Differential Equations

\( f(x) \cdot \sin 2 x+\sin x-\left(1+\cos ^2 x\right) f^{\prime}(x)=0 \)
\( \Rightarrow\left(1+\cos ^2 x\right) f^{\prime}(x)-(\sin 2 x) f(x)=\sin x \)
\( \Rightarrow \frac{d}{d x}\left(f(x)\left(1+\cos ^2 x\right)\right)=\sin x \)
\( \Rightarrow f(x)\left(1+\cos ^2 x\right)=-\cos x+c, \text { As } f(0)=0 \Rightarrow c=1 \)
\( \Rightarrow f(x)=\frac{1-\cos ^2 x}{1+\cos ^2 x} \Rightarrow f\left(\frac{\pi}{2}\right)=1\)

\( f(x) \cdot \sin 2 x+\sin x-\left(1+\cos ^2 x\right) f^{\prime}(x)=0 \)
\( \Rightarrow\left(1+\cos ^2 x\right) f^{\prime}(x)-(\sin 2 x) f(x)=\sin x \)
\( \Rightarrow \frac{d}{d x}\left(f(x)\left(1+\cos ^2 x\right)\right)=\sin x \)
\( \Rightarrow f(x)\left(1+\cos ^2 x\right)=-\cos x+c, \text { As } f(0)=0 \Rightarrow c=1 \)
\( \Rightarrow f(x)=\frac{1-\cos ^2 x}{1+\cos ^2 x} \Rightarrow f\left(\frac{\pi}{2}\right)=1\)

$x\frac{dy}{dx}-\sin 2y=x^3\left(2-x^3\right){\cos }^{2}y$

${\sec }^{2}y\frac{dy}{dx}-2\tan y\cdot \frac{1}{x}=x^2\left(2-x^3\right)$

$\tan y=t\Rightarrow {\sec }^{2}y\frac{dy}{dx}=\frac{dt}{dx}$

$\frac{dt}{dx}-\frac{2t}{x}=x^2\left(2-x^3\right)$(LDE)

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

$\therefore \frac{t}{x^2}=\int \frac{1}{x^2}{x}^2\left(2-x^3\right)dx+C$

$\frac{\tan y}{x^2}=2x-\frac{x^4}{4}+C$

$y\left(2\right)=0\Rightarrow 0=4-4+C\Rightarrow C=0$

$\tan y=2x^3-\frac{1}{4}{x}^6$

at $x=1\Rightarrow \tan y=2-\frac{1}{4}=\frac{7}{4}$

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

$$\begin{gathered} \Rightarrow \frac{dx}{dy}+\frac{x}{1+y^2}=\frac{2e^{{\tan }^{-1}y}}{1+y^2} \\ \text{I.F. }={\mathrm{e}}^{\int \frac{\mathrm{dy}}{1+{\mathrm{y}}^2}}={\mathrm{e}}^{{\tan }^{-1}\mathrm{y}} \\ {\mathrm{xe}}^{{\tan }^{-1}\mathrm{y}}=\int \frac{2{\left({\mathrm{e}}^{{\tan }^{-1}\mathrm{y}}\right)}^2\mathrm{dy}}{1+{\mathrm{y}}^2} \\ \text{Put }{\tan }^{-1}\mathrm{y}=\mathrm{t},\frac{\mathrm{dy}}{1+{\mathrm{y}}^2}=\mathrm{dt} \\ {\mathrm{xe}}^{{\tan }^{-1}\mathrm{y}}=\int 2{\mathrm{e}}^{2\mathrm{t}}\mathrm{dt} \\ {\mathrm{xe}}^{{\tan }^{-1}\mathrm{y}}={\mathrm{e}}^{2{\tan }^{-1}\mathrm{y}}+\mathrm{c} \\ \mathrm{x}={\mathrm{e}}^{{\tan }^{-1}\mathrm{y}}+{\mathrm{ce}}^{-{\tan }^{-1}\mathrm{y}} \\ \because \mathrm{y}=0,\mathrm{x}=1 \\ 1=1+\mathrm{c}\Rightarrow \mathrm{c}=0 \\ \mathrm{x}={\mathrm{e}}^{{\tan }^{-1}\mathrm{y}} \\ \text{at} \mathrm{y}=\frac{1}{\sqrt{3}} \Rightarrow \mathrm{x}={\mathrm{e}}^{\mathrm{\pi }/6} \end{gathered}$$

$\underset{t\to x}{\lim }\frac{t^{2}y\left(x\right)-x^{2}y\left(t\right)}{x-t}=3$

Using L'hospital rule :

