Differential Equations

$ \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} $

$\int_\limits0^x \sqrt{1-\left(y^{\prime}(t)\right)^2} d t=\int_\limits0^x y(t) d t$

Differentiating both side

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

$\therefore$

$\begin{aligned} & \left(x^2+y^2\right) d x-5 x y d y=0 \\ & \frac{d y}{d x}=\frac{x^2+y^2}{5 x y} \\ & \text { Let } y=v x \\ & \frac{d y}{d x}=v+x \frac{d v}{d x} \\ & V+x \frac{d v}{d x}=\frac{1+v^2}{5 v} \\ & x \frac{d v}{d x}=\frac{1+v^2-5 v^2}{5 v} \end{aligned}$

$\begin{aligned} & \frac{1}{8} \int \frac{8 \times 5 v d v}{1-4 v^2}=\int \frac{d x}{x} \\ & \frac{-5}{8} \ln \left|1-4 v^2\right|=\ln |x|+\ln c \\ & \Rightarrow \frac{5}{8} \ln \frac{\left|x^2-4 y^2\right|}{x^2}+\ln |x|=\ln c \\ & \left|\frac{x^2-4 y^2}{x^2}\right|^{5 / 8}|x|=c \\ & \frac{\left|x^2-4 y^2\right|^{5 / 8}}{|x|^{\frac{1}{4}}}=c \because y(1)=0 \Rightarrow c=1 \\ \end{aligned}$

$\begin{aligned} & \Rightarrow\left|x^2-4 y^2\right|^{5 / 8}=|x|^{\frac{1}{4}} \\ & \left|x^2-4 y^2\right|^5=x^2 \end{aligned}$

$\begin{aligned} & 2 y \frac{d y}{d x}+3=5 \frac{d y}{d x} \\ & 2 y d y+3 d x=5 d y \\ & y^2+3 x=5 y+\left.c\right|_{(0,1)} \\ & 1+0=5+c \\ & c=-4 \\ & y^2-5 y=-3 x-4 \\ & y^2-5 y+\frac{25}{4}-\frac{25}{4}=-3 x-4 \\ & \left(y-\frac{5}{2}\right)^2=-3 x+\frac{9}{4} \\ & \left(y-\frac{5}{2}\right)^2=-3\left(x-\frac{3}{4}\right) \\ & \left(\frac{3}{4}, \frac{5}{2}\right) \text { satisfies by } 2 x+3 y=9 \end{aligned}$

$\begin{aligned} & \sec y \frac{d y}{d x}+2 x \sin y=x^3 \cos y \\ & \Rightarrow \sec ^2 y \frac{d y}{d x}+2 x \tan y=x^3 \end{aligned}$

Let $z=\tan y$

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

$\begin{aligned} & \text { I.F. }=e^{x^2} \\ & \Rightarrow z . e^{x^2}=\int e^{x^2} \cdot x^3 d x+c \end{aligned}$

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

$\Rightarrow \tan (0) \cdot e=\frac{1}{2}(1 \cdot e-e)+c$

$\Rightarrow c=0$

$\Rightarrow \tan y=\frac{x^2-1}{2}$

$f(x)=\tan ^{-1}\left(\frac{x^2-1}{2}\right) \Rightarrow f(\sqrt{3})=\frac{\pi}{4}$

$\int_\limits0^a f(x) d x=e^{-a}+4 a^2+a-1$

Differentiate equation w.r.t. 'a'

$\begin{aligned} & f(a)=-e^{-a}+8 a+1 \\ & \Rightarrow f(x)=-e^{-x}+8 x+1 \end{aligned}$

And $y=c_1 f(x)+c_2$

$\begin{aligned} & y=c_1\left(-e^{-x}+8 x+1\right)+c_2 \\ & y^{\prime}=c_1\left(e^{-x}+8\right) \Rightarrow c_1=\frac{y^{\prime}}{e^{-x}+8} \end{aligned}$

