Differential Equations

$\because y=A t^2+\frac{B}{t}$

$\Rightarrow y^{\prime}=2 A t-B t^{-2}$ and $y^{\prime \prime}=2 A+2 B t^{-3}$

Substituting into the differential equation

$ \begin{aligned} & f(t)\left(2 A+2 B t^{-3}\right)+g(t)\left(2 A t-B t^{-2}\right) +h(t)\left(A t^2+B t^{-1}\right)=0 \\ & \Rightarrow A\left[2 f(t)+2 t g(t)+t^2 h(t)\right] +B\left[2 t^{-3} f(t)-t^{-2} g(t)+t^{-1} h(t)\right]=0 \end{aligned} $

the coefficients must be zero.

So, $\quad 2 f(t)+2 t g(t)+t^2 h(t)=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and $2 t^{-3} f(t)-t^{-2} g(t)+t^{-1} h(t)=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Multiply Eq. (ii) by $t^3$

$ 2 f(t)-t g(t)+t^2 h(t)=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

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

$ \begin{gathered} {\left[2 f(t)+2 t g(t)+t^2 h(t)\right]} \\ -\left[2 f(t)-t g(t)+t^2 h(t)\right]=0 \Rightarrow 3 t g(t)=0 \end{gathered} $

$\because t \neq 0$, so put $g(t)=0$ in Eq. (i)

$ 2 f(t)+t^2 h(t)=0 $

$\because \quad g(t)=0$, so $g(t) \cdot f(t)=0$

Hence, $2 f(t)+t^2 h(t)=g(t) \cdot f(t)$

$\because(2 x-y)^2 d y-2(2 x-y)^2 d x-2 d x=0$

put $v=2 x-y \Rightarrow d v=2 d x-d y$

$\Rightarrow d y=2 d x-d v$

So, $v^2(2 d x-d v)-2 v^2 d x-2 d x=0$

$\Rightarrow \quad 2 v^2 d x-v^2 d v-2 v^2 d x-2 d x=0$

$\Rightarrow-v^2 d v-2 d x=0 \Rightarrow v^2 d v+2 d x=0$

$\Rightarrow \int v^2 \cdot d v=-2 \int d x \Rightarrow \frac{v^3}{3}=-2 x+c$

$\Rightarrow \quad \frac{(2 x-y)^3}{3}=-2 x+c$

$\Rightarrow \quad(2 x-y)^3=-6 x+3 c$

$\Rightarrow \quad(2 x-y)^3+6 x=C$, where $C=3 x$

$x \log x d y=(x \log x-y) d x$

$ \begin{aligned} & \Rightarrow \quad x \log x \frac{d y}{d x}=x \log x-y \\ & \Rightarrow \quad \frac{d y}{d x}+\frac{y}{x \log x}=1 \end{aligned} $

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

$ \therefore \mathrm{IF}=e^{\int \frac{1}{x \log x} d x}=e^{\log |\log x|}=\log x $

Multiply IF both sides

$ \begin{array}{ll} & \log x \cdot \frac{d y}{d x}+\frac{y}{x}=\log x \\ \Rightarrow & \frac{d}{d x}(y \log x)=\log x \\ \Rightarrow & y \log x=\int \log x \cdot d x \\ \Rightarrow & y \log x=x \log x-x+C \\ \Rightarrow & \log x(y-x)+x=C \end{array} $

$ \begin{aligned} &\text { We have, }\left(x \sin \frac{y}{x}\right) d y\\ &\begin{aligned} & =\left(y \sin \frac{y}{x}-x\right) d x \\ \Rightarrow \quad \frac{d y}{d x} & =\frac{\frac{y}{x} \cdot \sin \frac{y}{x}-1}{\sin \frac{y}{x}} \end{aligned} \end{aligned} $

$ \begin{aligned} &=\frac{y}{x}-\frac{1}{\sin \frac{y}{x}}\\ &\text { Let } y=v x\\ &\frac{d y}{d x}=v+x \frac{d v}{d x} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad v+x \frac{d v}{d x}=v-\frac{1}{\sin v} \\ & \Rightarrow \quad x \frac{d v}{d x}=-\frac{1}{\sin v} \\ & \Rightarrow \quad \int-\sin v d v=\int \frac{d x}{x} \\ & \Rightarrow \quad \cos v=\ln x+C \\ & \Rightarrow \quad \cos \left(\frac{y}{x}\right)=\ln x+C \\ & \Rightarrow \quad \cos \frac{y}{x}=\log |x|+C \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \cos (x+y) d y=d x \\ & \qquad \frac{d x}{d y}=\cos (x+y) \\ & \text { Let } \quad x+y=t \\ & \Rightarrow \quad 1+\frac{d x}{d y}=\frac{d t}{d y} \Rightarrow \frac{d t}{d y}-1=\cos t \\ & \Rightarrow \quad \frac{d t}{d y}=1+\cos t \\ & \Rightarrow \int \frac{d t}{1+\cos t}=\int d y \\ & \Rightarrow \frac{1}{2} \int \sec ^2 \frac{t}{2} d t=\int d y \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \frac{1}{2} \cdot \frac{\tan \frac{t}{2}}{\frac{1}{2}}=y+C \\ & \Rightarrow \quad y=\tan \left(\frac{x+y}{2}\right)+C \end{aligned} $

We have, $A x^3+B x y=4$

Differentiate w.r.t $x$, we get

$ 3 A x^2+B y+B x \frac{d y}{d x}=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Differentiate w.r.t. $x$, we get

$ \begin{aligned} & 6 A x+B \frac{d y}{d x}+B \frac{d y}{d x}+B x \frac{d^2 y}{d x^2}=0 \\ & 6 A x+2 B \frac{d y}{d x}+B x \frac{d^2 y}{d x^2}=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

Multiply Eq. (i) 2 and Eq. (ii) by $x$ and subtract them

$ \begin{aligned} & 2 B y+2 B x \frac{d y}{d x}-2 B x \frac{d y}{d x}-B x^2 \frac{d^2 y}{d x^2}=0 \\ & \Rightarrow \quad 2 B y-B x^2 \frac{d^2 y}{d x^2}=0 \\ & \Rightarrow \quad 2 y-x^2 \frac{d^2 y}{d x^2}=0 \\ & \Rightarrow \quad x^2 \frac{d^2 y}{d x^2}-2 y=0 \\ & \Rightarrow \quad F(x)=x^2 \\ & \quad\,\,\,\,\,\,\,\,\, G(x)=0 \\ & \therefore F(1)+G(1)=1 \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & y=\tan (a x+b) \\ & \Rightarrow \tan ^{-1} y=a x+b \\ & \Rightarrow \frac{1}{1+y^2} \frac{d y}{d x}=a \\ & \Rightarrow \frac{1}{\left(1+y^2\right)} \frac{d^2 y}{d x^2}+\frac{d y}{d x} \frac{-1}{\left(1+y^2\right)^2} \cdot 2 y \frac{d y}{d x}=0 \\ & \Rightarrow\left(1+y^2\right) \frac{d^2 y}{d x^2}-2 y\left(\frac{d y}{d x}\right)^2=0 \\ & \Rightarrow\left(1+y^2\right) y_2-2 y y_1^2=0 \end{aligned} \end{aligned} $

