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ \text { and } \quad \frac{d^2 y}{d x^2} & =16 a e^{4 x}+b e^{-x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{aligned} $

On adding Eqs. (i) and (ii), we get

$ \begin{aligned} y+\frac{d y}{d x} & =5 a e^{4 x} \\ \Rightarrow \quad a e^{4 x} & =\frac{1}{5}\left(y+\frac{d y}{d x}\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

From Eqs. (i) and (iv), we get

$ \begin{aligned} y & =\frac{1}{5} y+\frac{1}{5} \frac{d y}{d x}+b e^{-x} \\ \Rightarrow b e^{-x} & =\frac{4}{5} y-\frac{1}{5} \frac{d y}{d x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v) \end{aligned} $

$ \begin{aligned} &\text { From Eqs. (iii), (iv) and (v), we get }\\ &\begin{aligned} & \quad \frac{d^2 y}{d x^2}=\frac{16}{5} y+\frac{16}{5} \frac{d y}{d x}+\frac{4}{5} y-\frac{1}{5} \frac{d y}{d x} \\ & \quad \frac{d^2 y}{d x^2}=4 y+3 \frac{d y}{d x} \Rightarrow \frac{d^2 y}{d x^2}-3 \frac{d y}{d x}-4 y=0 \\ & \therefore \quad f=\frac{d^2 y}{d x^2}-3 \frac{d y}{d x}-4 y \\ & \therefore \quad \frac{d f}{d x}=\frac{d^3 y}{d x^3}-3 \frac{d^2 y}{d x^2}-4 \frac{d y}{d x} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text {We know that, } \\ & \text { Length of subtangent }=\frac{y}{d y / d x} \\ & \Rightarrow \quad x+7 y^2=\frac{y}{d y / d x} \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{y}{x+7 y^2} \\ & \Rightarrow \quad \frac{d x}{d y}=\frac{x+7 y^2}{y} \\ & \Rightarrow \quad \frac{d x}{d y}-\frac{1}{y} x=7 y \end{aligned} $

$ \begin{aligned} &\therefore \mathrm{IF}=e^{\int-\frac{1}{y} d y}=e^{-\log y}=\frac{1}{y}\\ &\text { So, solution is given by }\\ &\begin{aligned} & x \cdot \frac{1}{y}=\int 7 y \cdot \frac{1}{y} d y+C \Rightarrow \frac{x}{y}=7 y+C \\ & x=7 y^2+c y \Rightarrow 7 y^2+c y-x=0 \\ \therefore \quad & f(x, y): 7 y^2+c y-x=0 \end{aligned} \end{aligned} $

We have,

$ \begin{aligned} & (y-x+1) d y-(y+x+2) d x=0 \\ & y d y-x d y+d y-y d x-x d x+2 d x=0 \\ & y d y+d y-(x d y+y d x)-x d x+2 d x=0 \end{aligned} $

On integrating, we get

$ \begin{aligned} & \frac{y^2}{2}+y-x y-\frac{x^2}{2}+2 x-C=0 \\ & f(x, y, c)=\frac{y^2}{2}-\frac{x^2}{2}-x y+2 x+y-C \\ & f(1,1, c)=0 \\ & \therefore \frac{1}{2}-\frac{1}{2}-1+2+1-C=0 \\ & \Rightarrow C=2 \end{aligned} $

We have, ${{dy} \over {dx}} + {{\tan y} \over x} = {{\tan y\sin y} \over {{x^2}}}$

Now, divide by tan y sin y both sides, we get

$\cot y\cos ec\,y{{dy} \over {dx}} + {{\cos ec\,y} \over x} = {1 \over {{x^2}}}$ ...... (i)

Put cosec y = z

$\therefore$ $ - \cos ec\,y\cot y{{dy} \over {dx}} = {{dz} \over {dx}}$

$\therefore$ Eq. (i) reduces to

${{dz} \over {dx}} - {z \over x} = - {1 \over {{x^2}}}$

$IF = {e^{\int { - {1 \over x}dx} }} = {e^{ - \log x}} = {1 \over x}$

Hence solution is

${z \over x} = \int { - {1 \over {{x^3}}}dx = {1 \over {2{x^2}}} + c} $

$ \Rightarrow {1 \over {x\sin y}} = {{1 - 2c{x^2}} \over {2{x^2}}}$

$ \Rightarrow 2x = \sin y{(1 - 2cx)^2}$

$ \Rightarrow 2x = \sin y(1 - 2c{x^2})$

We have, $(x{y^5} + 2y)dx = x\,dy$

$ \Rightarrow x{{dy} \over {dx}} - 2y = x{y^5}$

$ \Rightarrow {{dy} \over {dx}} - {{2y} \over x} = {y^5}$

$ \Rightarrow {y^{ - 5}}{{dy} \over {dx}} - {{2{y^{ - 4}}} \over x} = 1$ ..... (i)

Put, ${y^{ - 4}} = t$

$ \Rightarrow - 4{y^{ - 5}}{{dy} \over {dx}} = {{dt} \over {dx}}$

$ \Rightarrow {y^{ - 5}}{{dy} \over {dx}} = {{ - 1} \over 4}{{dt} \over {dx}}$ ..... (ii)

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

$ - {1 \over 4}{{dt} \over {dx}} - {{2t} \over x} = 1$

$ \Rightarrow {{dt} \over {dx}} + {{8t} \over x} = - 4$

Now, $IF = {e^{\int {{8 \over x}dx} }} = {e^{8\log x}} = {x^8}$

and the solution is

$t\,.\,{x^8} = \int {( - 4){x^8}dx + C} $

$ \Rightarrow {{{x^8}} \over {{y^4}}} = - {{4\,.\,{x^9}} \over 9} + C$

$ \Rightarrow 9{x^8} + 4{x^9}\,.\,{y^4} = 9{y^4}C$

We have,

Slope of tangent of the curve is ${{{{(x + 1)}^2} + y - 3} \over {x + 1}}$

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

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

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

This is a form of linear differential equations

$\therefore$ $IF = {e^{\int {{{ - 1} \over {x + 1}}dx} }} = {e^{\log {1 \over {1 + x}}}} = {1 \over {1 + x}}$

Solution of given differential equation is

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

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

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

$\because$ Curve passes through (2, 0)

$\therefore$ $0 = 2 + {3 \over 3} + c$

$ \Rightarrow c = - 3$

Hence, equation of curve is

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

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

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

$({y^2} - {x^3})dx - xydy = 0$

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

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

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

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

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

$ \Rightarrow {y^2} + 2{x^3} + c{x^2} = 0$

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

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

Integrating factor (I.F) = ${e^{ - {1 \over y}}}$

Now x.${e^{ - {1 \over y}}}$ = $\int {{e^{ - {1 \over y}}}{1 \over {{y^3}}}dy} $ by putting ${ - {1 \over y}}$ = t

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

$ \Rightarrow x{e^t} = - (t.{e^t} - {e^t}) + c$

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

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

It passes through (1, 1)

