Differential Equations

The given differential equation is

$\frac{d y}{d x} = y + 7$

This is a first order linear differential equation and can be solved using an integrating factor.

Rearrange the equation to the standard form of a linear differential equation :

$\frac{d y}{d x} - y = 7$

The integrating factor is $e^{-\int dx} = e^{-x}$.

Multiplying each side of the equation by the integrating factor gives :

$e^{-x} \frac{d y}{d x} - e^{-x} y = 7e^{-x}$

The left-hand side of the equation is the derivative of $(e^{-x}y)$ with respect to $x$. So we can write the equation as :

$\frac{d}{d x}(e^{-x}y) = 7e^{-x}$

Integrate both sides with respect to $x$ :

$e^{-x}y = -7e^{-x} + C$

Multiply both sides by $e^{x}$ to isolate $y$ :

$y = -7 + Ce^{x}$

So, the general solution to the differential equation is $y = -7 + Ce^{x}$.

Now, let's apply the initial conditions to find the particular solutions :

For $y_{1}(0)=0$, we substitute into the general solution and solve for $C$ :

$0 = -7 + C$

So, $C = 7$, and the solution for $y_{1}$ is $y_{1}(x) = 7e^{x} - 7$.

For $y_{2}(0)=1$, again substitute into the general solution:

$1 = -7 + C$

So, $C = 8$, and the solution for $y_{2}$ is $y_{2}(x) = 8e^{x} - 7$.

The two curves intersect when $y_{1}(x) = y_{2}(x)$. Setting these equal and solving for $x$ gives :

$7e^{x} - 7 = 8e^{x} - 7$

$e^{x} = 0$

But $e^{x} = 0$ has no solution, because the exponential function never equals zero.

So, the curves $y=y_{1}(x)$ and $y=y_{2}(x)$ do not intersect at any point.

$y = vx$

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

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

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

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

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

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

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

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

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

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

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

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

$\therefore$ Solution is

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

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

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

$x = 1$

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

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

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

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

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

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

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

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

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

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

Putting $x = e$ in (i),

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

Given,

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

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

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

Integrating both sides, we get

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

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

Given $y(1) = e$

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

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

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

$ \Rightarrow 0 - 1 = 1 + C$

$ \Rightarrow C = - 2$

$\therefore$ Equation (1) becomes,

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

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

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

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

Now,

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

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

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

$ = 0$

Given,

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

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

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

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

Let ${y^2} = t$

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

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

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

This is linear Differential Equation.

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

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

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

When, $x=1, y=1$

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

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

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

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

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

$ \begin{aligned} & \text { } y=x a \cos (\log x)+x b \sin (\log x) \\ & \frac{d y}{d x}=a \cos (\log x)+a x \frac{(-\sin (\log x))}{x} +b \sin (\log x)+b x \frac{\cos (\log x)}{x} \end{aligned} $

$ \begin{array}{r} \Rightarrow \frac{d y}{d x}=\frac{y}{x}+(-a \sin (\log g x))+b \cos (\log x) \\ \Rightarrow \frac{d^2 y}{d x^2}=\frac{d y}{d x} \cdot \frac{1}{x}+y\left(\frac{-1}{x^2}\right) -\frac{a \cos (\log x)}{x}-\frac{b \sin (\log x)}{x} \end{array} $

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

The given differential equation is

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

above equation is in the form of homogeneous equation.

So, we put $y=v x$

$ \begin{aligned} & \Rightarrow \quad \frac{d y}{d x}=v+x \frac{d v}{d x} \text { in Eq. (i), We get, } \\ & v+x \frac{d v}{d x}=\frac{-v^2 x^2}{x^2-x(v x)-v^2 x^2} \\ & \Rightarrow x \frac{d v}{d x}=\frac{-v^2}{1-v-v^2}-v \end{aligned} $

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

On integrating both sides, we get

$ \begin{aligned} & -\left[\frac{1}{2 \times 1} \log \left|\frac{v-1}{v+1}\right|+\log v\right]=(\log x)+\log C \\ & \Rightarrow-\log \left(\sqrt{\frac{v-1}{v+1}} \cdot v\right)=\log C x \\ & \Rightarrow-\log \sqrt{\frac{y-x}{y+x}} \cdot \frac{y}{x}=\log C x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

On putting $x=2, y=1$

$ \begin{aligned} -\log \sqrt{\frac{1}{3}} \cdot \frac{1}{2} & =\log 2 C \\ \Rightarrow \quad C & =\sqrt{3} \end{aligned} $

So, solution of above differential equation is (from Eq. (ii)