Given, $x y(y+2) d y+\left(y^3-1\right) d x=0$

$ \begin{aligned} & \left(y^3-1\right) \frac{d x}{d y}-x y(y+2)=0 \\ & \Rightarrow \frac{d x}{d y}=\frac{x y(y+2)}{y^3-1} \\ & \Rightarrow-\frac{d x}{x}=\frac{y(y+2)}{y^3-1} d y \\ & \Rightarrow-\int \frac{d x}{x}=\int \frac{y(y+2)}{(y-1)\left(y^2+y+1\right)} d y \end{aligned} $

Now, $\frac{y(y+2)}{(y-1)\left(y^2+y+1\right)}=\frac{A}{y-1}+\frac{B(2 y+1)+C}{y^2+y+1}$

$ \begin{aligned} &\begin{aligned} \Rightarrow y^2 & +2 y=A\left(y^2+y+1\right) \\ & +B(y-1)(2 y+1)+C(y-1) \end{aligned}\\ &\text { By comparing coefficients }\\ &1=A+2 B|2=A-B+C| 0=A-B-C \end{aligned} $

$ \begin{aligned} &\text { Solving, we get }\\ &\begin{gathered} 2=2 C \Rightarrow C=1 \\ A+2 B=1 \\ A-B=1 \\ \frac{-\,\,+\,\,-}{B=0 \Rightarrow A=1} \\ \Rightarrow-\ln x=\int \frac{1}{y-1}+\frac{1}{y^2+y+1} \\ =\int\left(\frac{1}{y-1}+\frac{1}{\left(y+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\right)dy \\ \Rightarrow-\ln x=\ln (y-1)+\frac{2}{\sqrt{3}} \tan ^{-1} \left(\frac{2 y+1}{\sqrt{3}}\right)+C \end{gathered} \end{aligned} $

$ \begin{aligned} & \Rightarrow C=\ln |x(y-1)|+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 y+1}{\sqrt{3}}\right) \\ & \Rightarrow \ln |x y-x|+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 y+1}{\sqrt{3}}\right)= C\end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} \left(1+\sin ^2 x\right) \frac{d y}{d x}+(\sin 2 x) y & = \cos x+\sin ^2 x \cos x \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \frac{d y}{d x}+\left(\frac{\sin 2 x}{1+\sin ^2 x}\right) y \\ & \quad=\frac{\cos x\left(1+\sin ^2 x\right)}{1+\sin ^2 x} \\ & \begin{aligned} \Rightarrow \frac{d y}{d x} & +\left(\frac{\sin 2 x}{1+\sin ^2 x}\right) y=\cos x \\ \therefore \quad & P=\frac{\sin 2 x}{1+\sin ^2 x}, Q=\cos x \end{aligned} \\ & \begin{aligned} \therefore \mathrm{IF}= & e^{\int P d x}=e^{\int \frac{2 \sin x \cos x}{1+\sin ^2 x} d x} \\ & =e^{\ln \left|1+\sin ^2 x\right|} \\ & =1+\sin ^2 x \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { ∴ Solution of differential equation }\\ &\begin{gathered} y(\text { IF })=\int Q(\text { IF }) d x+C \\ \Rightarrow y\left(1+\sin ^2 x\right) \\ =\int \cos x\left(1+\sin ^2 x\right) d x \\ =\int \cos x d x+\int \cos x \sin ^2 x d x \\ y\left(1+\sin ^2 x\right)=\sin x+\frac{\sin ^3 x}{3}+C \end{gathered} \end{aligned} $

Since, the slope of tangent to a curve at any point $(x, y)$ is $x+y$.

$ \therefore \quad \frac{d y}{d x}=x+y \Rightarrow \frac{d y}{d x}-y=x $

This is a linear differentiation equation

So, $P=-1$ and $Q=x$

Now, $\mathrm{IF}=e^{\int P d x}=e^{\int-d x}=e^{-x}$

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

Integrating both sides, we get

$ \begin{aligned} y \cdot e^{-x} & =\int x e^{-x} d x \\ & =-x e^{-x}-\int-e^{-x} d x \\ & =-x e^{-x}+\left(-e^{-x}\right)+c \end{aligned} $

where $C=$ constant of integration $\Rightarrow y e^{-x}=-x e^{-x}-e^{-x}+c$

$\Rightarrow y=-x-1+c e^x$, where $c$ is arbitrary constant.

Given, differential equation is

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

Integrating to both sides, we get

$ \begin{gathered} \int \frac{(y+1)}{y^2} d y=-\int \frac{(x+1)^2}{x^2} d x \\ \Rightarrow \quad \int\left(\frac{1}{y}+\frac{1}{y^2}\right) d y \\ =-\int\left(1+\frac{2}{x}+\frac{1}{x^2}\right) d x \\ \Rightarrow \ln |y|-\frac{1}{y}=-\left(x+2 \ln |x|-\frac{1}{x}\right)+C, \end{gathered} $

where $C=$ constant

$ \begin{array}{ll} \Rightarrow & \ln |y|+2 \ln |x|=\frac{1}{y}+\frac{1}{x}-x+C \\ \Rightarrow & \ln \left|x^2 y\right|=\frac{1}{y}+\frac{1}{x}-x+C \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

Given, $y(1)=2$, so we get

$ \begin{aligned} & \Rightarrow \quad \ln \left|1^2 \cdot 2\right|=\frac{1}{2}+\frac{1}{1}-1+C \\ & \Rightarrow \quad \ln |2|=\frac{1}{2}+C \Rightarrow c=\ln |2|-\frac{1}{2} \end{aligned} $

From, Eq. (i), we get

$ \begin{array}{ll} & \ln \left|x^2 y\right|=\frac{1}{y}+\frac{1}{x}-x+\ln |2|-\frac{1}{2} \\ \Rightarrow & \ln \left|x^2 y\right|-\ln |2|=\frac{1}{y}+\frac{1}{x}-x-\frac{1}{2} \\ \Rightarrow & \ln \left|\frac{1}{2} x^2 y\right|=\frac{1}{y}+\frac{1}{x}-x-\frac{1}{2} \end{array} $

Given, differential equation

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

Let $x=X+h, y=Y+k$, then

$ \begin{aligned} \frac{d Y}{d X} & =\frac{2(X+h)+(Y+k)-3}{2(Y+k)-(X+h)+3} \\ & =\frac{2 X+Y+2 h+k-3}{2 Y-X+2 k-h+3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Let $2 h+k-3=0$ and $-h+2 k+3=0$

