Differential Equations

$\begin{aligned} & y=\sin (\sin x), \text { then } y^{\prime}=\cos (\sin x) \cdot \cos x \\ & y^{\prime \prime}=-\sin x \cos (\sin x)-\cos ^2 x \sin (\sin x) \\ &=-\sin x \cos (\sin x)-\left(\cos ^2 x\right) y \\ &=-\sin x\left(\frac{y^{\prime}}{\cos x}\right)-\left(\cos ^2 x\right) y \\ & \Rightarrow y^{\prime \prime}+(\tan x) y^{\prime}+\left(\cos ^2 x\right) y=0 \\ & \Rightarrow \quad \quad f(x)=\tan x, g(x) \cos ^2 x \\ & \Rightarrow f(x) \cdot g(x)=\sin x \cos x=\frac{1}{2}(2 \sin x \cos x) \\ & \quad f(x) \cdot g(x)=\frac{1}{2} \sin 2 x\end{aligned}$

$\left.\left(e^x \tan y\right) d x+\left(1+e^x\right) \sec ^2 y\right) d y=0$ ...... (i)

Its of the form $M d x+N d y=0$

Now, $\frac{\partial M}{\partial y}=e^x \sec ^2 y$ and $\frac{\partial N}{\partial x}=e^x \sec ^2 y$

$\Rightarrow$ Eq. (i) is exact.

Thus, its solution will be

$\int M d x+\int N d y=C$

(treat $y$-constant) No terms of $x$ included

$\begin{aligned} \int e^x \tan y d x+\int \sec ^2 y d y & =C \\ e^x \tan y+\tan y & =C \\ \tan y\left(1+e^x\right) & =C \quad \text{......(ii)} \end{aligned}$

Since, Eq. (ii) passes through $(0, \pi / 4)$

$\tan \frac{\pi}{4}(1+1)=C \Rightarrow C=2$

From Eq. (ii), $\tan y\left(1+e^x\right)=2$

$\begin{aligned} 2 x \frac{d y}{d x}-y & =4 \\ \frac{d y}{d x}-\left(\frac{y}{2 x}\right) & =\left(\frac{4}{2 x}\right) \\ \mathrm{IF} & =e^{\int \frac{-1}{2 x} d x}=e^{\frac{\log x}{-2}}=\frac{1}{\sqrt{x}} \\ \left(\frac{1}{\sqrt{x}} y\right) & =\int \frac{2}{x \sqrt{x}} d x \Rightarrow \frac{y}{\sqrt{x}}=2\left(\frac{x^{-\frac{3}{2}+1}}{\frac{-3}{2}+1}\right)+C \\ \frac{y}{\sqrt{x}} & =2\left(\frac{x^{-\frac{1}{2}}}{-\frac{1}{2}}\right)+C \\ y & =C \sqrt{x}-4 \Rightarrow(y+4)^2=C^2 x \end{aligned}$

$\Rightarrow$ Represents a parabola.

Given, differential equation $\frac{d^2 y}{d x^2}+y=0$

A solution is function of $x$ which satisfies given differential equation.

$\begin{aligned} & \text { Option (a) } y=3 \sin x+4 \cos x \\ & \qquad \begin{aligned} \frac{d y}{d x} & =3 \cos x-4 \sin x \\ \frac{d^2 y}{d x^2} & =-3 \sin x-4 \cos x=-y \\ \Rightarrow \frac{d^2 y}{d x^2}+y & =0 \end{aligned} \end{aligned}$

Given the solution,

$ \begin{equation*} a x^2+b y^2=1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

Now differentiating Eq. (i) w.r.t. $x$, we get

$ 2 a x+2 b y \frac{d y}{d x}=0 \text { or } \frac{y}{x}\left(\frac{d y}{d x}\right)=-\frac{a}{b} $

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

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

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

This is the differential equation of order 2 and degree 1 obtained from the solution (1).

∴ Comparing with $\alpha x^2+\beta y^2=1$

$ \alpha=2, \beta=1 \Rightarrow 2 x^2+y^2=1 $

or   $ \frac{x^2}{\left(\frac{1}{2}\right)}+\frac{y^2}{1}=1 \Rightarrow a^2=\frac{1}{2}, b^2=1 $

Eccentricity, $e=\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-\frac{1 / 2}{1}}$

$ \begin{aligned} & =\sqrt{1-\frac{1}{2}} \\ \Rightarrow \quad e & =\frac{1}{\sqrt{2}} \end{aligned} $

Given, differential equation

$ \begin{equation*} x d y-y d x=\sqrt{x^2+y^2} d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

Putting, $y=v x$ and differentiating w.r.t. $x$, we get

$ \begin{equation*} d y=v d x+x d v \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{equation*} $

From Eq. (ii),

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

Integrating on both sides, we get

$ \begin{aligned} & \ln \left(v+\sqrt{1+v^2}\right)=\ln x+C \\ & \left\{\because \int \frac{d x}{\sqrt{1+x^2}}=\ln \left(x+\sqrt{1+x^2}\right)+C\right\} \end{aligned} $