$ \begin{aligned} & \sqrt{\frac{y-x}{y+x}} \cdot \frac{y}{x}=\frac{1}{\sqrt{3} x} \\ & 3 y^2(y-x)=y+x \\ & \quad-3 y^2(x-y)=x+y \end{aligned} $

On comparing it with given solution, we get

$ k=-3 $

Given, differential equation

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

This is linear differential equation

$ \begin{aligned} & P=1 / x, Q=x^2 \\ & \mathrm{IF}=e^{\int \frac{1}{x} d x}=e^{\ln x}=x \end{aligned} $

Solution $y \cdot x=\int x^2 \cdot x d x+C$

$ \Rightarrow \quad x y=\frac{x^4}{4}+C $

$y^2=4 a(x-a)$

Differentiating, $2 y \frac{d y}{d x}=4 a \Rightarrow y \frac{d y}{d x}=2 a$

Again differentiating,

$ y \frac{d^2 y}{d x^2}+\frac{d y}{d x} \cdot \frac{d y}{d x}=0 \Rightarrow y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2=0 $

Order $m=1$

Degree $n=2$

$ \therefore \quad m+n^2=1+(2)^2=1+4=5 $

$\frac{d y}{d x}=\frac{2 x+3 y}{3 x-2 y}$

On putting, $y=v x$

$ \begin{aligned} & \frac{d y}{d x}=v+x \frac{d v}{d x} \\ \Rightarrow & v+x \frac{d v}{d x}=\frac{2 x+3 v x}{2 x-2 v x} \\ \Rightarrow & v+x \frac{d v}{d x}=\frac{2+3 v}{3-2 v} \\ \Rightarrow & v-\frac{2+3 v}{3-2 v}=\frac{-x d v}{d x} \\ \Rightarrow & \frac{3 v-2 v^2-2-3 v}{3-2 v}=-x \frac{d v}{d x} \\ \Rightarrow & \frac{-2 v^2-2}{3-2 v}=-x \frac{d v}{d x} \\ \Rightarrow & \frac{-d x}{x}=\frac{3-2 v}{-2 v^2-2} d v \end{aligned} $

$ \begin{aligned} &\text { On integrating both sides, we get }\\ &\begin{aligned} &-\int \frac{d x}{x}=\int \frac{2 v-3}{2 v^2+2} d v \\ & \Rightarrow-\log |x|=\int \frac{2 v}{2 v^2+2} d v-\int \frac{3}{2 v^2+2} d v \\ & \Rightarrow-\log |x|=\frac{1}{2} \log \left|v^2+1\right|-\frac{3}{2} \tan ^{-1} v+C \\ & \Rightarrow-\log |x|=\frac{1}{2} \log \left|(y / x)^2+1\right| -\frac{3}{2} \tan ^{-1}(y / x)+C \\ & \Rightarrow \frac{3}{2} \tan ^{-1}(y / x)=\log \left(\frac{y^2+x^2}{x^2}\right)^{1 / 2}+\log x+C \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \frac{3}{2} \tan ^{-1}(y / x)=\log \left(y^2+x^2\right)^{1 / 2} -\log x+\log x+C \\ & \Rightarrow 3 \tan ^{-1}(y / x)=2 \log \left(y^2+x^2\right)^{1 / 2} \\ & \Rightarrow \quad 3 \tan ^{-1}(y / x)=\log \left(y^2+x^2\right)^{\frac{1}{2} \times 2} \\ & \Rightarrow \quad \tan ^{-1}(y / x)=\frac{1}{3} \log \left(y^2+x^2\right) \\ & \Rightarrow \quad y / x=\tan \left[\frac{1}{3} \log \left(y^2+x^2\right)\right] \\ & \Rightarrow \quad y=x \tan \left[\frac{1}{3} \log \left(y^2+x^2\right)\right] \end{aligned}\\ &\text { Thus, } f(x)=\frac{1}{3} \log \left(x^2+y^2\right) \end{aligned} $

$\left(x^2+2\right) d y+2 x y d x=e^x\left(x^2+2\right) d x$

Dividing on both sides by ' $d x^{\prime}$ ',

$ \begin{aligned} & \left(x^2+2\right) \frac{d y}{d x}+2 x y=e^x\left(x^2+2\right) \\ \Rightarrow & \left(x^2+2\right) \cdot \frac{d y}{d x}+y \cdot \frac{d}{d x}\left(x^2+2\right)=e^x\left(x^2+2\right) \\ \Rightarrow & \frac{d}{d x}\left[y \cdot\left(x^2+2\right)\right]=e^x\left(x^2+2\right) \end{aligned} $