Solving these two equations, we get $h=\frac{9}{5}$ and $k=\frac{-3}{5}$

So, Eq. (i) becomes,

$ \frac{d Y}{d X}=\frac{2 X+Y}{2 Y-X} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Let $Y=v X$

$ \Rightarrow \frac{d Y}{d X}=v+X \frac{d v}{d X^{\prime}} $

Substituting this into Eq. (ii), we get

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

Integrating both sides, we get

$ \int \frac{2 v-1}{-2 v^2+2 v+2} d v=\int \frac{d X}{X} $

Let $u=-2 v^2+2 v+2$

Then $d u=(-4 v+2) d v$

$ \therefore \frac{-1}{2} \int \frac{d u}{u}=\int \frac{d X}{X} \Rightarrow \frac{-1}{2} \ln |u|=\ln |X|+c_1, $

where $c_1=$ constant

$ \begin{aligned} & \Rightarrow \quad \frac{-1}{2} \ln \left|-2 v^2+2 v+2\right|=\ln |X|+c_1 \\ & \Rightarrow \quad \ln \left|-2 v^2+2 v+2\right|=-2 \ln |X|+c_2 \end{aligned} $

where $c_2=-2 c_1$

$ \begin{aligned} & \Rightarrow \quad \ln \left|X^2\left(-2 v^2+2 v+2\right)\right|=c_2 \\ & \Rightarrow \quad X^2\left(-2 v^2+2 v+2\right)=c, \text { where } c=e^{c_2} \\ & \Rightarrow \quad X^2\left(\frac{-2 Y^2}{X^2}+\frac{2 Y}{X}+2\right)=c \\ & \Rightarrow \quad\left(-2 Y^2+2 X Y+2 X^2\right)=c \end{aligned} $

Substitute $X$ and $Y$, we get

$ \begin{aligned} -2\left(y+\frac{3}{5}\right)^2+2\left(x-\frac{9}{5}\right)(y & \left.+\frac{3}{5}\right) +2\left(x-\frac{9}{5}\right)^2=c \end{aligned} $

$ \begin{aligned}\Rightarrow & -2\left(y^2+\frac{6}{5} y+\frac{9}{25}\right)+2\left(x y+\frac{3}{5} x-\frac{9}{5} y-\frac{27}{25}\right) +2\left(x^2-\frac{18}{5} x+\frac{81}{25}\right)=c \\ & \begin{array}{r} \Rightarrow \quad-2 y^2-\frac{12}{5} y-\frac{18}{25}+2 x y+\frac{6}{5} x-\frac{18}{5} y \end{array} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & -\frac{54}{25}+2 x^2-\frac{36}{5} x+\frac{162}{25}=c \\ \Rightarrow & -2 y^2+2 x y+2 x^2-\frac{30}{5} y-\frac{30}{5} x+\frac{90}{25}=c \\ \Rightarrow & \quad 2 x^2+2 x y-2 y^2-6 x-6 y=c \\ \Rightarrow & \quad x^2+x y-y^2-3 x-3 y=c \end{aligned}\\ &\text { Thus, the general solution is }\\ &x^2+x y-y^2-3 x-3 y+c=0 \end{aligned} $

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

$ \begin{gathered} y(e)=0 \\ \Rightarrow \quad \frac{d y}{d x}+\frac{1}{x \log x} y=\frac{\log x^2}{x \log x} \\ \Rightarrow \quad \frac{d y}{d x}+\frac{1}{x \log x} y=\frac{2 \log x}{x \log x}=\frac{2}{x} \end{gathered} $

This is a linear differential equation of the form $\frac{d y}{d x}+P(x) y=Q(x)$,

Where $P(x)=\frac{1}{x \log x}$ and $Q(x)=\frac{2}{x}$

Now, IF $=e^{\int \frac{1}{x \log x} d x}$

Let $u=\log x$.

Then $d u=\frac{1}{x} d x$

So, $\mathrm{IF}=e^{\int \frac{1}{u} d u}=e^{\log u}=u=\log x$

So, the solution is given by

$y \cdot \mathrm{IF}=\int Q(x) \cdot \mathrm{IF} d x+c$

$y \log x=\int \frac{2}{x} \log x d x+c$

Let $v=\log x$

Then, $d v=\frac{1}{x} d x$

So, $y \log x=\int 2 v d v+c=v^2+c$

$\Rightarrow y \log x=v^2+c=(\log x)^2+c$

Given, $y(e)=0$

So, $y \log x=(\log x)^2+c$

$\Rightarrow 0 \cdot \log (e)=(\log e)^2+c$

$ \Rightarrow \quad 0=1+c \Rightarrow c=-1 $

$\therefore$ The particular solution is $y \log x=(\log x)^2-1$

$ \Rightarrow \quad y=\frac{(\log x)^2-1}{\log x} $

So, $y\left(e^2\right)=\frac{\left(\log e^2\right)^2-1}{\log \left(e^2\right)}$

$ \begin{aligned} & =\frac{(2 \log e)^2-1}{2 \log (e)}\,\,\,\,\, \left[\because \log _e=1\right]\\ & =\frac{2^2-1}{2}=\frac{4-1}{2}=\frac{3}{2} \\ \therefore y\left(e^2\right) & =\frac{3}{2} \end{aligned} $

$ \begin{aligned} & \text { } x \frac{d^2 y}{d x^2}=\left(1+\left(\frac{d^2 y}{d x^2}\right)^2\right)^{\frac{-1}{2}} \\ & \Rightarrow \quad\left[x \frac{d^2 y}{d x^2}\right]^2=\left[\left(1+\left(\frac{d^2 y}{d x^2}\right)^2\right)^{\frac{-1}{2}}\right]^{-2} \\ & \Rightarrow \quad \frac{1}{x^2\left(\frac{d^2 y}{d x^2}\right)^2}=1+\left(\frac{d^2 y}{d x^2}\right)^2 \\ & \Rightarrow \quad 1=x^2\left(\frac{d^2 y}{d x^2}\right)^2+x^2\left(\frac{d^2 y}{d x^2}\right)^4 \\ & \Rightarrow \quad x^2\left(\frac{d^2 y}{d x^2}\right)^4+x^2\left(\frac{d^2 y}{d x^2}\right)^2-1=0 \end{aligned} $

so, order $=k=2$

and degree $=l=4$

Thus, quadratic equation,

$ \begin{aligned} & (x-2)(x-4)=0 \\ \Rightarrow \quad & x^2-6 x+8=0 \end{aligned} $