$y^{\prime \prime}=-c_1 e^{-x}$

put value of $c_1$

$\begin{aligned} & \frac{d^2 y}{d x^2}=\frac{-\frac{d y}{d x} \cdot e^{-x}}{\left(e^{-x}+8\right)}=\frac{\frac{d y}{d x}}{\left(1+8 e^x\right)} \\ & \Rightarrow\left(1+8 e^x\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=1 \end{aligned}$

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

Let $e^{\tan x}=t$

$\begin{aligned} & e^{\tan x} \cdot \sec ^2 x d x=d t \\ & \int \frac{d y}{1+y^2}=-\int \frac{d t}{1+t^2} \end{aligned}$

$\begin{aligned} & \tan ^{-1} y=-\tan ^{-1} t+c \\ & \tan ^{-1} y=-\tan ^{-1}\left(e^{\tan x}\right)+c \\ & \text { at } x=0, y=1 \\ & \tan ^{-1}(1)=-\tan ^{-1}(1)+c \\ & \frac{\pi}{4}=-\frac{\pi}{4}+c \\ & c=\frac{\pi}{2} \\ & \tan ^{-1} y=-\tan ^{-1}\left(e^{\tan x}\right)+\frac{\pi}{2} \end{aligned}$

Now, at $x=\frac{\pi}{4}$

$\begin{aligned} & \tan ^{-1} y=-\tan ^{-1}(e)+\frac{\pi}{2} \\ & \tan ^{-1} y=\cot ^{-1} e=\tan ^{-1} \frac{1}{e} \\ & \Rightarrow y=\frac{1}{e} \end{aligned}$

$\begin{aligned} & \frac{d y}{d x}=\frac{(2+\alpha) x-\beta y+2}{\beta x-2 \alpha y-(\beta \gamma-4 \alpha)} \\ & \beta x d y-2 \alpha y d y-(\beta \gamma-4 \alpha) d y \\ & =2 x d x+\alpha x d x-\beta y d x+2 d x \\ & \beta \int(x d y+y d y)-\alpha y^2-(\beta \gamma-4 x) y=x^2+\frac{\alpha x^2}{2}+2 x \\ & \beta x y-\alpha y^2-(\beta \gamma-4 \alpha) y=x^2+\frac{\alpha x^2}{2}+2 x \\ & \left(1+\frac{\alpha}{2}\right) x^2+\alpha y^2-\beta x y+2 x+(\beta \gamma-4 \alpha) y=0 \\ & \because \text { this represents circle passing through origin } \\ & \Rightarrow \beta=0 \text { and } 1+\frac{\alpha}{2}=\alpha \\ & \Rightarrow \alpha=2 \end{aligned}$

$\begin{aligned} & \therefore C: 2 x^2+2 y^2+2 x-8 y=0 \\ & x^2+y^2+x-4 y=0 \\ & \text { Radius }=\sqrt{\frac{1}{4}+4-0} \\ & \quad=\frac{\sqrt{17}}{2} \end{aligned}$

$\begin{aligned} & (2 x \log x) \frac{d y}{d x}+2 y=\frac{3}{x} \log x \\ & \frac{d y}{d x}+\frac{y}{x \log x}=\frac{3}{2 x^2} \\ & I F=e^{\int \frac{1}{x \log x}}=e^{\log (\log x)}=\log x \\ & y \times \log x=\int \log x \times \frac{3}{2 x^2} d x+C \\ & y \log x=\frac{3}{2}\left[\frac{-\log x}{x}-\frac{1}{x}\right]+C \\ & y\left(e^{-1}\right)=0 \\ & \Rightarrow 0=\frac{3}{2}[e-e]+C \\ & C=0 \\ & y \log x=\frac{-3}{2}\left[\frac{\log x+1}{x}\right] \\ & y(e) \rightarrow y(e) \times 1=\frac{-3}{2}\left[\frac{1+1}{e}\right]=-\frac{3}{e} \end{aligned}$

To determine $ y(0) $, we start by solving the differential equation given:

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

First, we rewrite it in the standard form for a linear differential equation:

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

Next, we find the integrating factor (I.F.):