$ \therefore $ c = $-{1 \over e}$

Equation of the curve is

$x = 1 + {1 \over y} - {e^{{1 \over y} - 1}}$ It passes through (k, 2)

$ \therefore $ k = $1 + {1 \over 2} - {e^{{1 \over 2} - 1}}$ = $3 \over 2$ - $1 \over{\sqrt e}$

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

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

y. sec x = $\int {(2x + {x^2}\tan x)\sec \,x\,dx} $

$ \Rightarrow y\sec x = {x^2}\sec x + \lambda $

$ \Rightarrow $ y(0) = 0 + $\lambda $ = 1 $ \Rightarrow $ $\lambda $ = 1

$ \Rightarrow $ y = x² + cos x

$ \Rightarrow $ y' = 2x – sinx

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

$ \Rightarrow y'\left( { - {\pi \over 4}} \right) = - {\pi \over 2} + {1 \over {\sqrt 2 }}$

$ \therefore $ $y'\left( {{\pi \over 4}} \right) - y'\left( { - {\pi \over 4}} \right) = \pi - \sqrt 2 $

${{dy} \over {dx}} + y{\sec ^2}x = se{c^2}x\,tanx$
$ \to $ This is linear differential equation

$IF = {e^{\int {{{\sec }^2}xdx} }} = {e^{\tan x}}$

Now solution is

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

$ \therefore $ Let tanx = t

sec²xdx = dt

$y.{e^{\tan x}} = \int {\mathop {{e^t}}\limits_{||} } \mathop t\limits_| dt$

ye^tanx = te^t – e^t + c

ye^tanx = (tanx – 1)e^tanx + c

y = (tanx – 1) + c · e^–tanx

Given y(0) = 0

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

$y\left( { - {\pi \over 4}} \right) = - 1 - 1 + e = - 2 + e$

$\cos x{{dy} \over {dx}} - y\sin x = 6x$

$ \Rightarrow $ ${{dy} \over {dx}} - y\tan x = 6x.\sec x$

This is a linear differential equation.

$ \therefore $ I.F = ${e^{\int { - \tan xdx} }}$ = $\left| {\cos x} \right|$

As 0 < x < ${\pi \over 2}$, so I.F = $\cos x$

So solution is

y(I.F) = $\int {6x.\sec x\left( {I.F} \right)} dx$

$ \Rightarrow $ y cosx = $\int {6x.\sec x\cos x} dx$

$ \Rightarrow $ y cosx = $6\int x dx$

$ \Rightarrow $ y cosx = 3x² + C

As $y\left( {{\pi \over 3}} \right) = 0 $

$ \therefore $ $ \left( 0 \right) \times \left( {{1 \over 2}} \right) = {{3{\pi ^2}} \over 9} + C $

$\Rightarrow C = {{ - {\pi ^2}} \over 3}$

$ \Rightarrow y\cos x = 3{x^2} - {{{\pi ^2}} \over 3}$

For $y\left( {{\pi \over 6}} \right)$

$y{{\sqrt 3 } \over 2} = {{3{\pi ^2}} \over {36}} - {{{\pi ^2}} \over 3}$

$ \Rightarrow $ ${{\sqrt 3 y} \over 2} = {{ - 3{\pi ^2}} \over {12}} $

$\Rightarrow y = {{ - {\pi ^2}} \over {2\sqrt 3 }}$

$x{{dy} \over {dx}} + 2y$ = x²

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

$ \therefore $ I.F = ${e^{\int {{2 \over x}dx} }}$ = x²

$ \therefore $ The solution is

yx² = $\int {{x^3}dx} $

$ \Rightarrow $ yx² = ${{{x^4}} \over 4} + C$ .....(1)

As y(1) = 1

$ \therefore $ when x = 1 then y = 1.

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

1 = ${1 \over 4} + C$

$ \Rightarrow $ C = ${3 \over 4}$

$ \therefore $ Required solution

yx² = ${{{x^4}} \over 4} + {3 \over 4}$

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

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

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

I.F = ${e^{\int {{{2x} \over {1 + {x^2}}}dx} }} = {e^{\ln \left( {1 + {x^2}} \right)}} = 1 + {x^2}$

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

Integrating both sides we get,

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

When x = 0 then y = 0

So, 0 = 0 + C

$ \Rightarrow $ C = 0

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

Put x = 1,

y.(2) = ${\pi \over 4}$

$ \Rightarrow $ y = ${\pi \over 8}$ = y(1)

Given,

$\sqrt ay(1)$ = $\pi \over 32$

$ \Rightarrow $ $\sqrt a. {\pi \over 8}$ = $\pi \over 32$

$ \Rightarrow $ $\sqrt a = {1 \over 4}$

$ \Rightarrow $ $a = {1 \over {16}}$

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

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

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

$ \therefore $  y.x² = $\int {x.{x^2}} dx + C$

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

This curve passes through (1, $-$ 2) $ \Rightarrow $ C= $-$ ${9 \over 4}$

$ \therefore $  yx² = ${{{x^4}} \over 4} - {9 \over 4}$

Now check option(s), Which is satisly by option (ii)

${{dy} \over {dx}} = {y \over x} = \ell nx$

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

$xy = \int {x\ell nx + C} $

$\ell nx{{{x^2}} \over 2} - \int {{1 \over x}.{{{x^2}} \over 2}} $

$xy = {x \over 2}\ell nx - {{{x^2}} \over 4} + C,$

for   $2y\left( 2 \right) = 2\ell n2 - 1$

$ \Rightarrow $  $C = 0$

$y = {x \over 2}\ell nx - {x \over 4}$

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

x $-$ y = t

$ \Rightarrow $ ${{dy} \over {dx}} = 1 - {{dt} \over {dx}}$

$ \Rightarrow $  1 $-$ ${{dt} \over {dx}}$ = t² $ \Rightarrow $  $\int {{{dt} \over {1 - {t^2}}}} $ = $\int {1dx} $

$ \Rightarrow $  ${1 \over 2}\ell n\left( {{{1 + t} \over {1 - t}}} \right) = x + \lambda $

$ \Rightarrow $  ${1 \over 2}\ell n\left( {{{1 + x - y} \over {1 - x + y}}} \right) = x + \lambda $ given y(1) = 1

$ \Rightarrow $  ${1 \over 2}\ell n(1) = 1 + \lambda \Rightarrow \lambda = - 1$

$ \Rightarrow $  $\ell n\left( {{{1 + x - y} \over {1 - x + y}}} \right)$ = 2(x $-$ 1)

$ \Rightarrow $  $ - \ell n\left( {{{1 - x + y} \over {1 + x - y}}} \right)$ = 2(x $-$ 1)

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

I.F. $ = {e^{\int {\left( {{{2x + 1} \over x}} \right)dx} }} = {e^{\int {\left( {2 + {1 \over x}} \right)dx} }} = {e^{2x + \ell nx}} = {e^{2x}}.x$