$y d x=\left(x+y^3 \cos y\right) d y$

$ \begin{array}{ll} \Rightarrow & \frac{d x}{d y}-\frac{x}{y}=y^2 \cos y \\ \therefore & \mathrm{IF}=e^{\int-\frac{1}{y} d y} \\ & =e^{\ln (y)^{-1}}=\frac{-1}{y} \\ \therefore & \frac{1}{y}\left(\frac{d y}{d y}\right)-\frac{x}{y^2}=y \cdot \cos y \\ \Rightarrow & \int \frac{d}{d y}\left(\frac{x}{y}\right) d y=\int y \cdot \cos y d y \\ \Rightarrow & \frac{x}{y}=y \sin y-\int \sin y d y \\ & \quad x=y \sin y+\cos y+C \\ \Rightarrow & x=y^2 \sin y+y \cos y+c y \end{array} $

Put point ( $0, \pi$ ) into it

$ \begin{array}{ll} \Rightarrow & 0=0-\pi+c \pi \Rightarrow c=1 \\ \therefore & x=y^2 \sin y+y \cos y+y \end{array} $

$ \begin{aligned} & \Rightarrow \quad x=y^2 \sin y+y(1+\cos y) \\ & \Rightarrow \quad x=y^2 \sin y+2 y \cos ^2\left(\frac{y}{2}\right) \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \begin{array}{l} (x-(x+y) \log (x+y)) d x+x d y=0 \\ \Rightarrow \frac{d y}{d x}=\frac{(x+y) \log (x+y)-x}{x} \\ =\frac{(x+y)}{x} \log (x+y)-1 \\ \text { put. } \quad u=x+y \Rightarrow \frac{d u}{d x}=\frac{d y}{d x}+1 \\ \Rightarrow \quad \frac{d y}{d x}=\frac{d u}{d x}-1 \\ \Rightarrow \quad \frac{d u}{d x}-1=\frac{u}{x} \log (u)-1 \\ \Rightarrow \quad \int \frac{d u}{u \log u}=\int \frac{1}{x} d x \\ \Rightarrow \quad \log (\log u)=\log x+\log (c) \\ \Rightarrow \quad \log (\log u)=\log (c x) \\ \Rightarrow \quad \log u=c x \\ \Rightarrow \quad \log (x+y)=c x \quad \quad(\because u=x+y) \end{array} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, differential equation is }\\ &\sec (x-y+1) d y=d x \end{aligned} $

$ \Rightarrow \quad \frac{d y}{d x}=\cos (x-y+1) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Put $x-y=z$

$ \Rightarrow 1-\frac{d y}{d x}=\frac{d z}{d x} \Rightarrow \frac{d y}{d x}=1-\frac{d z}{d x} $

So, Eq. (i),

$ \begin{aligned} & \Rightarrow \quad 1-\frac{d z}{d x}=\cos (z+1) \\ & \Rightarrow \quad \frac{d z}{d x}=1-\cos (z+1)=2 \sin ^2\left(\frac{z+1}{2}\right) \\ & \Rightarrow \quad d x=\frac{1}{2} \operatorname{cosec}^2\left(\frac{z+1}{2}\right) d z \end{aligned} $

On integrating,

$ x=\frac{1}{2} \times 2\left(-\cot \left(\frac{z+1}{2}\right)\right)+C $

$ \Rightarrow \quad x+\cot \left(\frac{x-y+1}{2}\right)=C $

Given, $y^2=4 a(x+a)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \begin{aligned} & 2 y \frac{d y}{d x}=4 a(1+0)=4 a \\ \Rightarrow \quad & y \frac{d y}{d x}=2 a \Rightarrow a=\frac{y}{2} \frac{d y}{d x} \end{aligned} $

$ \begin{aligned} & \text { From Eq. (1) } \\ & \qquad \begin{array}{l} y^2=4 \times \frac{y}{2} \cdot \frac{d y}{d x}\left(x+\frac{y}{2} \frac{d y}{d x}\right) \\ \Rightarrow \quad y^2=2 y \frac{d y}{d x} \times x+y^2\left(\frac{d y}{d x}\right)^2 \\ \Rightarrow \quad y=2 x \frac{d y}{d x}+y\left(\frac{d y}{d x}\right)^2 \end{array} \end{aligned} $

$ \begin{aligned} &\text { Given, differential equation is }\\ &\begin{aligned} \frac{d y}{d x} & =\frac{2 x y-4 x+y-2}{2 x y+x-4 y-2} \\ & =\frac{2 x(y-2)+1(y-2)}{2 y(x-2)+1(x-2)} \\ & =\frac{(2 x+1)(y-2)}{(2 y+1)(x-2)} \\ & \Rightarrow \int \frac{2 y+1}{y-2} d y=\int \frac{2 x+1}{x-2} d x \\ & \Rightarrow \int \frac{2(y-2)+5}{y-2} d y=\int \frac{2(x-2)+5}{x-2} d x \\ & \Rightarrow \int 2 d y+\int \frac{5}{y-2} d y=\int 2 d x+\int \frac{5}{x-2} d x \\ & \Rightarrow 2 y+5 \log (y-2)=2 x+5 \log (x-2)+c \\ & \Rightarrow 2(y-x)+5 \log \left|\frac{y-2}{x-2}\right|=C \end{aligned} \end{aligned} $

General equation of circle passing through the origin and having centres on the $X$-axis.

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

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \quad \frac{d y}{d x}=\frac{x+y}{x-y} \\ & \Rightarrow \quad y=v x \Rightarrow v=\frac{y}{x} \\ & \Rightarrow \quad \frac{d y}{d x}=v+x \frac{d v}{d x} \\ & \Rightarrow \quad v+x \frac{d v}{d x}=\frac{x+v x}{x-v x} \\ & \Rightarrow \quad v+x \frac{d v}{d x}=\frac{1+v}{1-v} \\ & \Rightarrow \quad x \frac{d v}{d x}=\frac{1+v}{1-v}-v \\ & \Rightarrow \quad=\frac{1+v-v+v^2}{1-v} \\ & \Rightarrow \quad x \frac{d v}{d x}=\frac{1+v^2}{1-v} \\ & \Rightarrow \quad \frac{1-v}{1+v^2} d v=\frac{1}{x} d x \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Integrating both sides, we get }\\ &\begin{aligned} & \Rightarrow \quad \int \frac{1-v}{1+v^2} d v=\int \frac{1}{x} d x \\ & \Rightarrow \quad \log x=\int \frac{1}{1+v^2} d v-\int \frac{v}{1+v^2} d v \\ & \Rightarrow \quad \log x=\tan ^{-1} v-\frac{1}{2} \int \frac{1}{t} d t \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { where, } 1+v^2=t\\ &\begin{aligned} 2 v d v & =d t \\ v d v & =\frac{1}{2} d t \end{aligned}\\ &\begin{aligned} & \Rightarrow \log x=\tan ^{-1} v-\frac{1}{2} \log \left|1+v^2\right|+C \\ & \Rightarrow \log x=\tan ^{-1} \frac{y}{x}-\frac{1}{2} \log \left|1+\frac{y^2}{x^2}\right|+C \\ & \Rightarrow \log x=\tan ^{-1} \frac{y}{x}-\frac{1}{2} \log \left|\frac{y^2+x^2}{x^2}\right|+C \\ & \Rightarrow \tan ^{-1}\left(\frac{y}{x}\right)=\log \left(c \sqrt{x^2+y^2}\right) \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \frac{d y}{d x}+\frac{\sec x}{\cos x+\sin x} y=\frac{\cos x}{1+\tan x} \\ & \begin{aligned} & \text { IF }=e^{\int \frac{\sec x}{\cos x+\sin x} d x}=e^{\int \frac{\sec ^2 x}{1+\tan x} d x} \\ &=e^{\log (1+\tan x)}=1+\tan x \\ & \Rightarrow y \cdot(1+\tan x) \\ &=\int(1+\tan x) \cdot \frac{\cos x}{1+\tan x} d x \\ &=\int \cos x d x \\ &=\sin x+C \\ & \Rightarrow \sec x(\cos x+\sin x) y=\sin x+C \end{aligned} \end{aligned} \end{aligned} $