$ \text{I.F.} = e^{\int \frac{1}{1 + x^2} dx} = e^{\tan^{-1} x} $

Multiply through by the integrating factor:

$ y \cdot e^{\tan^{-1} x} = \int e^{\tan^{-1} x} \cdot \frac{e^{\tan^{-1} x}}{1 + x^2} dx $

This simplifies to:

$ y \cdot e^{\tan^{-1} x} = \int \frac{e^{2 \tan^{-1} x}}{1 + x^2} dx $

Make the substitution $ \tan^{-1} x = t $, then $ \frac{1}{1 + x^2} dx = dt $:

$ \int e^{2t} dt = \frac{e^{2t}}{2} + \frac{C}{2} $

Rewrite in terms of $ x $:

$ y \cdot e^{\tan^{-1} x} = \frac{e^{2 \tan^{-1} x}}{2} + \frac{C}{2} $

Use the initial condition $ y(1) = 0 $:

$ 0 = \frac{e^{2 \cdot \tan^{-1} 1}}{2} + \frac{C}{2} $

Since $ \tan^{-1} 1 = \frac{\pi}{4} $:

$ 0 = \frac{e^{\pi/2}}{2} + \frac{C}{2} \implies C = -e^{\pi/2} $

Thus, the solution is:

$ y \cdot e^{\tan^{-1} x} = \frac{e^{2 \tan^{-1} x} - e^{\pi/2}}{2} $

Evaluating $ y(0) $:

$ y(0) = y \cdot 1 = \frac{1 - e^{\pi/2}}{2} $

Therefore,

$ y(0) = \frac{1}{2} (1 - e^{\pi/2}) $

Hence, the correct answer is:

Option B

$ \frac{1}{2}\left(1-e^{\pi / 2}\right) $

Equation of circle passing through origin & having centre at the line $y=x$ is

$\begin{aligned} & (x-t)^2+(y-t)^2=2 t^2 \\ & x^2+y^2+t^2+t^2-2 t x-2 t y=2 t^2 \\ & x^2+y^2=2 t(x+y) \end{aligned}$

Now differentiate

$\begin{aligned} & 2 x+2 y y^{\prime}=2 t\left(1+y^{\prime}\right) \\ & t=\frac{x+y y^{\prime}}{1+y^{\prime}} \end{aligned}$

Now, $x^2+y^2=2\left(\frac{x+y y^{\prime}}{1+y^{\prime}}\right)(x+y)$

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

$\begin{aligned} & \frac{d y}{d x}+2 y=\sin 2 x \\ & \text { IF }=e^{2 d x}=e^{2 x} \\ & y \cdot e^{2 x}=\int e^{2 x} \sin 2 x d x+c \\ & =\frac{e^{2 x}}{8}(2 \sin 2 x-2 \cos 2 x)+c \\ & y(0)=\frac{3}{4} \\ & \frac{3}{4}=\frac{1}{8}(-2)+c \Rightarrow c=1 \\ & \text { Put } x=\frac{\pi}{8} \\ & y=\frac{1}{8} \times 2\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)+e^{-\pi / 4} \\ & y=e^{-\pi / 4} \end{aligned}$

$\frac{d y}{d x}+\frac{y\left(2 x^3+8 x\right)}{\left(x^2+4\right)^2}=\frac{2}{\left(x^2+4\right)^2}$

$\mathrm{IF}=e^{\int \frac{2 x^3+8 x}{\left(x^2+4\right)^2} d x}$

$\text { Let }\left(x^2+4\right)^2=t \quad \Rightarrow 2\left(x^2+4\right)(2 x) d x=d t$

$\begin{gathered} =e^{\int \frac{d t}{2 t}}=e^{\log \sqrt{t}}=\sqrt{t}=\left(x^2+4\right) \\ \therefore \quad y\left(x^2+4\right)=\int \frac{2}{x^2+4}+c \\ \Rightarrow y\left(x^2+4\right)=\tan ^{-1}\left(\frac{x}{2}\right)+c \\ y(0)=0 \end{gathered}$