Integrating on both sides, we get

$ \begin{aligned} & \int \frac{d}{d x}\left[y \cdot\left(x^2+2\right)\right] \cdot d x=\int e^x \cdot\left(x^2+2\right) \cdot d x \\ & \begin{array}{r} \Rightarrow \quad\left(x^2+2\right) y=\left(x^2+2\right) \\ \quad \int e^x \cdot d x-\int\left[\frac{d}{d x}\left(x^2+2\right) \cdot \int e^x \cdot d x\right] d x \end{array} \\ & =\left(x^2+2\right) e^x-\int 2 x \cdot e^x \cdot d x \\ & =\left(x^2+2\right) e^x-2 \int x \cdot e^x d x \\ & =\left(x^2+2\right) e^x-2\left[x \cdot \int e^x d x\right. \left.\quad-\left[\int \frac{d}{d x}(x) \cdot \int e^x d x\right] d x\right] \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\left(x^2+2\right) e^x-2\left[x \cdot e^x-\int e^x \cdot d x\right] \\ & =\left(x^2+2\right) e^x-2\left[e^x(x-1)\right] \\ & =e^x\left[x^2+2-2 x+2\right] \\ & =e^x\left(x^2-2 x+4\right) \end{aligned}\\ &\text { Hence, }\left(x^2+2\right) y=e^x\left(x^2-2 x+4\right)+C \end{aligned} $

Given,

$ \begin{aligned} & (3 x-4 y)(d x-3 d y)+(6 d x-4 d y)=0 \\ & \Rightarrow(3 x-4 y+6) d x+(-9 x+12 y-4) d y=0 \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{3 x-4 y+6}{9 x-12 y+4} \end{aligned} $

Let $\quad 3 x-4 y=z$

$ \therefore \quad 3-4 \frac{d y}{d x}=\frac{d z}{d x} \Rightarrow \frac{d y}{d x}=\frac{1}{4}\left(3-\frac{d z}{d x}\right) $

Now, $\frac{1}{4}\left(3-\frac{d z}{d x}\right)=\frac{z+6}{3 z+4}$

$ \begin{aligned} \Rightarrow \quad 3-\frac{d z}{d x} & =\frac{4 z+24}{3 z+4} \\ \Rightarrow \quad \frac{d z}{d x} & =3-\frac{4 z+24}{3 z+4} \\ & =\frac{9 z+12-4 z-24}{3 z+4} \\ & =\frac{5 z-12}{3 z+4} \end{aligned} $

$ \begin{aligned} &\Rightarrow \frac{3 z+4}{5 z-12} d z=d x \Rightarrow \frac{1}{5}\left[3+\frac{56}{5 z-12}\right] d z=d x\\ &\text { On integrating both sides, we get }\\ &\frac{1}{5}\left[3 z+\frac{56}{5} \ln |5 z-12|\right]=x+A \end{aligned} $

$ \begin{array}{ll} \Rightarrow {\left[3(3 x-4 y)+\frac{56}{5} \ln |5(3 x-4 y)-12|\right]} \\ =5 x+5 A \\ \Rightarrow 9 x-12 y+\frac{56}{5} \ln |15 x-20 y-12| \\ =5 x+5 A \\ \Rightarrow 45 x-60 y+56 \ln |15 x-20 y-12| \\ \Rightarrow \begin{array}{l} 25 x+25 A \\ 20 x-60 y+56 \ln |15 x-20 y-12| \\ =25 A \end{array} \\ \Rightarrow 5 x-15 y+14 \ln |15 x-20 y-12|=C \\ {\left[\text { where, } C=\frac{25}{4} A\right]} \end{array} $

$ \begin{aligned} & \text { Given, }(\sec x+\tan x) \frac{d y}{d x}+\left(\sec ^2 x\right. \\ & +\sec x \tan x) y=1 \\ & \Rightarrow \quad(\sec x+\tan x)\left[\frac{d y}{d x}+y \cdot \sec x\right]=1 \\ & \Rightarrow \quad \frac{d y}{d x}+y \cdot \sec x=\frac{1}{(\sec x+\tan x)} \end{aligned} $

Eq. (i) is the linear differential equation of the form $\frac{d y}{d x}+P y=Q$.

where $P=\sec x$ and $Q=\frac{1}{\sec x+\tan x}$

$ \text { } \begin{array}{rlrl} & \text { Now, IF } =e^{\int P d x}=e^{\int \sec x d x} \\ & = e^{\log |\sec x+\tan x|} \\ \Rightarrow & \text { IF } =\sec x+\tan x \end{array} $