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

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad x \frac{d v}{d x}=\frac{2-v-v^2}{1-v^2}-v \\ & \Rightarrow \quad x \frac{d v}{d x}=\frac{2-v-v^2-v+v^3}{1-v^2} \\ & \Rightarrow \quad \frac{1-v^2}{v^3-v^2-2 v+2} d v=\frac{1}{x} d x \\ & \Rightarrow \quad \int \frac{1-v^2}{\left(v^2-2\right)(v-1)} d v=\int \frac{1}{x} d x \\ & \Rightarrow \quad \int\left(\frac{-v}{v^2-2}-\frac{1}{v^2-2}\right) d v=\frac{1}{x} d x \\ & \Rightarrow-\frac{1}{2} \log \left(v^2-2\right)-\frac{1}{2 \sqrt{2}} \log \frac{v-\sqrt{2}}{v+\sqrt{2}} -\log x+C \end{aligned}\\ &\text { Put } v=\frac{y}{x}\\ &\begin{aligned} \Rightarrow \sqrt{2} \log \left|\frac{y^2-2 x^2}{x^2}\right| & +\log \left|\frac{y-\sqrt{2} x}{y+\sqrt{2} x}\right| +2 \sqrt{2} \log |x|=C \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } y=a x+\frac{1}{a} \\ & \Rightarrow \frac{d y}{d x}=a \Rightarrow \frac{d^2 y}{d x^2}=0 \\ & \Rightarrow x \frac{d y}{d x}+\frac{d x}{d y}=y=0 \\ & \Rightarrow \quad x\left(\frac{d y}{d x}\right)^2+1-y \frac{d y}{d x}=0 \\ & \quad r=2, m=1 \\ & \text { Now, } \frac{d y}{d x}=\frac{y}{2 x}, y(1)=\sqrt{3} \\ & \Rightarrow \int \frac{1}{y} d y=\int \frac{1}{2 x} d x \\ & \Rightarrow \log y=\frac{1}{2} \log (x)+\log C \\ & \Rightarrow \log \sqrt{3}=\log C \Rightarrow C=\sqrt{3} \\ & \Rightarrow \log y=\log \sqrt{x}+\log \sqrt{3} \\ & \Rightarrow \quad y=\sqrt{3 x} \\ & \therefore \quad y^2=3 x \end{aligned} \end{aligned} $

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

$ \begin{aligned} & \text { } \frac{d y}{d x}+x y=4 x-2 y+8 \\ & \Rightarrow \frac{d y}{d x}+y(x+2)=4 x+8 \\ & \text { IF }=e^{\int(x+2) d x}=e^{\frac{(x+2)^2}{2}} \\ & y \cdot e^{\frac{(x+2)^2}{2}}=\int 4(x+2) e^{\frac{(x+2)^2}{2}} d x \\ & \text { Put } \frac{(x+2)^2}{2}=t \Rightarrow(x+2) d x=d t \\ & \therefore y \cdot e^{\frac{(x+2)^2}{2}}=4 \int e^t d t+C \\ & y \cdot e^{\frac{(x+2)^2}{2}}=4 e^{\frac{(x+2)^2}{2}}+C \\ & y=4+C e^{-\frac{(x+2)^2}{2}} \end{aligned} $

$\because C$ is constant

$ \therefore y=4-C e^{-\frac{(x+2)^2}{2}} $

$ \begin{aligned} &\\ &\begin{gathered} \text { }\left(x+2 y^3\right) \frac{d y}{d x}-y=0 \\ \left(x+2 y^3\right) \frac{d y}{d x}=y \\ \frac{d x}{d y}=\frac{x+2 y^3}{y} \\ \frac{d x}{d y}-\frac{x}{y}=2 y^2 \\ \mathrm{IF}=e^{\int-\frac{1}{y} d y}=e^{-\ln y}=\frac{1}{y} \\ x \cdot \frac{1}{y}=\int 2 y^2 \times \frac{1}{y} d y+C \\ \Rightarrow \quad \frac{x}{y}=\int 2 y d y+C \\ \Rightarrow \quad \frac{x}{y}=y^2+C \\ \therefore \quad x=y^3+C y \end{gathered} \end{aligned} $

Given, $\frac{d y}{d x}+\frac{x+y+1}{x-3 y+5}=0$

$ \begin{aligned} & \Rightarrow x d y-3 y d y+5 d y+x d x+y d x+d x=0 \\ & \Rightarrow(x d y+y d x)+5 \cdot d y-3 y d y+x d x +d x=0 \\ & \Rightarrow d(x y)+5 d y-3 y d y+x d x+d x=0 \end{aligned} $

Now, integrating because every term is variable separable.

$ x y+5 y-\frac{3 y^2}{2}+\frac{x^2}{2}+x=C $

Clearly, this equation is obtained by solving option

$ 3(y-1)^2-2(x+2)(y-1)-(x+2)^2=C $

Let the equation of parabola having axis $x=1$ be

$ (x-1)^2=4 a y $

Now, differenting w.r.t. $x$, we get

$ \begin{aligned} & & 2(x-1) & =4 a y^{\prime} \\ \Rightarrow & & 2(x-1) & =\frac{(x-1)^2}{y} \cdot y^{\prime} \\ \Rightarrow & & 2 y & =(x-1) y^{\prime} \end{aligned} $

Again, differentiating

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

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

$ \mathrm{IF}=e^{\int \frac{1}{x} d x}=e^{\ln x}=x $

Solution of differential equations is

$ \begin{aligned} & & y \cdot x & =\int e^x \cdot \frac{1}{x} \cdot x d x \\ & \Rightarrow & x y & =e^x+c \\ & \therefore & y & =\frac{e^x+c}{x} \end{aligned} $