So,  $y\left( {x{e^{2x}}} \right) = \int {{e^{ - 2x}}.x{e^{2x}} + C} $

$ \Rightarrow xy{e^{2x}} = \int {xdx + C} $

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

It passes through $\left( {1,{1 \over 2}{e^{ - 2}}} \right)$ we get C $=$ 0

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

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

$ \Rightarrow f(x)$ is decreasing in $\left( {{1 \over 2},1} \right)$

$y\left( {{{\log }_e}2} \right) = {{\left( {{{\log }_e}2} \right){e^{ - 2({{\log }_e}2)}}} \over 2}$

$ = {1 \over 8}{\log _e}2$

$f'(x) = 7 - {3 \over 4}{{f\left( x \right)} \over x}\,\,\,\left( {x > 0} \right)$

Given f(1) $ \ne $ 4    $\mathop {\lim }\limits_{x \to {0^ + }} \,xf\left( {{1 \over x}} \right)\, = ?$

${{dy} \over {dx}} + {3 \over 4}{y \over x} = 7$  (This is LDE)

IF $ = {e^{\int {{3 \over {4x}}dx} }} = {e^{{3 \over 4}\ln \left| x \right|}} = {x^{{3 \over 4}}}$

$y.{x^{{3 \over 4}}} = \int {7.{x^{{3 \over 4}}}} dx$

$y.{x^{{3 \over 4}}} = 7.{{{x^{{7 \over 4}}}} \over {{7 \over 4}}} + C$

$f(x) = 4x + C.{x^{ - {3 \over 4}}}$

$f\left( {{1 \over 4}} \right) = {4 \over x} + C.{x^{{3 \over 4}}}$

$\mathop {\lim }\limits_{x \to {0^ + }} xf\left( {{1 \over x}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {4 + C.{x^{{7 \over 4}}}} \right) = 4$

(x² $-$ y²) dx + 2xy dy = 0

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

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

Solving we get,

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

ln(v² + 1) = $-$ ln x + C

(y² + x²) = Cx

1 + 1 = C $ \Rightarrow $ C = 2

y² + x² = 2x

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

I.F. = ${e^{3\int {{{\sec }^2}xdx} }} = {e^{3\tan x}}$

or   $y.e{}^{3\tan x} = \int {{{\sec }^2}x.{e^{3\tan x}}} $

or   $y.{e^{3\tan x}} = {1 \over 3}{e^{3\tan x}} + C$

Given

$y\left( {{\pi \over 4}} \right) = {4 \over 3}$

$ \therefore $   ${4 \over 3}.{e^3} = {1 \over 3}{e^3} + C$

$ \therefore $   C = e³

If f(xy) = f(x) f(y) $\forall $ x, y $ \in $ R and f(0) $ \ne $ 0

put x = y = 0

$ \Rightarrow $  f(0) = [f(0)]²

$ \Rightarrow $  f(0) = 1

put y = 0 $ \Rightarrow $ f(0) = f(x) f(0)

$ \Rightarrow $ f(x) = 1

given that ${{dy} \over {dx}}$ = f(x)

$ \therefore $  ${{dy} \over {dx}}$ = 1 $ \Rightarrow $ y = x + k

given that y(0) = 1

$ \therefore $  k = 1

hence y = x + 1

y$\left( {{1 \over 4}} \right)$ + y$\left( {{3 \over 4}} \right)$ = $\left( {{1 \over 4} + 1} \right)$ + $\left( {{3 \over 4} + 1} \right)$ = 3

Given,

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

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

This is a linear differential equation.

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

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

$ = {x^2}$

$ \therefore $  Solution is,

$y \cdot {x^2} = \int {x \cdot {x^2}dx} $

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

given  $y\left( 1 \right) = 1$

$ \therefore $  $1.1 = {4 \over 4} + C$

$ \Rightarrow $  $C = {3 \over 4}$

$ \therefore $  Equation is

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

$ \therefore $  $y\left( {{1 \over 2}} \right)$   means   $x = {1 \over 2}$

$ \therefore $  $y \cdot {\left( {{1 \over 2}} \right)^2} = {1 \over 4} \times {\left( {{1 \over 2}} \right)^4} + {3 \over 4}$

$ \Rightarrow $  ${y \over 4} = {1 \over {64}} + {3 \over 4}$

$ \Rightarrow $   ${y \over 4} = {{1 + 48} \over {64}}$

$ \Rightarrow $  y = ${{49} \over {16}}$

Let a point P(h, k) on the curve y = y(x), so equation of tangent to the curve at point P is

$y - k = {\left( {{{dy} \over {dx}}} \right)_{h,k}}(x - h)$ ....(i)

Now, the tangent (i) intersect the Y-axis at Y_p, so coordinates Y_p is $\left( {0,k - h{{dy} \over {dx}}} \right)$,

where ${{{dy} \over {dx}}}$ = ${\left( {{{dy} \over {dx}}} \right)}$_(h,k)

So, PY_p = 1 (given)

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

$ \Rightarrow {{dy} \over {dx}} = \pm {{\sqrt {1 - {x^2}} } \over x}$

[on replacing h by x]

$ \Rightarrow dy = \pm {{\sqrt {1 - {x^2}} } \over x}dx$

On puting x = sin$\theta $, dx = cos$\theta $d$\theta $, we get

$dy = \pm {{\sqrt {1 - {{\sin }^2}\theta } } \over {\sin \theta }}\cos \theta d\theta $

$ = \pm {{{{\cos }^2}\theta } \over {\sin \theta }}d\theta $

$ = \pm (\cos ec\theta - \sin \theta )d\theta $

$ \Rightarrow y = \pm [1n(\cos ec\theta - \cot \theta ) + \cos \theta ] + C$

$ \Rightarrow y = \pm \left[ {1n\left( {{{1 - \cos \theta } \over {\sin \theta }}} \right) + \cos \theta } \right] + C$

$ \Rightarrow y = \pm \left[ {1n\left( {{{1 - \sqrt {1 - {{\sin }^2}\theta } } \over {\sin \theta }}} \right) + \sqrt {1 - {{\sin }^2}\theta } } \right] + C$

$ \Rightarrow y = \pm \left[ {1n\left( {{{1 - \sqrt {1 - {x^2}} } \over x}} \right) + \sqrt {1 - {x^2}} } \right] + C$

[$ \because $ x = sin$\theta $]

$ = \pm \left[ { - 1n{{1 + \sqrt {1 - {x^2}} } \over x} + \sqrt {1 - {x^2}} } \right] + C$

[on rationalization]

$ \because $ The curve is in the first quadrant so y must be positive, so

$y = 1n\left( {{{1 + \sqrt {1 - {x^2}} } \over x}} \right) - \sqrt {1 - {x^2}} + C$

As curve passes through (1, 0), so

$0 = 0 - 0 + c \Rightarrow c = 0$, so required curve is

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

and required differential eaquation is

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

$ \Rightarrow xy' + \sqrt {1 - {x^2}} = 0$

Hence, options (a) and (c) are correct.