∴ Solution is $y \times$ IF $=\int$ IF $\times Q d x+C$

$ \begin{aligned} & \Rightarrow y(\sec x+\tan x)=\int 1 d x+C \\ & \Rightarrow y(\sec x+\tan x)=x+C \end{aligned} $

we have, $y=A e^x+B \sin 2 x$

$ \begin{aligned} d y / d x & =A e^x+2 B \cos 2 x \\ d^2 y / d x^2 & =A e^x-4 B \sin 2 x \end{aligned} $

On substituting these values in

$ \begin{aligned} & \text { option (a) } \\ & \begin{array}{r} \text { LHS }=(\cos 2 x-\sin 2 x)\left(A e^x-4 B \sin 2 x\right) \\ +4 \sin 2 x\left(A e^x+2 B \cos 2 x\right) \\ -4(\sin 2 x+\cos 2 x)\left(A e^x+B \sin 2 x\right) \\ =A e^x(\cos 2 x)-4 B \sin 2 x \cos 2 x-A e^x \sin 2 x \\ +4 B \sin ^2 2 x+4 A e^x \sin 2 x+8 B \sin 2 x \cos 2 x \\ -4 A e^x \sin 2 x-4 B \sin ^2 2 x-4 A e^x \cos 2 x \\ -4 B \sin 2 x \cos 2 x \neq 0 \end{array} \end{aligned} $

No other option is also satisfying the equation.

$\frac{d y}{d x}=\sin (x-y)+\cos (x-y)$

$ \begin{aligned} & \Rightarrow 1-\frac{d y}{d x}=\frac{d v}{d x} \Rightarrow \frac{d y}{d x}=1-\frac{d v}{d x} \\ & \therefore 1-\frac{d v}{d x}=\sin v+\cos v \\ & \Rightarrow(1-\sin v-\cos v)=\frac{d v}{d x} \\ & \Rightarrow d x=\frac{d v}{1-\sin v-\cos v} \\ & \Rightarrow d x=\frac{d v}{1-\frac{2 \tan \frac{v}{2}}{1+\tan ^2 \frac{v}{2}}-\frac{1-\tan ^2 \frac{v}{2}}{1+\tan ^2 \frac{v}{2}}} \end{aligned} $

$ \begin{aligned} & \Rightarrow d x=\frac{\sec ^2 \frac{v}{2} d v}{1+\tan ^2 \frac{v}{2}-2 \tan \frac{v}{2}-1+\tan ^2 \frac{v}{2}} \\ & \Rightarrow \quad d x=\frac{\sec ^2 \frac{v}{2} d v}{2 \tan ^2 \frac{v}{2}-2 \tan \frac{v}{2}} \\ & \Rightarrow \quad \int d x=\int \frac{\sec ^2 \frac{v}{2} d v}{2 \tan ^2 \frac{v}{2}-2 \tan \frac{v}{2}} \\ & \begin{aligned} \text { Let } & \tan \frac{v}{2}=t \\ \Rightarrow & \sec \mathrm{c}^2 \frac{v}{2} d v=2 d t \\ \Rightarrow & x+C=\int \frac{d t}{t(t-1)}=\int\left(\frac{-1}{t}+\frac{1}{t-1}\right) d t \\ & =\log |(t-1)|-\log |t|=\log \left|\frac{(t-1)}{t}\right| \\ & =\log \left|\frac{\tan \frac{v}{2}-1}{\tan \frac{v}{2}}\right| \\ & x+C=\log \left|\frac{\tan \frac{x-y}{2}-1}{\tan \frac{x-y}{2}}\right| \end{aligned} \end{aligned} $

$ \begin{aligned} &\text {Given, differential equation }\\ &\begin{aligned} & x^2 d y-\left(x y-y^2\right) d x=0 \\ & \quad \frac{d y}{d x}=\frac{x y-y^2}{x^2} \quad \quad[\text { put } y=v x] \\ & \Rightarrow \quad \frac{d y}{d x}=v+x \frac{d v}{d x} \\ & \therefore \quad v+x \frac{d v}{d x}=\frac{x^2 v-x^2 v^2}{x^2} \\ & \Rightarrow \quad x \frac{d v}{d x}=v-v^2-v \Rightarrow x \frac{d v}{d x}=-v^2 \\ & \Rightarrow \quad \frac{d x}{x}=-\frac{1}{v^2} d v \Rightarrow \log x=\frac{1}{v}+C \\ & \Rightarrow \quad \log x=\frac{x}{y}+C \Rightarrow y \log x=x+C y \end{aligned} \end{aligned} $

Now, equation of family of parabola whose axis is the $X$-axis i.e $y^2=4 a(x+\alpha)$ where, $a$ and $\alpha$ are constant.