$ \text { } \begin{aligned} \left(x \sin \frac{y}{x}\right) d y & =\left(y \sin \frac{y}{x}-x\right) d x \\ \frac{d y}{d x} & =\frac{y \sin \frac{y}{x}-x}{x \sin \frac{y}{x}} \end{aligned} $

Put $y=v x$

$ \begin{gathered} \frac{d y}{d x}=v+x \frac{d v}{d x} \\ \Rightarrow v+x \frac{d v}{d x}=\frac{v \sin v-1}{\sin v} \\ \Rightarrow x \frac{d v}{d x}=\frac{v \sin v-1-v \sin v}{\sin v} \\ =-\frac{1}{\sin v} \\ \Rightarrow \quad \int \sin v d v+\int \frac{d x}{x}=0 \\ \Rightarrow \quad-\cos v+\ln x=C \\ \Rightarrow \quad \ln x-\cos \frac{y}{x}=C \end{gathered} $

To determine which option can form a differential equation of order two, consider each geometric configuration:

Option A: All circles passing through the origin

The general equation for a circle passing through the origin is:

$ x^2 + y^2 - 2hx - 2py = 0 \,\,\,\,\,\,\,\,\,\,(i) $

In this equation, there are two arbitrary constants, $ h $ and $ p $. The presence of two arbitrary constants implies that the differential equation derived from this circle's equation would be of order two, since each arbitrary constant contributes to the order of the differential equation.

Conclusion

Among the given options, the circles passing through the origin allow for a differential equation of order two. This is due to the presence of two arbitrary constants in the circle's equation, which dictate the order of the corresponding differential equation.

To find the differential equation that has $ ax + by = 1 $ as its general solution, follow these steps:

Differentiate the Equation: Start by differentiating both sides of $ ax + by = 1 $ with respect to $ x $. This gives:

$ a + b \frac{dy}{dx} = 0 $

Solve for $\frac{dy}{dx}$: From the differentiated equation, solve for $\frac{dy}{dx}$:

$ \frac{dy}{dx} = \frac{-a}{b} $

Differentiate Again: Differentiate the expression $\frac{dy}{dx} = \frac{-a}{b}$ with respect to $ x $:

$ \frac{d^2y}{dx^2} = 0 $

Therefore, the differential equation that has $ ax + by = 1 $ as its general solution is $\frac{d^2y}{dx^2} = 0$.

Given differential equation

$ \begin{array}{ll} e^x y d x+e^x d y+x d x=0 \\ \Rightarrow \left(e^x y+x\right) d x+e^x d y=0 \\ \Rightarrow e^x \frac{d y}{d x}+e^x y+x=0 \\ \Rightarrow \frac{d y}{d x}+y=-\frac{x}{e^x} \end{array} $

which is linear differential equation

$ \therefore \mathrm{IF}=e^{\int P d x}=e^{\int d x}=e^x $

$\therefore$ Required solution

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

The differential equation for a family of hyperbolas with centers at the origin and axes along the coordinate axes is derived from the standard hyperbola equation:

$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $

Differentiating this equation with respect to $ x $, we get:

$ \frac{2x}{a^2} - \frac{2y}{b^2} \cdot \frac{dy}{dx} = 0 $

This simplifies to:

$ \Rightarrow \frac{2x}{a^2} = \frac{2y}{b^2} \cdot \frac{dy}{dx} \Rightarrow \frac{y}{x} \cdot \frac{dy}{dx} = \frac{b^2}{a^2} $

Differentiating both sides with respect to $ x $ again, we obtain:

$ \frac{y}{x} \cdot \frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot \frac{x \cdot \frac{dy}{dx} - y \cdot 1}{x^2} = 0 $

Which can be rewritten as:

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

This results in the differential equation:

$ xyy_2 + y_1^2x - yy_1 = 0 $

To solve the given differential equation:

$ (x y + y^2) \, dx - (x^2 - 2 x y) \, dy = 0 $

we start by rewriting it in the form of a differential equation:

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

Using the substitution $ y = ux $, where $ u $ is a function of $ x $, we differentiate $ y $ with respect to $ x $:

$ \frac{dy}{dx} = u + x \frac{du}{dx} $

Substituting these into the differential equation provides:

$ u + x \frac{du}{dx} = \frac{u + u^2}{1 - 2u} $

Rearranging terms, we get:

$ x \frac{du}{dx} = \frac{3u^2}{1 - 2u} $

This leads to:

$ \frac{1 - 2u}{3u^2} \, du = \frac{dx}{x} $

Integrating both sides yields:

$ \int \left(\frac{1}{3u^2} - \frac{2}{3u}\right) \, du = \int \frac{dx}{x} $

Calculating the integrals gives:

$ -\frac{1}{3u} - \frac{2}{3} \log u = \log x + \log c $

Substituting back $ u = \frac{y}{x} $:

$ -\frac{x}{3y} - \frac{2}{3} \log \frac{y}{x} = \log xc $

After simplification, the general solution is:

$ c x y^2 e^{\frac{x}{y}} = 1 $

To solve the differential equation $ (1+\tan y)(dx-dy)+2x \, dy = 0 $, we'll start by rewriting the equation:

$ dx + \tan y \, dx - dy - \tan y \, dy + 2x \, dy = 0. $

This simplifies to:

$ -dy(1+\tan y-2x) + dx(1+\tan y) = 0. $

Rewriting the equation, we separate variables:

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

Alternatively, in terms of $ \frac{dx}{dy} $:

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

Now, this can be rewritten as:

$ \frac{dx}{dy} + 2 \frac{x}{1+\tan y} = 1. $

This is a first-order linear differential equation of the form:

$ \frac{dx}{dy} + Px = Q $

where $ P = \frac{2}{1+\tan y} $ and $ Q = 1 $.

To solve it, we calculate the integrating factor (IF):

$ \mathrm{IF} = e^{\int P \, dy} = e^{\int \frac{2}{1+\tan y} \, dy}. $

Evaluating the integral, we get:

$ \mathrm{IF} = (\sin y + \cos y) e^y. $

The general solution formula is:

$ x(\mathrm{IF}) = \int Q \cdot (\mathrm{IF}) \, dy + C. $

Substitute the values:

$ x(\sin y+\cos y) e^y = \int 1 \cdot (\sin y+\cos y) e^y \, dy + C. $

Splitting the integral:

$ = \int \sin y \, e^y \, dy + \int \cos y \, e^y \, dy + C. $

Therefore:

$ x(\sin y+\cos y) e^y = \frac{e^y}{2}(\sin y - \cos y) + \frac{e^y}{2}(\cos y + \sin y) + C. $

Simplifying gives:

$ x(\sin y+\cos y) e^y = e^y \cdot \sin y + C. $

This simplifies to:

$ e^y[x \cos y + x \sin y - \sin y] = C. $

Hence, the general solution is:

$ e^y(x \cos y + x \sin y - \sin y) = C. $

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

Let $y=v x$, then $\frac{d y}{d x}=v+x \frac{d v}{d x}$

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

$ \begin{aligned} & \Rightarrow \log \left|v+\sqrt{1+v^2}\right|=\log x+\log c \\ & \Rightarrow \log \left|\frac{y}{x}+\sqrt{1+\frac{y^2}{x^2}}\right|=\log c x \\ & \Rightarrow \log \left|\frac{y+\sqrt{x^2+y^2}}{x}\right|=\log c x \\ & \Rightarrow \log \left|y+\sqrt{\left(x^2+y^2\right)}\right|-\log x=\log c x \\ & \Rightarrow \log \left|y+\sqrt{x^2+y^2}\right|=\log c^2 \\ & \text { or } y+\sqrt{y^2+x^2}=\alpha^2 \end{aligned} $

$x\left(\frac{d^2 y}{d x^2}\right)^{1 / 2}=\left(1+\frac{d y}{d x}\right)^{4 / 3}$

Simplifying, on squaring both sides, we get

$ \begin{aligned} \left(x \frac{d^2 y}{d x^2}\right)^{1 / 2 \times 2} & =\left(1+\frac{d y}{d x}\right)^{4 / 3 \times 2} \\ x^2\left(\frac{d^2 y}{d x^2}\right) & =\left(1+\frac{d y}{d x}\right)^{8 / 3} \end{aligned} $

On cubic both sides, we get

$ \begin{aligned} & \left(x^2 \frac{d^2 y}{d x^2}\right)^3=\left(1+\frac{d y}{d x}\right)^{8 / 3 \times 3} \\ & x^6\left(\frac{d^2 y}{d x^2}\right)^3=\left(1+\frac{d y}{d x}\right)^8 \end{aligned} $

Clearly, here the highest derivative with respect to $x$ is 2 , so order is 2 .

Moreover the highest power of highest derivative is 3 i.e. $\left(\frac{d^2 y}{d x^2}\right)^3$, so degree is 3 .

$ \begin{aligned} & \text { Sum }=\text { order }+ \text { degree }=2+3 \\ & \text { Sum }=5 \end{aligned} $

We have,

$ y=A \cos 3 x+B \sin 3 x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

On differentiate $y$ w.r.t $x$, we get

$ \begin{aligned} & \frac{d y}{d x}=-3 A \sin 3 x+3 B \cos 3 x \\ \Rightarrow \quad & \frac{d^2 y}{d x^2}=-9 A \cos 3 x-9 B \sin 3 x \\ \Rightarrow \quad & \frac{d^2 y}{d x^2}=-9(A \cos 3 x+B \sin 3 x) \\ \Rightarrow \quad & \frac{d^2 y}{d x^2}=-9 y \quad \text { [from Eq. (i)] } \\ \Rightarrow \quad & \frac{d^2 y}{d x^2}+9 y=0 \end{aligned} $

We start with the given differential equation:

$ \cos x \frac{d y}{d x} - y \sin x = 6x, \left(0 < x < \frac{\pi}{2}\right) $

and the initial condition:

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

Rewriting the differential equation:

$ \frac{d y}{d x} - y \frac{\sin x}{\cos x} = \frac{6x}{\cos x} $

This simplifies to:

$ \frac{d y}{d x} - y \tan x = 6x \sec x $

This is in the form of the linear differential equation:

$ \frac{d y}{d x} + Ay = B $

where $ A = -\tan x $ and $ B = 6x \sec x $.

The integrating factor (IF) is found using:

$ \mathrm{IF} = e^{\int \operatorname{Adx}} = e^{-\int \tan x \, dx} = e^{-\log \sec x} = \frac{1}{\sec x} $

Applying the integrating factor, the solution becomes:

$ y \cdot \frac{1}{\sec x} = \int 6x \sec x \cdot \frac{1}{\sec x} \, dx $

Simplifying, we get:

$ \frac{y}{\sec x} = 3x^2 + C $

where $ C $ is a constant. Using the initial condition $ y\left(\frac{\pi}{3}\right)=0 $:

$ 0 = 3\left(\frac{\pi}{3}\right)^2 + C \implies C = -\frac{\pi^2}{3} $

Thus, substituting back, we have:

$ \frac{y}{\sec x} = 3x^2 - \frac{\pi^2}{3} $

Solving at $ x = \frac{\pi}{6} $:

$ y = \sec x \cdot \left(3x^2 - \frac{\pi^2}{3}\right) $

Calculating:

$ \sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}} $

$ 3\left(\frac{\pi}{6}\right)^2 = \frac{\pi^2}{12} $

Therefore,

$ y = \frac{2}{\sqrt{3}} \cdot \left(\frac{\pi^2}{12} - \frac{\pi^2}{3}\right) = \frac{2}{\sqrt{3}} \cdot \left(-\frac{\pi^2}{4}\right) $

Simplifying gives:

$ y\left(\frac{\pi}{6}\right) = \frac{-\pi^2}{2\sqrt{3}} $

We have,

$ \frac{d y}{d x}=\frac{y+x \tan \left(\frac{y}{x}\right)}{x} $

Put $y=v x$

$ \begin{array}{ll} \Rightarrow & \frac{d}{d x} \cdot(v x)=\frac{v x+x \tan \left(\frac{v x}{x}\right)}{x} \\ \Rightarrow & v+x \frac{d v}{d x}=\frac{x(v+\tan v)}{x} \\ \Rightarrow & \frac{x d v}{d x}=v+\tan v-v \\ \Rightarrow & x \frac{d v}{d x}=\tan v \Rightarrow \frac{d v}{\tan v}=\frac{d v}{x} \\ \Rightarrow & \frac{\cos v d v}{\sin v}=\frac{d x}{x} \end{array} $

$ \begin{aligned} &\text { On taking integration both sides, we get }\\ &\begin{aligned} & \Rightarrow \quad \int \frac{\cos v d v}{\sin v}=\int \frac{d x}{x} \\ & \text { Let } \quad \sin v=u \\ & \ \quad \,\,\,\,\,\,\,\,\cos v d v=d u \\ & \Rightarrow \quad \int \frac{1}{u} d u=\int \frac{d x}{x} \\ & \Rightarrow \quad \log u=\log x+\log C \\ & \Rightarrow \quad \log (\sin v)=\log x+\log C \\ & \Rightarrow \quad \log \left(\frac{\sin v}{x}\right)=\log C \Rightarrow \frac{\sin v}{x}=C \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{array}{ll} \Rightarrow & \sin v=x C \\ \Rightarrow & \sin \frac{y}{x}=x C \end{array}\\ &\begin{array}{ll} \Rightarrow & v^2=1.44 \\ \Rightarrow & v=1.2 \mathrm{~m} / \mathrm{s} \end{array} \end{aligned} $

To find the differential equation by eliminating the parameters $ a $ and $ b $ from the equation $ y = a e^{2x} + bx e^{2x} $, we take the derivatives with respect to $ x $ and express them in terms of $ y $.