Equation of ellipse,

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

As ellipse passes through (0, 3)

$\therefore\,\,\,$ ${{{0^2}} \over {{a^2}}} + {{{3^2}} \over {{b^2}}} = 1$

$ \Rightarrow $    b² = 9

$\therefore\,\,\,$ Equation of ellipse becomes,

${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 9} = 1$

Differentiating w.r.t    x, we get,

${{2x} \over {a{}^2}}$ + ${{2y} \over 9}$ . ${{dy} \over {dx}} = 0$

$ \Rightarrow $    ${x \over {{a^2}}}$ = $-$ ${y \over 9}.{{dy} \over {da}}$

$ \Rightarrow $    ${x \over {{a^2}}} = - {y \over 9}.y'......$ (1)

We got earlier,

${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 9}$ = 1

$ \Rightarrow $   ${x \over {{a^2}}}.x + {{{y^2}} \over 9} = 1$

putting value of equation (1) here,

$ {-{y\,y'} \over 9}.x + {{{y^2}} \over 9} = 1$

$ \Rightarrow $    $-$ xyy' + y² = 9

$ \Rightarrow $   xyy' $-$ y² + 9 = 0

Given,

sin x ${{dy} \over {dx}} + y\cos y = 4x$

$ \Rightarrow \,\,\,\,{{dy} \over {dx}}\,$ + y cot x = 4x cosec x

This is a linear differential equation of form,

${{dy} \over {dx}}\,$ + py = Q

Where p = cot x and Q = 4x cosec x

So, Integrating factor (I. F)

= ${e^{\int {pdx} }}$

= ${e^{\int {\cot dx} }}$

= ${e^{\ln \left| {\sin \,x} \right|}}$

= sin x as $x \in \left( {0,\pi } \right)$

Solution of the differential equation is

y sin x = $\int {} $ 4x cosecx sinx dx + c

$ \Rightarrow \,\,\,\,$ y sinx = $\int {4x\,dx\, + c} $

$ \Rightarrow \,\,\,\,$ y sinx = 4.${{{x^2}} \over 2} + c$

$ \Rightarrow \,\,\,\,$ y sinx = 2x² + c . . . . . (1)

Given that, $y\left( {{\pi \over 2}} \right) = 0$

$\therefore\,\,\,$ x = $y\left( {{\pi \over 2}} \right) = 0$ and y = 0

Put this x = ${\pi \over 2}$ and y = 0 at equation (1)

0.1 = 2. $\left( {{\pi \over 2}} \right)$² + c

$ \Rightarrow \,\,\,$ c =$ - {{{\pi ^2}} \over 2}$

So, differential equation is

y sin x = 2x² $-$ ${{{\pi ^2}} \over 2}\,\,.....(2)$

Now we have to find y $\left( {{\pi \over 6}} \right).$

So, put x = ${{\pi \over 6}}$ at equation (2)

y . sin ${{\pi \over 6}}$ = 2 ${\left( {{\pi \over 6}} \right)^2} - {{{\pi ^2}} \over 2}$

$ \Rightarrow \,\,\,\,\,\,\,y.{1 \over 2}$ = 2. ${{{\pi ^2}} \over {36}} - {{{\pi ^2}} \over 2}$

$ \Rightarrow \,\,\,\,\,{y \over 2} = {{{\pi ^2}} \over {18}} - {{{\pi ^2}} \over 2}$

$ \Rightarrow \,\,\,\,{y \over 2} = {{{\pi ^2} - 9{\pi ^2}} \over {18}}$

$ \Rightarrow \,\,\,\,\,y = - {{8{\pi ^2}} \over 9}$

(x² $-$ y²) dx + 2xydy = 0

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

Let y = vx

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

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

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

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

After intergrating, we get

$\ln \left| {{v^2} + 1} \right|$ = $-$ ln$\left| x \right|$ + lnc

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

As curve passes through the point (1, 1), so 1 + 1 = c

$ \Rightarrow $  c = 2

x² + y² $-$ 2x = 0, which is a circle of radius one.

When x $ \in $ [0, 1], then ${{dy} \over {dx}}$ + 2y = 1 

$ \Rightarrow $ y = ${1 \over 2}$ + C₁e^$-$2x

$ \because $ y(0) = 0 $ \Rightarrow $ y(x) = ${1 \over 2}$ $-$ ${1 \over 2}$e^$-$2x

Here, y(1) = ${1 \over 2}$ $-$ ${1 \over 2}$ e^$-$2 = ${{{e^2} - 1} \over {2{e^2}}}$

When $x \notin \left[ {0,1} \right]$, then ${{dy} \over {dx}}$ + 2y = 0 $ \Rightarrow $ y = c₂ e^$-$2x

$ \because $  y(1) = ${{{e^2} - 1} \over 2}$ $ \Rightarrow $ ${{{e^2} - 1} \over 2}$ = c²e^$-$2 $ \Rightarrow $ C₂ = ${{{e^2} - 1} \over 2}$

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

It is given that

$2x = {y^{1/5}} + {y^{ - 1/5}}$

$ \Rightarrow 2x = {y^{1/5}} + 1/{y^{1/5}}$

Therefore, $2x = a + {1 \over a} \Rightarrow {a^2} - 2ax + 1 = 0$

$a = {{2x \pm \sqrt {4{x^2} - 4} } \over 2}$

$ \Rightarrow a = {{2x \pm 2\sqrt {{x^2} - 1} } \over 2}$

$ \Rightarrow a = x \pm \sqrt {{x^2} - 1} $

$ \Rightarrow {y^{1/5}} = x \pm \sqrt {{x^2} - 1} $

$ \Rightarrow y = {(x \pm \sqrt {{x^2} - 1} )^5}$

Therefore,

${{dy} \over {dx}} = 5{(x \pm \sqrt {{x^2} - 1} )^4}\left( {1 \pm {{2x} \over {2\sqrt {{x^2} - 1} }}} \right)$

$ = 5{(x + \sqrt {{x^2} - 1} )^4}\left( {{{\sqrt {{x^2} - 1} \pm x} \over {\sqrt {{x^2} - 1} }}} \right)$

$ \Rightarrow {{dy} \over {dx}} = {{ - 5y} \over {\sqrt {{x^2} - 1} }}$ ...... (1)

$ \Rightarrow {{{d^2}y} \over {d{x^2}}} = {{\left[ {\sqrt {{x^2} - 1} \left( { - 5{{dy} \over {dx}}} \right) - 5( - 5y){1 \over 2}{{2x} \over {\sqrt {{x^2} - 1} }}} \right]} \over {({x^2} - 1)}}$

Therefore, $({x^2} - 1){{{d^2}y} \over {d{x^2}}} = - 5\sqrt {{x^2} - 1} {{dy} \over {dx}} + 5y{x \over {\sqrt {{x^2} - 1} }}$

$ \Rightarrow ({x^2} - 1){{{d^2}y} \over {d{x^2}}} = 25y - x{{dy} \over {dx}}$

$ \Rightarrow ({x^2} - 1){{{d^2}y} \over {d{x^2}}} + 1x{{dy} \over {dx}} - 25y = 0$

Therefore, $\lambda$ = 1, k = $-$25; hence,

$\lambda + k = - 24$

Given,

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

$ \Rightarrow $$\,\,\,$ y ${{dx} \over {dy}}$ = x + 3y²

$ \Rightarrow $$\,\,\,$ ${{dx} \over {dy}}$ $-$ ${x \over y} = 3y$

If   =    ${e^{ - \int {{1 \over y}dy} }}$ = ${e^{ - \ln y}}$ = ${1 \over y}$

$\therefore\,\,\,$ Soluation is ,

x . ${1 \over y}$ = $\int {3y.{1 \over y}dy} $

$ \Rightarrow $$\,\,\,$ ${x \over y}$ = 3y + c

This curve passing through (1, 1)

$\therefore\,\,\,$ 1 = 3 + c

$ \Rightarrow $$\,\,\,$ c = $-$ 2

$\therefore\,\,\,$ Curve is, x = 3y² $-$ 2y

Now put every point in this equation, and see which point satisfy this equation.