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

$ \begin{aligned} & \ln \left(\frac{y}{x}+\sqrt{1+\left(\frac{y}{x}\right)^2}\right)=\ln x+C \\ & \ln \left(\frac{y}{x}+\frac{1}{x} \sqrt{x^2+y^2}\right)=\ln x+C \\ & \ln \left\{\frac{1}{x}\left(y+\sqrt{x^2+y^2}\right)\right\}-\ln x=C \\ & \ln \left\{\frac{y+\sqrt{x^2+y^2}}{x^2}\right\}=C \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

$ \left[\because \ln a-\ln b=\ln \frac{a}{b}\right] $

Given that, $y=1, x=\sqrt{3}$

$ \begin{array}{ll} \therefore & \ln \left\{\frac{1+\sqrt{(\sqrt{3})^2+1^2}}{(\sqrt{3})^2}\right\}=C \\ \Rightarrow & \ln \left\{\frac{1+2}{3}\right\}=C \\ & \ln 1=C \text { or } C=0 \quad[\because \ln 1=0] \end{array} $

Putting the value of $C$ in Eq. (iii), we get

$ \ln \left\{\frac{y+\sqrt{x^2+y^2}}{x^2}\right\}=0 $

or $\quad \frac{y+\sqrt{x^2+y^2}}{x^2}=e^0=1 \quad\left[\because e^0=1\right]$

$ \begin{aligned} & y+\sqrt{x^2+y^2}=x^2 \\ &or\,\, x^2-y=\sqrt{x^2+y^2} \end{aligned} $

Squaring on both sides, we get

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

Given differential equation,

$ \sin x \frac{d y}{d x}+y \cos x=e^{2 x}, x \in(0, \pi) $

or $\frac{d y}{d x}+y \cot x=\frac{e^{2 x}}{\sin x}$

This is linear differential equation and general form is given by

$ \frac{d y}{d x}+P(x) y=Q(x) $

Here $P(x)=\cot x, Q(x)=\frac{e^{2 x}}{\sin x}$

Integrating factor,

$ \begin{aligned} & \mathrm{IF}=e^{\int P(x) d x}=e^{\int \cot x d x}=e^{\ln \sin x} \\ & \mathrm{IF}=\sin x \end{aligned} $

Solution of linear differential equation

$ \begin{align*} y(\mathrm{IF}) & =\int Q(x) \text { IF } d x \\ y(\sin x) & =\int \frac{e^{2 x}}{\sin x} \cdot \sin x d x \\ y \sin x & =\int e^{2 x} d x \\ \Rightarrow \quad y \sin x & =\frac{e^{2 x}}{2}+C\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{align*} $

{where, $C=$ integrating constant}

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

$ \therefore \quad 0 \times \sin \frac{\pi}{2}=\frac{e^{2 \times \pi / 2}}{2}+C \Rightarrow C=-\frac{e^\pi}{2} $

Putting the value of $c$ in Eq. (i), we get

$ \begin{aligned} & y \sin x=\frac{e^{2 x}}{2}-\frac{e^\pi}{2} \\ & \text { At, } x=\frac{\pi}{6}, y\left(\frac{\pi}{6}\right) \sin \frac{\pi}{6}=\frac{e^{2 \times \frac{\pi}{6}}}{2}-\frac{e^\pi}{2} \\ & y\left(\frac{\pi}{6}\right) \times \frac{1}{2}=\frac{1}{2}\left(e^{\pi / 3}-e^\pi\right) \\ & \Rightarrow y\left(\frac{\pi}{6}\right)=e^{\pi / 3}-e^\pi \end{aligned} $

$ \begin{aligned} &\text { Given differential equation, }\\ &\begin{aligned} & \frac{d^2 y}{d x^2}+y+\left(\frac{d y}{d x}-\frac{d^3 y}{d x^3}\right)^{3 / 2}=0 \\ & \quad\left(\frac{d y}{d x}-\frac{d^3 y}{d x^3}\right)^{3 / 2}=-\left(\frac{d^2 y}{d x^2}+y\right) \end{aligned} \end{aligned} $

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

Clearly, order and degree is 3 and 3 respectively.

$ \begin{aligned} & \text { We have, } \frac{d y}{d x}=\frac{2 x-3 y+4}{3 x+2 y-7} \\ & \begin{aligned} \Rightarrow 3 x d y+2 y d y-7 d y & \\ & =2 x d x-3 y d x+4 d x \\ \Rightarrow 3 x d y+3 y d x-7 d y+2 y d y & -2 x d x-4 d x =0 \end{aligned} \end{aligned} $

$ \begin{aligned} &\Rightarrow 3 d(x y)+(2 y-7) d y-(4+2 x) d x=0\\ &\text { On integrating, we get }\\ &\begin{array}{r} 3 x y+y^2-7 y-4 x-x^2=C \\ \Rightarrow x^2-y^2-3 x y+4 x+7 y+C=0 \end{array} \end{aligned} $