$\begin{aligned} & \Rightarrow 0=0+c \quad \Rightarrow c=0 \\ & \text { put } x=2 \\ & y(8)=\frac{\pi}{4} \quad \Rightarrow \quad y=\frac{\pi}{32} \\ \end{aligned}$

$\begin{aligned} & \left(x^4+2 x^3+3 x^2+2 x+2\right) d y-\left(2 x^2+2 x+3\right) d x=0 \\ & \int d y=\int\left(\frac{2 x^2+2 x+3}{x^4+2 x^3+3 x^2+2 x+2} d x\right. \\ & \int d y=\int \frac{1}{x^2+1} d x+\int \frac{}{x^2+2 x+2} d x \\ & y=\tan ^{-1}(x)+\tan ^{-1}(1+x)+C \\ & y(-1)=\tan ^{-1}(-1)+\tan ^{-1}(1-1)+C \\ & y(-1)=-\frac{\pi}{4}+C=\left(\frac{-\pi}{4}\right)-\{\text { given }\} \\ & \Rightarrow C=0 \\ & \text { So, } y(x)=\tan ^{-1}(x)+\tan ^{-1}(1+x) \\ & y(0)=\tan ^{-1}(0)+\tan ^{-1}(1+0) \\ & y(0)=\frac{\pi}{4} \end{aligned}$

$\begin{aligned} & \frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-80) \\ & \int_\limits{160}^{\mathrm{T}} \frac{\mathrm{dT}}{(\mathrm{T}-80)}=\int_\limits0^{\mathrm{t}}-\mathrm{Kdt} \\ & {[\ln |\mathrm{T}-80|]_{160}^{\mathrm{T}}=-\mathrm{kt}} \\ & \ln |\mathrm{T}-80|-\ln 80=-\mathrm{kt} \\ & \ln \left|\frac{\mathrm{T}-80}{80}\right|=-\mathrm{kt} \\ & \mathrm{T}=80+80 \mathrm{e}^{-\mathrm{kt}} \\ & 120=80+80 \mathrm{e}^{-\mathrm{k} .15} \\ & \frac{40}{80}=\mathrm{e}^{-\mathrm{k} 15}=\frac{1}{2} \\ & \therefore \mathrm{T}(45)=80+80 \mathrm{e}^{-\mathrm{k} .45} \\ & =80+80\left(\mathrm{e}^{-\mathrm{k} .15}\right)^3 \\ & =80+80 \times \frac{1}{8} \\ & =90 \end{aligned}$

$\begin{aligned} & \frac{d y}{d x}=\frac{\sin x+y \cos x}{\sin x \cdot \cos x\left(\frac{1}{\cos x}-\sin x \cdot \frac{\sin x}{\cos x}\right)} \\ & =\frac{\sin x+y \cos x}{\sin x\left(1-\sin ^2 x\right)} \\ & \frac{d y}{d x}=\sec ^2 x+y \cdot 2(\operatorname{cosec} 2 x) \\ & \frac{d y}{d x}-2 \operatorname{cosec}(2 x) \cdot y=\sec ^2 x \\ & \frac{d y}{d x}+p \cdot y=Q \end{aligned}$

$\text { I.F. }=\mathrm{e}^{\int p \mathrm{dx}}=\mathrm{e}^{\int-2 \operatorname{cosec}(2 \mathrm{x}) \mathrm{dx}}$