$\underset{t\to x}{\lim }\frac{2ty\left(x\right)-x^2{y}^{\text{'}}\left(t\right)}{-1}=3$

$x^2{y}^{\text{'}}\left(x\right)-2xy\left(x\right)=3$

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

$\text{I.F. }=e^{-\int \frac{2}{x}dx}=e^{-2{\log }_{e}x}=1/x^2$

$y\cdot \frac{1}{x^2}=\int \frac{3}{x^4}dx$

$\frac{y}{x^2}=-\frac{1}{x^3}+c$

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

$y\left(1\right)=2=c-1\Rightarrow c=3$

$y\left(x\right)=3x^2-\frac{1}{x}$

$y\left(2\right)=12-\frac{1}{2}\Rightarrow 2f\left(2\right)=23$

$$\begin{gathered} ydy=(xdy-ydx)\sin \left(\frac{x}{y}\right) \\ \frac{dy}{y}=\left(\frac{xdy-ydx}{y^2}\right)\sin \left(\frac{x}{y}\right) \\ \frac{dy}{y}=\sin \left(\frac{x}{y}\right)d\left(-\frac{x}{y}\right) \\ \Rightarrow \ln y=\cos \frac{x}{y}+C \\ \text{Since}, x\left(1\right)=\frac{\pi }{2} \\ \Rightarrow 0=\cos \frac{\pi }{2}+C\Rightarrow C=0 \\ \Rightarrow \ln y=\cos \frac{x}{y} \\ \text{ but }y=2 \\ \Rightarrow \cos \frac{x}{2}=\ln 2 \\ \cos x=2{\cos }^2\frac{x}{2}-1=2(\ln 2{)}^2-1 \end{gathered}$$

\(\dfrac{dy}{dx}=\dfrac{y^3(e^{5x}+1)}{e^x(1+y^4)}\)

\(\dfrac{1+y^4}{y^3}\,dy=(e^{4x}+e^{-x})\,dx\)

\(\int\left(y+\dfrac1{y^3}\right)dy=\int(e^{4x}+e^{-x})dx\)

\(\dfrac{y^2}{2}-\dfrac1{2y^2}=\dfrac{e^{4x}}4-e^{-x}+C\)

\(y(0)=\dfrac1{\sqrt2}\)

\(\dfrac{1/2}{2}-\dfrac1{2(1/2)}=\dfrac14-1+C\)

\(\dfrac14-1=\dfrac14-1+C\)

\(C=0\)

\(\dfrac{y^2}{2}-\dfrac1{2y^2}=\dfrac{e^{4x}}4-e^{-x}\)

At \(x=\ln2\),

\(\dfrac{y^2}{2}-\dfrac1{2y^2}=4-\dfrac12=\dfrac72\)

\(y^2-\dfrac1{y^2}=7\)

\(y^4-7y^2-1=0\)

\(y^2=\dfrac{7+\sqrt{53}}2\)

\(y=\sqrt{\dfrac{7+\sqrt{53}}2}\)

\(d y=y\left(2+\log _e x\right) d x\)

\(\frac{d y}{y}=\left(2+\log _e x\right) d x\)

\(\int \frac{d y}{y}=\int\left(2+\log _e x\right) d x\)

\(\log _e y=2 x+x \log _e x-x+C\)

\(\log _e y=x+x \log _e x+C\)

Given curve passes through \((1, e)\), so \(x=1, y=e\)

\(\log _e e=1+1 \log _e 1+C\)

\(1=1+0+C \Rightarrow C=0\)

\(\log _e y=x+x \log _e x\)

At \(x=e, f(e)=y\)

\(\log _e f(e)=e+e \log _e e=e+e=2 e\)

\(f(e)=e^{2 e}\)

\(\begin{aligned}
& \frac{d y}{d x}+\left(\frac{x}{1+x^2}\right) y=\frac{\sqrt{x}}{\sqrt{1+x^2}}; \quad y(0)=0 \\
& \text { I.f} =e^{\int \frac{x}{1+x^2} d x} e^{\frac{1}{2} \ln \left(1+x^2\right)} \\
& I.f=\sqrt{1+x^2} \\
& y \sqrt{1+x^2}=\int \sqrt{x} d x \\
& y \sqrt{1+x^2}=\frac{2}{3} x^{\frac{3}{2}}+C \\
& x=0, \quad y=0 \\
& c=0 \\
& y \sqrt{1+x^2}=\frac{2}{3} x^{\frac{3}{2}} \\
& x=1 \\
& y \cdot \sqrt{2}=\frac{2}{3} \\
& y=\frac{2}{3 \sqrt{2}}
\end{aligned}\)