We have, $\frac{d y}{d x}=\frac{x+y+1}{y-x+1}$

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

On integrating,

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

$ \begin{aligned} & \text { We have, } y=e^{a x}(\cos b x+\sin b x) \\ & \begin{aligned} \frac{d y}{d x} & =e^{a x}(-\sin b x \cdot b+\cos b x \cdot b) \\ & +a \cdot e^{a x}(\cos b x+\sin b x) \end{aligned} \\ & \begin{aligned} \frac{d y}{d x} & =b e^{a x}(\cos b x-\sin b x)+a y \ldots \text { (i) } \\ \Rightarrow \frac{d^2 y}{d x^2} & =b\left[e^{a x}(-\sin b x \cdot b-\cos b x \cdot b)\right. \\ & \left.+a \cdot e^{a x}(\cos b x-\sin b x)\right]+a \frac{d y}{d x} \end{aligned} \\ & \Rightarrow \frac{d^2 y}{d x^2} \end{aligned} $

$ \begin{array}{r} {\left[-b e^{a x}(\sin b x+\cos b x)+a e^{a x}(\cos b x-\sin b x)\right]} \\ +a \frac{d y}{d x}=0 \end{array} $

$ \begin{aligned} & \Rightarrow \quad \frac{d^2 y}{d x^2}=b \\ & {\left[(-b)(y)+a e^{a x}(\cos b x-\sin b x)\right]+a \frac{d y}{d x}} \\ & \Rightarrow \quad \frac{d^2 y}{d x^2}-a \frac{d y}{d x}+b^2 y \\ & \quad-a b e^{a x}(\cos b x-\sin b x)=0 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \frac{d^2 y}{d x^2}-a \frac{d y}{d x}+b^2 y-a\left(\frac{d y}{d x}-a y\right)=0 \\ & \Rightarrow \frac{d^2 y}{d x^2}-2 a \frac{d y}{d x}+\left(b^2+a^2\right) y=0 \end{aligned}\\ &\text { On comparing with }\\ &\begin{aligned} & \frac{d^2 y}{d x^2}-k \frac{d y}{d x}+L y=0 \\ & \quad K=2 a, L=b^2+a^2 \\ & \therefore \quad L+b k=b^2+a^2+2 a b=(a+b)^2 \end{aligned} \end{aligned} $

Given,

$ \begin{aligned} & c f^{\prime}(c)=2 f(c)-2 c^3=x f^{\prime}(x)=2 f(x)-2 x^3 \\ & \text { Put, } \quad f(x)=y \Rightarrow f^{\prime}(x)=\frac{d y}{d x} \\ & \because \quad x \frac{d y}{d x}-2 y=-2 x^3 \\ & \qquad \frac{d y}{d x}-\frac{2 y}{x}=-2 x^2 \\ & \text { Here IF }=e^{\int \frac{-2}{x} d x}=e^{-2 \log x}=\frac{1}{x^2} \end{aligned} $

∴ Solution of diff erential equation is

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

$ \begin{aligned} & \because \quad f(x)=-2 x^3+c x^2 \\ & \text { Given, } \quad f(5)=f(2) \\ & \because \quad-2(5)^3+c(5)^2=-2\left(2^3\right)+c(2)^2 \end{aligned} $

$ \begin{aligned} -250+25 c & =-16+4 c \\ 21 c & =234 \\ c & =\frac{78}{7} \\ \because \quad f(x) & =-2 x^3+\frac{78}{7} x^2 \end{aligned} $

Given,

$ c f^{\prime}(c)=2 f(c)-2 c^3=x f^{\prime}(x)=2 f(x)-2 x^3 $

Put,

$ f(x)=y \Rightarrow f^{\prime}(x)=\frac{d y}{d x} $

$ \begin{array}{r} \because \frac{d y}{d x}-2 y=-2 x^3 \\ \frac{d y}{d x}-\frac{2 y}{x}=-2 x^2 \end{array} $

Here IF $=e^{\int \frac{-2}{x} d x}=e^{-2 \log x}=\frac{1}{x^2}$

∴ Solution of diff erential equation is

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

$ \begin{aligned} & \because \quad f(x)=-2 x^3+c x^2 \\ & \text { Given, } \quad f(5)=f(2) \\ & \because \quad-2(5)^3+c(5)^2=-2\left(2^3\right)+c(2)^2 \end{aligned} $

$ \begin{aligned} -250+25 c & =-16+4 c \\ 21 c & =234 \\ c & =\frac{78}{7} \\ \because \quad f(x) & =-2 x^3+\frac{78}{7} x^2 \end{aligned} $

We have, $y=a x^2+b x+c$

Here, Number of arbitrary constants are three as $a, b$ and $c$

On differentiating both sides w.r.t.' $x$ '

$ \frac{d y}{d x}=2 a x+b $

Again, $\frac{d^2 y}{d x^2}=2 a$

Again, differentiate, w.r.t. ' $x$ '

$ \frac{d^3 y}{d x^3}=0 $

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} $