Following this method you can see ($-$ ${1 \over 3}$, ${1 \over 3}$) point satisfy this equation.

$\left( {2 + \sin x} \right){{dy} \over {dx}} + \left( {y + 1} \right)\cos x = 0$

$ \Rightarrow $ ${d \over {dx}}\left( {2 + \sin x} \right)\left( {y + 1} \right) = 0$

On integrating, we get

(2 + sin x) (y + 1) = C

At x = 0, y = 1 we have

(2 + sin 0) (1 + 1) = C

$ \Rightarrow $ C = 4

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

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

Now $y\left( {{\pi \over 2}} \right) = {4 \over {2 + \sin {\pi \over 2}}} - 1$

= ${4 \over 3} - 1$ = ${1 \over 3}$

${{dy} \over {dx}} = {1 \over {8\sqrt x + \sqrt {9 + \sqrt x } \sqrt {4 + \sqrt {9 + \sqrt x } } }}$

$ \Rightarrow y = \sqrt {4 + \sqrt { + \sqrt x } } + c$

Now, $y(0) = \sqrt 7 + c$

$ \Rightarrow c = 0$

$y(256) = \sqrt {4 + \sqrt {9 + \sqrt {16} } } = \sqrt {4 + 5} = 3$

$g(x) = \int_{\sin x}^{\sin (2x)} {{{\sin }^{ - 1}}} (t)\,dt$

g'(x) = 2cos 2x sin^$-$1(sin2 x) $-$ cos x sin^$-$1(sin x)

$g'\left( {{\pi \over 2}} \right) = - 2{\sin ^{ - 1}}(0) = 0$

$g'\left( { - {\pi \over 2}} \right) = - 2{\sin ^{ - 1}}(0) = 0$

No option is matching.

Given,

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

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

Now, let

y² $=$ t

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

$ \therefore $    New equation,

${{dt} \over {dx}} + t\sec x = \tan x$

$ \therefore $   I.F $=$ ${e^{\int {\sec xdx} }}$

$=$   ${e^{\ln \left( {\sec x + \tan x} \right)}}$

$=$   sec x + tan x

$ \therefore $   Solution is,

t(sec x + tan x) $=$ $\int {\tan x} $ (sec x + tan x) dx

$ \Rightarrow $   t(sec x + tan x) $=$ sec x + tan x $-$ x + c

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

$ \Rightarrow $   y² $=$ 1 $-$ ${x \over {\sec x + \tan x}} + c$

Given,

y(0) $=$ 1

$ \therefore $   1 $=$ 1 $-$ 0 + c

$ \Rightarrow $   c $=$ 0

$ \therefore $   y² $=$ 1 $-$ ${x \over {\sec x + \tan x}}$

$\mathop {\lim }\limits_{t \to x} {{{t^2}f\left( x \right) - {x^2}f\left( t \right)} \over {t - x}} = 1$

It is in ${0 \over 0}$ form

So, applying L' Hospital rule,

$\mathop {\lim }\limits_{t \to x} {{2tf\left( x \right) - {x^2}f'\left( x \right)} \over 1} = 1$

$ \Rightarrow $   2xf(x) $-$ x²f '(x) = 1

$ \Rightarrow $   f '(x) $-$ ${2 \over x}$f(x) $=$ ${1 \over {{x^2}}}$

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

$ \therefore $   Solution of equation,

f(x)${1 \over {{x^2}}}$ = $\int {{1 \over x}\left( { - {1 \over {{x^2}}}} \right)} \,dx$

$ \Rightarrow $   ${{f(x)} \over {{x^2}}} = {1 \over {3{x^3}}} + C$

Given that,

f(1) = 1

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

$ \Rightarrow $   C $=$ ${2 \over 3}$

$ \therefore $   f(x) $=$ ${2 \over 3}$ x² + ${1 \over {3x}}$

$ \therefore $   f$\left( {{3 \over 2}} \right) = {2 \over 3} \times {\left( {{3 \over 2}} \right)^2} + {1 \over 3} \times {2 \over 3} = {{31} \over {18}}$

$y\left( {1 + xy} \right)dx = xdy$

${{xdy - ydx} \over {{y^2}}} = xdx \Rightarrow \int { - d\left( {{x \over y}} \right) = \int {xdx} } $

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

as $\,\,\,y\left( 1 \right) = - 1 \Rightarrow C = {1 \over 2}$

Hence, $y = {{ - 2x} \over {{x^2} + 1}} \Rightarrow f\left( {{{ - 1} \over 2}} \right) = {4 \over 5}$

Options (A) and (B): The given differential equation is

$ \left[x^2+4 x+4+y(x+2)\right] \frac{d y}{d x}-y^2=0 \quad(x>0) $

which is further simplified as follows:

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

Substituting $x+2=t$, we get

$ \frac{d x}{d y}=\frac{d t}{d y} $

Now,

$ (x+2)^2+y(x+2)-y^2 \frac{d x}{d y}=0 $

That is,

$ \begin{aligned} &t^2+y t-y^2 \frac{d t}{d y} =0 \\\\ &y^2 \frac{d t}{d y}-y t-t^2 =0 \\\\ &\frac{1}{t^2} \frac{d t}{d y}-\frac{1}{y t} =\frac{1}{y^2} \end{aligned} $

Let $\frac{1}{t}=z$; therefore,

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

Now,

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

$ \Rightarrow \frac{y}{(x+2)}=-\ln |y|+\ln c ~~~~~...(1) $

which passes through the point $(1,3)$. Therefore, from Eq. (1), we get

$ \begin{aligned} &\frac{z}{3} =-\ln 3+c \Rightarrow c=\ln 3 e \\\\ &\frac{y}{x+2} =-\ln |y|+\ln 3 e=\ln \left(\frac{3 e}{|y|}\right) \\\\ &\frac{3 e}{|y|} =e^{y /(x+2)} \\\\ &3 e =|y| e^{y /(x+2)} \end{aligned} $

Substituting $y=(x+2)$, we get $3 e=|x+2| e^1$

$ |x+2|=3 \Rightarrow x+2=-3,3 \Rightarrow x=-5,1 $

Therefore, $x=1$ (since $x \neq-5)$.