Let $2 \mathrm{x}=\mathrm{t}$

$2 \frac{\mathrm{dx}}{\mathrm{dt}}=1$

$\mathrm{dx}=\frac{\mathrm{dt}}{2}$

$=\mathrm{e}^{-\int \operatorname{cosec}(\mathrm{t}) \mathrm{dt}}$

$=\mathrm{e}^{-\ln \left|\tan \frac{t}{2}\right|}$

$=\mathrm{e}^{-\ln |\tan x|}=\frac{1}{|\tan x|}$

$\begin{aligned} & y(\text { IF })=\int Q(I F) d x+c \\ & \Rightarrow y \frac{1}{|\tan x|}=\int \sec ^2 x \cdot \frac{1}{|\tan x|}+c \\ & y \cdot \frac{1}{|\tan x|}=\int \frac{d t}{|t|}+c \quad \text { for } \tan x=t \\ & y \cdot \frac{1}{|\tan x|}=\ln |t|+c \\ & y=|\tan x|(\ln |\tan x|+c) \\ & \text { Put } x=\frac{\pi}{4}, y=2 \\ & 2=\ln 1+c \Rightarrow c=2 \\ & y=|\tan x|(\ln |\tan x|+2) \\ & y\left(\frac{\pi}{3}\right)=\sqrt{3}(\ln \sqrt{3}+2) \end{aligned}$

$\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{\mathrm{x}}{\mathrm{y}}\left(\ln \left(\frac{\mathrm{x}}{\mathrm{y}}\right)+1\right)$

Let $\frac{x}{y}=t \Rightarrow x=t y$

$\begin{aligned} & \frac{d x}{d y}=t+y \frac{d t}{d y} \\ & t+y \frac{d t}{d y}=t(\ln (t)+1) \end{aligned}$

$\mathrm{y} \frac{\mathrm{dt}}{\mathrm{dy}}=\mathrm{t} \ln (\mathrm{t}) \Rightarrow \frac{\mathrm{dt}}{\mathrm{t} \ln (\mathrm{t})}=\frac{\mathrm{dy}}{\mathrm{y}}$

$\Rightarrow \int \frac{\mathrm{dt}}{\mathrm{t} \cdot \ln (\mathrm{t})}=\int \frac{\mathrm{dy}}{\mathrm{y}}$

$\Rightarrow \int \frac{d p}{p}=\int \frac{d y}{y} \quad$ let $\ln t=p$

$\frac{1}{\mathrm{t}} \mathrm{dt}=\mathrm{dp}$

$\begin{aligned} & \Rightarrow \ln p=\ln y+c \\ & \ln (\ln t)=\ln y+c \\ & \ln \left(\ln \left(\frac{x}{y}\right)\right)=\ln y+c \\ & \text { at } x=e, y=1 \\ & \ln \left(\ln \left(\frac{e}{1}\right)\right)=\ln (1)+c \Rightarrow c=0 \end{aligned}$

$\begin{aligned} & \ln \left|\ln \left(\frac{x}{y}\right)\right|=\ln y \\ & \left|\ln \left(\frac{x}{y}\right)\right|=e^{\ln y} \\ & \left|\ln \left(\frac{x}{y}\right)\right|=y \end{aligned}$

$\frac{d y}{d x}=2(x-1) \sin x+\left(x^2-2 x\right) \cos x$

Now both side integrate

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

Differential equation :-

$\begin{aligned} & x \cos \frac{y}{x} \frac{d y}{d x}=y \cos \frac{y}{x}+x \\ & \cos \frac{y}{x}\left[x \frac{d y}{d x}-y\right]=x \end{aligned}$

Divide both sides by $\mathrm{x}^2$

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

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

$\begin{aligned} & \cos \mathrm{t}\left(\frac{\mathrm{dt}}{\mathrm{dx}}\right)=\frac{1}{\mathrm{x}} \\ & \cos \mathrm{t~dt}=\frac{1}{\mathrm{x}} \mathrm{dx} \end{aligned}$

Integrating both sides

$\begin{aligned} & \sin \mathrm{t}=\ln |\mathrm{x}|+\mathrm{c} \\ & \sin \frac{\mathrm{y}}{\mathrm{x}}=\ln |\mathrm{x}|+\mathrm{c} \end{aligned}$

Using $\mathrm{y}(1)=\frac{\pi}{3}$, we get $\mathrm{c}=\frac{\sqrt{3}}{2}$

So, $\alpha=\sqrt{3} \Rightarrow \alpha^2=3$

$\begin{aligned} & \frac{d y}{d x}-\left(\frac{\sin 2 x}{1+\cos ^2 x}\right) y=\sin x \\ & \text { I.F. }=1+\cos ^2 x \\ & y \cdot\left(1+\cos ^2 x\right)=\int(\sin x) d x \\ & =-\cos x+C \\ & x=0, C=1 \\ & y\left(\frac{\pi}{2}\right)=1 \end{aligned}$