Now, Differentiate on both sides w.r.t.

$ \begin{aligned} 2 y \frac{d y}{d x} & =4 a \Rightarrow y \frac{d y}{d x}=2 a \\ \Rightarrow y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2 & =0 \end{aligned} $

Thus, order 2 and degree 1 .

$ \begin{aligned} &\text { Given, differential equation is }\\ &\begin{aligned} & \left(x \sin \frac{y}{x}\right) d y=\left(y \sin \frac{y}{x}-x\right) d x \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{y \sin \frac{y}{x}-x}{x \sin \frac{y}{x}} \\ & \text { Let } \quad y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \\ & \Rightarrow \quad v+x \frac{d v}{d x}=\frac{v x \sin v-x}{x \sin v} \\ & \Rightarrow \quad v+x \frac{d v}{d x}=\frac{v \sin v-1}{\sin v} \\ & \Rightarrow \quad x \frac{d v}{d x}=\frac{v \sin v-1}{\sin v}-v=-\frac{1}{\sin v} \\ & \Rightarrow \quad \int \sin v d v=-\int \frac{d x}{x}-C \\ & \Rightarrow \quad-\cos v=-\log |x|-C \\ & \Rightarrow \quad \cos \left(\frac{y}{x}\right)=\log |x|+C \end{aligned} \end{aligned} $

The given differential equation is

$ \begin{aligned} & \begin{array}{l} \Rightarrow \quad\left(2 x-10 y^3\right) d y+y d x=0 \\ \Rightarrow \quad \frac{d x}{d y}=-\frac{2 x}{y}+10 y^2 \end{array} \\ & \Rightarrow \quad \frac{d x}{d y}+\frac{2 x}{y}=10 y^2 \\ & \text { Now, IF }=e^{\int \frac{2}{y} d y}=e^{2 \log y}=y^2 \end{aligned} $

solution is

$ \begin{aligned} x \cdot y^2 & =\int\left(10 y^2\right) \cdot\left(y^2\right) d y+C \\ & =\int 10 \cdot y^4 d y+C \\ & =2 y^5+C \\ \Rightarrow x y^2-2 y^5 & =C \end{aligned} $

Let $a$ be any arbitrary constant

$ \begin{aligned} P F & =P A \\ \Rightarrow \quad \sqrt{x^2+y^2} & =(x-a) \end{aligned} $

$ \begin{aligned} \Rightarrow \quad x^2+y^2 & =x^2+a^2-2 a x \\ y^2 & =a^2-2 a x \\ 2 y \frac{d y}{d x} & =-2 a \\ a & =-y \frac{d y}{d x} \end{aligned} $

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

$ \begin{aligned} y^2 & =y^2\left(\frac{d y}{d x}\right)^2+2 x y \frac{d y}{d x} \\ m & =1, n=2 \\ m n-m+n & =3 \end{aligned} $

$\begin{aligned} & \ \frac{d y}{d x}+y f^{\prime}(x)-f(x) f^{\prime}(x)=0 \\ & \Rightarrow \frac{d y}{d x}+y \frac{d f(x)}{d x}-f(x) \frac{d f(x)}{d x}=0\end{aligned}$

$\begin{aligned} & \begin{array}{l}\Rightarrow \frac{d y}{d f(x)}+y=f(x) \\ \text { Integrating factor }=e^{\int d f(x)}=e^{f(x)} \\ e^{f(x)} y=\int f(x) e^{f(x)} d f(x) \\ e^{f(x)} y=f(x) \int e^{f(x)} d f(x) \\ \quad-\int\left(\int e^{f(x)} d x \cdot \frac{d f(x)}{d f(x)}\right) d f(x)\end{array} \\ & \begin{aligned} e^{f(x)} y & =f(x) e^{f(x)}-\int e^{f(x)} d f(x)+c\end{aligned} \\ & \begin{aligned} e^{f(x)} y & =f(x) e^{f(x)}-e^{f(x)}+c \\ y & =f(x)-1+c e^{-f(x)}\end{aligned}\end{aligned}$

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

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

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

$\therefore$ Solution

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

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

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

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

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

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

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

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

Solution of differential equation

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

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

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

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

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

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

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

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

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

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

Differentiate w.r.t x

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

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

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

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

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

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

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

Let ${y \over x} = v$

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

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

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

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

$ \Rightarrow C = 0$

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

At $x = 2$,

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

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

OR $y = {3 \over 2}$

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

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

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

Integrating factor

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

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

Solution of differential equation

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

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

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

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

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

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

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

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