That is, the solution curve intersects $y=(x+2)$ exactly at one point and not at two points. Therefore, option (A) is correct and option (B) is incorrect.

Option (C): We have

$ \frac{3 e}{\left|(x+2)^2\right|}=e^{(x+2)} $

which meets at two points for $x<0$ and for $x>0$, there is no intersection point.

Hence, option $(\mathrm{C})$ is incorrect.

Option (D): We have

$ \frac{3 e}{(x+3)^2}=e^{\frac{(x+3)^2}{(x+2)}}=e^{\frac{(x+2)^2+1+2(x+2)}{(x+2)}}=e^{2+\frac{1}{(x+2)}+(x+2)} $

Therefore, there is no intersection point for $x>0$. Hence option (D) is correct.

$f'(x) = 2 - {{f(x)} \over x}$

$ \Rightarrow f'(x) = {1 \over x}f(x) = 2$ is linear differential equation.

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

Thus the solution is given by

$x\,.\,f(x) = \int {2x\,dx + \lambda } $

i.e., $xf(x) = {x^2} + \lambda $

As f(1) $\ne$ 1, we have $\lambda$ $\ne$ 0

$\therefore$ $f(x) = x + {\lambda \over x}$, $\lambda$ $\ne$ 0

Thus, $f'(x) = 1 - {\lambda \over {{x^2}}}$, $\lambda$ $\ne$ 0

Now, $\mathop {\lim }\limits_{x \to {0^ + }} f'\left( {{1 \over x}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} (1 - \lambda {x^2}) = 1$

$\mathop {\lim }\limits_{x \to {0^ + }} xf\left( {{1 \over x}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} x\left( {{1 \over x} + \lambda x} \right) = \mathop {\lim }\limits_{x \to {0^ + }} (1 + \lambda {x^2}) = 1$

$\mathop {\lim }\limits_{x \to {0^ + }} {x^2}f'(x) = \mathop {\lim }\limits_{x \to {0^ + }} {x^2}\left( {1 - {\lambda \over {{x^2}}}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} ({x^2} - \lambda ) = - \lambda $

Again, $\mathop {\lim }\limits_{x \to {0^ + }} f(x) \to \infty $

Hence the function is not bounded.

Note that $\lambda$ can be $-$ve or +ve.

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

$I.F. = {e^{\int {{1 \over {x\log x}}dx} }} = {e^{\log \left( {\log x} \right)}} = \log x$

$y.\log x = \int {2\,\log xdx + c} $

$y\log x = 2\left[ {x\log x - x} \right] + c$

Put $x=1,y.0=-2+c$ $ \Rightarrow c = 2$

Put $x=e$

$y\log e = 2e\left( {\log e - 1} \right) + c \Rightarrow y\left( e \right) = c = 2$

Given: $\left(1+e^x\right) y^{\prime}+y e^x=1$

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

$\frac{d y}{d x}+\frac{e^x}{1+e^x} y=\frac{1}{1+e^x}$, which is liner differential equation in $y$.

Comparing above equation with $\frac{d y}{d x}+\mathrm{P} y=Q$, we get $Q$, we get

$\begin{aligned} \quad \mathrm{P} & =\frac{e^x}{1+e^x} \text { and } \mathrm{Q}=\frac{1}{1+e^x} \\ \text { So, I.F. } & =e^{\int \mathrm{P} \cdot d x} \\ \text { I.F. } & =e^{\int \frac{e^x}{1+e^x} d x} \\ \text { I.F. } & =e^{\ln \left(1+e^x\right)} \end{aligned}$

$\left\{\because \int \frac{f^{\prime}(x)}{f(x)} d x=\ln f(x)\right\}$

$\text { I.F }=1+e^x$

So, solution of given differential equation is given by

$y \text { (I.F.) }=\int \text { Q.(I.F.) } d x$

$\begin{aligned} y \cdot\left(1+e^x\right) & =\int \frac{1}{1+e^x} \cdot\left(1+e^x\right) d x \\ y\left(1+e^x\right) & =\int 1 d x \\ y\left(1+e^x\right) & =x+\mathrm{C} \\ \because \quad y(0) & =2 \\ \Rightarrow \quad 2\left(1+e^0\right) & =0+\mathrm{C} \\ c & =4 \end{aligned}$

So, $y(x)=\frac{x+4}{1+e^x}\quad \text{... (i)}$

Put $x=-4$ in the equation, we get

$y(-4)=0$

Put $x=-2$ in the above equation (i), we get

$y(-2)=\frac{2}{1+e^{-2}} \neq 0$

For critical points, $y^{\prime}=0$

From e.q, (i), $y\left(1+e^x\right)=x+4$

Differentiating the above equation w.r.t. $x$, we get

$\begin{aligned} y^{\prime}\left(1+e^x\right)+y\left(e^x\right) & =1 \\ 0\left(1+e^x\right)+y e^x & =1 \quad \left\{\because y^{\prime}=0\right\} \\ y e^x-1 & =0 \end{aligned}$

$\quad \begin{aligned} \text { Now, } \text { let } g(x) & =y e^x-1 \\ g(x) & =\frac{(x+4) e^x}{1+e^x}-1 \\ g(x) & =\frac{(x+3) e^x-1}{1+e^x} \end{aligned}$

$\text { Now, } \quad g(-1)=\frac{2 e^{-1}-1}{1+e^{-1}}=\frac{2-e}{1+e}<0 \quad \{\because e=2 \cdot 7 \cdot 8\}$

$\text { And, } \quad g(0)=\frac{3 e^0-1}{1+e^0}=1>0$

So, there exists one value of $x$ in $(-1,0)$ for which

$\begin{aligned} g(x) & =0 \\ \Rightarrow \quad y^{\prime} & =0 \end{aligned}$

There exist a critical point of $y(x)$ in $(-1,0)$

Hint :

(i) Use solution of liner differential equation

$\begin{aligned} \frac{d y}{d x}+\mathrm{P} y & =\mathrm{Q} \text { is given by } \\ y \text { (I.F.) } & =\int \mathrm{Q} \cdot \text { (I.F) } d x, \text { where I.F }=e^{\int p \cdot d x} \end{aligned}$

(ii) For critical points, $y^{\prime}=0$

Let equation of circle whose centre lie on straight line $y=x$ be

$(x-k)^2+(y-k)^2=r^2~~~~...(i)$

Differentiating the above equation w.r.t. $x$, we get

$ \begin{aligned} & 2(x-k)+2(y-k) y^{\prime}=0 \\\\ \Rightarrow & x+y y^{\prime}=k\left(1+y^{\prime}\right)~~~~~...(ii) \\\\ \Rightarrow & k=\frac{x+y y^{\prime}}{1+y^{\prime}} \end{aligned} $