$\begin{aligned} & \left(x^2-4\right) d y-\left(y^2-3 y\right) d x=0 \\ & \Rightarrow \int \frac{d y}{y^2-3 y}=\int \frac{d x}{x^2-4} \\ & \Rightarrow \frac{1}{3} \int \frac{y-(y-3)}{y(y-3)} d y=\int \frac{d x}{x^2-4} \\ & \Rightarrow \frac{1}{3}(\ln |y-3|-\ln |y|)=\frac{1}{4} \ln \left|\frac{x-2}{x+2}\right|+C \\ & \Rightarrow \frac{1}{3} \ln \left|\frac{y-3}{y}\right|=\frac{1}{4} \ln \left|\frac{x-2}{x+2}\right|+C \end{aligned}$

$\begin{aligned} & \text { At } \mathrm{x}=4, \mathrm{y}=\frac{3}{2} \\ & \therefore \mathrm{C}=\frac{1}{4} \ln 3 \\ & \therefore \frac{1}{3} \ln \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\frac{1}{4} \ln \left|\frac{\mathrm{x}-2}{\mathrm{x}+2}\right|+\frac{1}{4} \ln (3) \\ & \text { At } \mathrm{x}=10 \\ & \frac{1}{3} \ln \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\frac{1}{4} \ln \left|\frac{2}{3}\right|+\frac{1}{4} \ln (3) \\ & \ln \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\ln 2^{3 / 4}, \forall \mathrm{x}>2, \frac{\mathrm{dy}}{\mathrm{dx}}<0 \\ & \text { as } \mathrm{y}(4)=\frac{3}{2} \Rightarrow \mathrm{y} \in(0,3) \\ & -\mathrm{y}+3=8^{1 / 4} \cdot \mathrm{y} \\ & \mathrm{y}=\frac{3}{1+8^{1 / 4}} \end{aligned}$

$\begin{aligned} & \frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{ax}=0 \\ & \frac{\mathrm{dx}}{\mathrm{x}}=-\mathrm{adt} \\ & \int \frac{\mathrm{dx}}{\mathrm{x}}=-\mathrm{a} \int \mathrm{dt} \\ & \ln |\mathrm{x}|=-\mathrm{at}+\mathrm{c} \\ & \mathrm{at} \mathrm{t}=0, \mathrm{x}=2 \\ & \ln 2=0+\mathrm{c} \\ & \ln \mathrm{x}=-\mathrm{at}+\ln 2 \\ & \frac{\mathrm{x}}{2}=\mathrm{e}^{-\mathrm{at}} \\ & \mathrm{x}=2 \mathrm{e}^{-\mathrm{at}} \quad \text{.... (i)} \end{aligned}$

$\begin{aligned} & \frac{d y}{d t}+b y=0 \\ & \frac{d y}{y}=-b d t \\ & \ln |y|=-b t+\lambda \\ & t=0, y=1 \\ & 0=0+\lambda \\ & y=e^{-b t} \quad \text{..... (ii)} \end{aligned}$

According to question

$\begin{aligned} & 3 \mathrm{y}(1)=2 \mathrm{x}(1) \\ & 3 \mathrm{e}^{-\mathrm{b}}=2\left(2 \mathrm{e}^{-\mathrm{a}}\right) \\ & \mathrm{e}^{\mathrm{a}-\mathrm{b}}=\frac{4}{3} \end{aligned}$

$\begin{aligned} & \text { For } x(t)=y(t) \\ & \begin{array}{r} 2 \mathrm{e}^{-a t}=e^{-b t} \\ 2=e^{(a-b) t} \\ 2=\left(\frac{4}{3}\right)^t \\ \log _{\frac{4}{3}} 2=t \end{array} \end{aligned}$