First Derivative:

$ y' = \frac{d}{dx} (a e^{2x} + bx e^{2x}) = 2a e^{2x} + (2bx e^{2x} + b e^{2x}) $

$ = 2a e^{2x} + 2bx e^{2x} + b e^{2x} = 2y + b e^{2x} $

Second Derivative:

$ y'' = \frac{d}{dx} (2a e^{2x} + 2bx e^{2x} + b e^{2x}) $

$ = 4a e^{2x} + 2b e^{2x} (2x + 1) + b (2 e^{2x}) $

$ = 4a e^{2x} + 4bx e^{2x} + 2b e^{2x} + 2b e^{2x} $

$ = 4a e^{2x} + 4bx e^{2x} + 4b e^{2x} $

$ = 4(a e^{2x} + bx e^{2x}) + 4b e^{2x} $

$ = 4y + 4b e^{2x} $

Constructing the Differential Equation:

Substitute into the expression $ y'' - 4y' + 4y $:

$ y'' - 4y' + 4y = [4y + 4b e^{2x}] - 4[2y + b e^{2x}] + 4y $

$ = 4y + 4b e^{2x} - 8y - 4b e^{2x} + 4y $

$ = 0 $

The expression simplifies to zero, confirming that the differential equation $ y'' - 4y' + 4y = 0 $ is satisfied, supporting the correctness of option (D).

Given the general solution $ y = a^3 e^{b^2 x + c} $, where $ a $ and $ c $ are arbitrary constants and $ b $ is a fixed constant, we are tasked with identifying the order of the differential equation related to this solution.

To find the order, let's derive the expression for $ y $:

$ \begin{aligned} y' &= \frac{d}{dx}\left(a^3 e^{b^2 x + c}\right) \\ &= a^3 b^2 e^{b^2 x + c}. \end{aligned} $

Notice that we can express this in terms of $ y $:

$ y' = b^2 y. $

This differential equation, $ y' - b^2 y = 0 $, is a first-order linear differential equation. The order of a differential equation is determined by the highest derivative present. In this case, since the highest derivative is $ y' $, the order is 1.

To solve the differential equation $(x + 2y^3) \frac{dy}{dx} = y$, we begin by rearranging terms:

$ \frac{x + 2y^3}{y} = \frac{dx}{dy} $

This can be rewritten as:

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

This gives us a form where:

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

We recognize this as a linear first-order differential equation, which can be solved using an integrating factor. The integrating factor ($\text{IF}$) is derived as follows:

$ \text{IF} = e^{\int -\frac{1}{y} \, dy} = e^{-\ln|y|} = \frac{1}{y} $

Apply the integrating factor to both sides of the differential equation:

$ \frac{x}{y} = \int \left(2y^2 \cdot \frac{1}{y}\right) \, dy $

This integral becomes:

$ \frac{x}{y} = 2 \int y \, dy $

Calculate the integral:

$ \frac{x}{y} = 2 \left(\frac{y^2}{2}\right) + C $

Simplifying gives:

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

Therefore, multiplying both sides by $y$ results in:

$ x = y(y^2 + C) $

Thus, the solution to the differential equation is:

$ x = y(y^2 + C) $

Given differential equation is

$ \begin{aligned} \frac{d^3 y}{d x^3}=\left[1+\left(\frac{d y}{d x}\right)^2\right]^{\frac{5}{2}} \\ \Rightarrow \quad\left(\frac{d^3 y}{d x^3}\right)^2=\left[1+\left(\frac{d y}{d x}\right)^2\right]^5 \end{aligned} $

Highest derivative is third order and power of highest order derivative is 2 .

Hence, order is 3 and degree is 2 .

To find the integrating factor of the given differential equation, we start with the equation:

$ \sin x \frac{d y}{d x} - y \cos x = 1 $

Rewriting the equation, we have:

$ \frac{d y}{d x} - \frac{\cos x}{\sin x} y = \frac{1}{\sin x} $

This simplifies to:

$ \frac{d y}{d x} - \cot x \cdot y = \operatorname{cosec} x $

The standard form for integrating factors is:

$ I F = e^{\int -\cot x \, dx} $

Calculating the integral, we find:

$ I F = e^{-\log |\sin x|} $

This can further be expressed as:

$ I F = e^{\log (\sin x)^{-1}} = \frac{1}{\sin x} $

Therefore, the integrating factor for the given differential equation is:

$ \frac{1}{\sin x} = \operatorname{cosec} x $

To solve the given differential equation:

$ \left(x \sin \frac{y}{x}\right) d y = \left(y \sin \frac{y}{x} - x\right) d x $

we start by expressing it in differential form:

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

Introduce a new variable $ v $ such that $ y = vx $. This implies:

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

Substitute $ y = vx $ into the differential equation:

$ v + x \frac{d v}{d x} = \frac{vx \sin v - x}{x \sin v} $

Simplify:

$ x \frac{d v}{d x} = \frac{vx \sin v - x - vx \sin v}{x \sin v} = \frac{-x}{x \sin v} = \frac{-1}{\sin v} $

This results in the following separable equation:

$ \sin v \, d v = -\frac{d x}{x} $

Integrate both sides:

$ \int \sin v \, d v = -\int \frac{d x}{x} $

The integration yields:

$ -\cos v = -\log x + C $

Rearranging gives:

$ \log x - \cos v = C $

Substituting back $ v = \frac{y}{x} $, we arrive at the solution:

$ \cos \left(\frac{y}{x}\right) = \log_e x + C $

To determine the sum of the order and degree of the given differential equation:

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

First, rewrite the equation in a form where the degree is clear:

$ \left(\frac{d^4 y}{d x^4}\right)^2 = \left(C + \left(\frac{d y}{d x}\right)^2\right)^3 $

From this, we observe:

The order of the differential equation is the highest derivative present, which is 4.

The degree of the differential equation is the power of the highest derivative when the equation is free of fractions and any radical signs. Here, the degree is 2 (as the highest derivative, $ \frac{d^4 y}{d x^4} $, is squared).

Therefore, the sum of the order and degree is $ 4 + 2 = 6 $.

To solve the differential equation:

$ (x+y) y \, dx + (y-x) x \, dy = 0 $

we can start by rewriting the equation:

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

Substitute $ y = vx $, where $ v $ is a function of $ x $. Then, the derivative $ \frac{dy}{dx} $ becomes:

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

Substitute $ y = vx $ into the differential equation:

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

Simplify the equation:

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

Rearranging gives:

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

Next, separate variables and integrate:

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

This can be rewritten as:

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

Integrate both sides:

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

Substitute back $ v = \frac{y}{x} $:

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

Which simplifies to the equation:

$ -\frac{x}{y} = \log (cxy) $

Thus, the general solution is:

$ x + y \log (cxy) = 0 $