Differentiating the eq. (ii) w.r.t. $x$, we get

$ \begin{aligned} & \Rightarrow 1+y y^{\prime \prime}+\left(y^{\prime}\right)^2=k y^{\prime \prime} \quad\left\{\because(u v)^{\prime}=u v^{\prime}+v u^{\prime}\right\} \\\\ & \Rightarrow 1+y y^{\prime \prime}+\left(y^{\prime}\right)^2=\left(\frac{x+y y^{\prime}}{1+y^{\prime}}\right) y^{\prime \prime} \\\\ & \Rightarrow 1+y y^{\prime \prime}+\left(y^{\prime}\right)^2+y^{\prime}+y y^{\prime} y^{\prime \prime}+\left(y^{\prime}\right)^3 \\\\ & =x y^{\prime \prime}+y y^{\prime} y^{\prime \prime} \\\\ & \Rightarrow y^{\prime \prime}(y-x)+\left(y^{\prime}\right)^2\left(1+y^{\prime}\right)+1+y^{\prime}=0 \\\\ & \Rightarrow y^{\prime \prime}(y-x)+y^{\prime}\left(y^{\prime}+\left(y^{\prime}\right)^2+1\right)+1=0 \end{aligned} $

Comparing the above equation with $p y^{\prime \prime}+\mathrm{Q} y^{\prime}$ $+1=0$, we get

$ \begin{aligned} & P=y-x \text { and } \mathrm{Q}=y^{\prime}+\left(y^{\prime}\right)^2+1 \\\\ \therefore & P+Q=1-x+y+y^{\prime}+\left(y^{\prime}\right)^2 \end{aligned} $

Given differential equation is

${{dp\left( t \right)} \over {dt}} = {1 \over 2}p\left( t \right) - 200$

By separating the variable, we get

$dp\left( t \right) = \left[ {{1 \over 2}p\left( t \right) - 200} \right]dt$

$ \Rightarrow {{dp\left( t \right)} \over {{1 \over 2}p\left( t \right) - 200}} = dt$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ Integrating on both the sides,

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int {{{d\left( {p\left( t \right)} \right)} \over {{1 \over 2}p\left( t \right) - 200}}} = \int {dt} $

Let ${1 \over 2}p\left( t \right) - 200 = s \Rightarrow {{dp\left( t \right)} \over 2} = ds$

So, $\int {{{d\,p\left( t \right)} \over {\left( {{1 \over 2}p\left( t \right) - 200} \right)}}} = \int {dt} $

$ \Rightarrow \int {{{2ds} \over s} = \int {dt} } \Rightarrow 2\log s = t + c$

$ \Rightarrow 2\log \left( {{{p\left( t \right)} \over 2} - 200} \right) = t + c$

$ \Rightarrow {{p\left( t \right)} \over 2} - 200 = {e^{{1 \over 2}}}k$

Using given condition $p\left( t \right) = 400 - 300{e^{t/2}}$

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

The above differential equation is a linear differential equation.

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

$\begin{aligned} & \Rightarrow \text { I.F. }=e^{\frac{1}{2} \int \frac{-2 x d x}{1-x^2}} \\ & \Rightarrow \text { I.F. }=e^{\frac{1}{2} \ln \left(1-x^2\right)} \quad\left(\int \frac{f^{\prime}(x) d x}{f(x)}=\ln f(x)+c\right) \\ & \Rightarrow \text { I.F. }=e^{\ln \sqrt{1-x^2}} \\ & \Rightarrow \text { I.F. }=\sqrt{1-x^2} \end{aligned}$

Now, the solution of the given differential equation is

$\begin{aligned} & \Rightarrow y \cdot \text { (I.F. })=\int\left(\frac{x^4+2 x}{\sqrt{1-x^2}}\right)(\text { L.F. }) d x+\text { C } \\ & \Rightarrow y \sqrt{1-x^2}=\int\left(\frac{x^4+2 x}{\sqrt{1-x^2}}\right) \sqrt{1-x^2} d x+\text { C } \\ & \Rightarrow y \sqrt{1-x^2}=\frac{x^5}{5}+2 \times \frac{x^2}{2}+\text { C } \\ & \Rightarrow y=f(x)=\frac{\frac{x^5}{5}+x^2+\mathrm{C}}{\sqrt{1-x^2}} \end{aligned}$

Given, $f(0)=0$

$\begin{array}{ll} \Rightarrow & 0=\frac{0+0+C}{\sqrt{1-0}} \\ \Rightarrow & C=0 \\ \therefore & f(x)=\frac{\frac{x^5}{5}+x^2}{\sqrt{1-x^2}}=\frac{x^5}{5 \sqrt{1-x^2}}+\frac{x^2}{\sqrt{1-x^2}} \end{array}$

Let $\mathrm{I}=\int_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) d x$

$\Rightarrow \mathrm{I}=\int_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}\left(\frac{x^5}{5 \sqrt{1-x^2}}+\frac{x^2}{\sqrt{1-x^2}}\right) d x$

Here, $\frac{x^5}{5 \sqrt{1-x^2}}$ is an odd function and $\frac{x^2}{\sqrt{1-x^2}}$ is an even function.

$\begin{aligned} & \therefore \mathrm{I}=0+2 \int_0^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1-x^2}} d x \\ & \Rightarrow \mathrm{I}=2 \int_0^{\frac{\sqrt{3}}{2}}\left(\frac{x^2-1+1}{\sqrt{1-x^2}}\right) d x \\ & \Rightarrow \mathrm{I}=2 \int_0^{\frac{\sqrt{3}}{2}}\left(-\sqrt{1-x^2}+\frac{1}{\sqrt{1-x^2}}\right) d x \\ & \Rightarrow \mathrm{I}=2\left[-\left(\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1} x\right)+\sin ^{-1} x\right]_0^{\frac{\sqrt{3}}{2}} \\ & \Rightarrow \mathrm{I}=2\left[-\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1} x\right]_0^{\frac{\sqrt{3}}{2}} \\ & \Rightarrow \mathrm{I}=2\left[-\frac{\sqrt{3}}{4} \times \frac{1}{2}+\frac{1}{2} \times \frac{\pi}{3}\right] \\ & \Rightarrow \mathrm{I}=\frac{\pi}{3}-\frac{\sqrt{3}}{4} \end{aligned}$

Hint:

(i) The solution of linear differential equation $\frac{d y}{d x}+\mathrm{P}(x) \cdot y=\mathrm{Q}(x)$ is

$y \text { (I.F.) } =\int \mathrm{Q}(x) \cdot(\text { I.F. }) d x+\text { C }$

Where, $\quad \text { L.F. } =e^{\int p(x) d x}$

(ii) Recall the property

$\int_{-a}^a f(x) d x=\left\{\begin{array}{l} 2 \int_0^a f(x) d x, \text { When } f(x) \text { is an even } \\ \text { function. } \\ 0, \text { When } f(x) \text { is an odd function. } \end{array}\right.$

Given, Rate of change is ${{dp} \over {dx}} = 100 - 12\sqrt x $

$ \Rightarrow dP = \left( {100 - 12\sqrt x } \right)dx$

By intergrating $\int {dP = \int {\left( {100 - 12\sqrt x } \right)} } dx$

$\int {dP} = \int {\left( {100 - 12\sqrt x } \right)} dx$

$P = 100x - 8{x^{3/2}} + C$

Given, when $x=0$ then $P=2000$

$ \Rightarrow C = 2000$

Now when $x$$=25$

then $P = 100 \times 25 - 8 \times {\left( {25} \right)^{3/2}} + 2000$

$\,\,\,\,\,\,\,\,\,\,$ $=4500-1000$

$ \Rightarrow P = 3500$

Given, the slope of curve at $(x,y)$ is

$\begin{aligned} & \frac{y}{x}+\sec \left(\frac{y}{x}\right), x>0 . \\ \therefore \quad & \frac{d y}{d x}=\frac{y}{x}+\sec \left(\frac{y}{x}\right) \quad \text{... (i)} \end{aligned}$

Put $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$ in the equation (i)

$\begin{aligned} & \Rightarrow \quad v+x \frac{d v}{d x}=\frac{v x}{x}+\sec \left(\frac{v x}{x}\right) \\ & \Rightarrow \quad v+x \frac{d v}{d x}=v+\sec (v) \\ & \Rightarrow \quad x \frac{d v}{d x}=\sec v \\ & \Rightarrow \quad \cos v d v=\frac{d x}{x} \\ & \Rightarrow \quad \int \cos v d v=\int \frac{d x}{x} \\ & \Rightarrow \quad \sin v=\ln x+c \\ & \Rightarrow \quad \sin \left(\frac{y}{x}\right)=\ln x+c \quad \text{... (i)} \end{aligned}$

Put $x=1$ and $y=\frac{\pi}{6}$ in the equation (ii)

$\begin{aligned} \Rightarrow & & \sin \frac{\pi}{6} & =\log 1+c \\ \Rightarrow & & \frac{1}{2} & =0+c \\ \Rightarrow & & c & =\frac{1}{2} \end{aligned}$

Put $c=\frac{1}{2}$ in the equation (ii)

$\Rightarrow \quad \sin \frac{y}{x}=\log x+\frac{1}{2}$

Hints:

(i) the slope of a curve at $(x, y)$ is equal to $\frac{d y}{d x}$

(ii) For the solution of homogeneous differential equation $\frac{d y}{d x}=f\left(\frac{y}{x}\right)$ put $y=v x$ i.e. $\frac{d y}{d x}=v+x \frac{d v}{d x}$.

Given differential equation is

${{dp\left( t \right)} \over {dt}} = 0.5p\left( t \right) - 450$

$ \Rightarrow {{dp\left( t \right)} \over {dt}} = {1 \over 2}p\left( t \right) - 450$

$ \Rightarrow {{dp\left( t \right)} \over {dt}} = {{p\left( t \right) - 900} \over 2}$

$ \Rightarrow 2{{dp\left( t \right)} \over {dt}} = \left[ {900 - p\left( t \right)} \right]$

$ \Rightarrow 2{{dp\left( t \right)} \over {900 - p\left( t \right)}} = - dt$

Integrate both sides, we get

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 2\int {{{dp\left( t \right)} \over {900 - p\left( t \right)}}} = \int {dt} $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Let\,\,900 - p\left( t \right) = u$

$ \Rightarrow - dp\left( t \right) = du$

$\therefore$ $\,\,\,\,\,\,$ We have,

$2\int {{{du} \over u}} = \int {dt \Rightarrow 2\,\ln \,u = t + c} $

$ \Rightarrow 2\ln \left[ {900 - p\left( t \right)} \right] = t + c$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ when $t = 0,p\left( 0 \right) = 850$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $2\ln \left( {50} \right) = c$

$ \Rightarrow 2\left[ {\ln \left( {{{900 - p\left( t \right)} \over {50}}} \right)} \right] = t$

$ \Rightarrow 900 - p\left( t \right) = 50{e^{{t \over 2}}}$

$ \Rightarrow p\left( t \right) = 900 - 50{e^{{t \over 2}}}$

Let $p\left( {{t_1}} \right) = 0$

$0 = 900 - 50{e^{{{{t_1}} \over 2}}}$

$\therefore$ $\,\,\,\,\,\,{t_1} = 2\ln 18$

Given, $\frac{d y}{d x}-y \tan x=2 x \sec x$

Comparing with the liner differential form

$\begin{gathered} \frac{d y}{d x}+\mathrm{P} y=\mathrm{Q} \\ \Rightarrow \mathrm{P}=-\tan x \text { and } \mathrm{Q}=2 x \sec x \end{gathered}$

Now, Integrating factor (I.F) $=e^{\int P d x}$

$\begin{aligned} & =e^{\int-\tan x d x} \\ & =e^{-\log \sec x} \\ & =\cos x \end{aligned}$

$\begin{aligned} & \text { The solution will be } y \text { (I.F) }=\int Q \text { (I.F) } d x \\ & \Rightarrow \quad y \cos x=\int 2 x \sec x \cdot \cos x d x \\ & \Rightarrow \quad y \cos x=2 \frac{x^2}{2}+c \\ & \Rightarrow \quad y=x^2 \sec x+c \sec x \\ \end{aligned}$

Now, $y(0)=0$

$\begin{aligned} & \Rightarrow \quad 0=0^2 \sec 0+\mathrm{c} \mathrm{sec} 0 \\ & \Rightarrow \quad c=0 \\ & \therefore \quad y=x^2 \sec x \\ & y^{\prime}=x^2 \sec x \cdot \tan x+2 x \sec x \quad \text{[using Product Rule in Differentiation]}\\ \end{aligned}$

Now,

$\begin{aligned} y\left(\frac{\pi}{4}\right) & =\left(\frac{\pi}{4}\right)^2 \sec \frac{\pi}{4}=\frac{\pi^2}{16} \sqrt{2}=\frac{\pi^2}{8 \sqrt{2}} \\ y^{\prime}\left(\frac{\pi}{4}\right) & =\left(\frac{\pi}{4}\right)^2 \sec \frac{\pi}{4} \tan \frac{\pi}{4}+2 \cdot \frac{\pi}{4} \sec \frac{\pi}{4} \\ & =\frac{\pi^2}{16} \cdot \sqrt{2} \cdot 1+\frac{\pi}{2} \cdot \sqrt{2} \\ & =\frac{\pi^2}{8 \sqrt{2}}+\frac{\pi}{\sqrt{2}} \end{aligned}$

$\begin{aligned} y\left(\frac{\pi}{3}\right) & =\left(\frac{\pi}{3}\right)^2 \sec \frac{\pi}{3} \\ & =\frac{\pi^2}{9} \cdot 2=\frac{2 \pi^2}{9} \\ y^{\prime}\left(\frac{\pi}{3}\right) & =\left(\frac{\pi}{3}\right)^2 \sec \frac{\pi}{3} \cdot \tan \frac{\pi}{3}+2 \cdot \frac{\pi}{3} \cdot \sec \frac{\pi}{3} \\ & =\frac{\pi^2}{9} \cdot 2 \cdot \sqrt{3}+\frac{2 \pi}{3} \cdot 2 \\ & =\frac{2 \pi^2}{3 \sqrt{3}}+\frac{4 \pi}{3} \end{aligned}$