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

\(Q(x)=2+e^{-2 x}\)

\(P(x)=\frac{3 x^2\left(2+e^{-2 x}\right)+2 e^{-2 x}\left(x^3+2\right)}{\left(x^3+2\right)\left(2+e^{-2 x}\right)}\)

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

Integrating factor \(=e^{\int P(x) d x}=e^{\int \frac{3 x^2}{x^3+2} d x+\int \frac{2 e^{-2 x}}{2+e^{-2 x}} d x}=\frac{x^3+2}{2+e^{-2 x}}\)

\(\Rightarrow \frac{y \cdot\left(x^3+2\right)}{2+e^{-2 x}}=\int\left(x^3+2\right) d x+C=\frac{x^4}{4}+2 x+C\)

\(\because y(0)=\frac{3}{2} \Rightarrow C=1\)

\(\Rightarrow \frac{y \cdot\left(x^3+2\right)}{2+e^{-2 x}}=\frac{x^4}{4}+2 x+1\)

Now,

\(\Rightarrow \frac{y(1) \cdot 3}{2+e^{-2}}=\frac{13}{4}\)

\(\Rightarrow y(1)=\frac{13}{12}\left(2+e^{-2}\right)\)

\( \Rightarrow \alpha=\frac{13}{12}\)

$\left(1+y^2\right)e^{\tan x}dx+{\cos }^{2}x\left(1+e^{2\tan x}\right)dy=0$

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

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

$\text{for }x=0,y=1,{\tan }^{-1}\left(1\right)+{\tan }^{-1}1=C$

$C=\frac{\pi }{2}$

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

$\text{Put }x=\pi ,{\tan }^{-1}e+{\tan }^{-1}y=\frac{\pi }{2}$

${\tan }^{-1}y={\cot }^{-1}e$

$y=\frac{1}{e}$

$\left(1+y^2\right)e^{\tan x}dx+{\cos }^{2}x\left(1+e^{2\tan x}\right)dy=0$

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

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

$\text{for }x=0,y=1,{\tan }^{-1}\left(1\right)+{\tan }^{-1}1=C$

$C=\frac{\pi }{2}$

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

$\text{Put }x=\pi ,{\tan }^{-1}e+{\tan }^{-1}y=\frac{\pi }{2}$

${\tan }^{-1}y={\cot }^{-1}e$

$y=\frac{1}{e}$

$$\begin{gathered} 2\cos x \frac{dy}{dx}=\sin 2x-4y.\sin x \\ 2\cos x\frac{dy}{dx}+4y\sin x=2\sin x \cos x \\ \frac{dy}{dx}+2y\tan x =\sin x \\ Now I.F = e^{\int 2\tan x dx}=e^{\ln se{c}^{2}x}=se{c}^{2}x \\ For General Solution \\ y. IF = \int Q. IF dx + c \\ y. se{c}^{2}x=\int \frac{\sin x}{{\cos }^{2}x}dx+c \\ let, \cos x = t \Rightarrow \sin x dx = -dt \\ y. se{c}^{2}x=-\int \frac{1}{t^2}dt+c \\ y. se{c}^{2}x=sec x +c \\ y=\frac{1}{secx}+\frac{c}{se{c}^{2}x} \\ y = \cos x+{\cos }^{2}x. c \\ y\left(\frac{\mathrm{\pi }}{3}\right)=0 \left(Given\right) \\ 0=\frac{1}{2}+\frac{c}{4} \Rightarrow c= -2 \\ So, y = \cos x-2{\cos }^{2}x \\ y\left(\frac{\mathrm{\pi }}{4}\right)=\frac{1}{\sqrt{2}}-1...........................\left(i\right) \\ and y\text{'}=-\sin x+4\sin x \cos x \\ y\text{'}\left(\frac{\mathrm{\pi }}{4}\right)=-\frac{1}{\sqrt{2}}+2...................\left(ii\right) \\ Adding equation \left(i\right) \& \left(ii\right), we get \\ y\text{'}\left(\frac{\mathrm{\pi }}{4}\right)+y\left(\frac{\mathrm{\pi }}{4}\right)=1 \end{gathered}$$

Given \( \frac{dy}{dx}=2x(x+y)^3-x(x+y)-1 \).

Put \(u=x+y\).

Then \( \frac{du}{dx}=1+\frac{dy}{dx} \).

So \( \frac{du}{dx}=1+2xu^3-xu-1 \).

Thus \( \frac{du}{dx}=2xu^3-xu=xu(2u^2-1) \).

Now put \(z=u^2\), then \( \frac{dz}{dx}=2u\frac{du}{dx} \).

So \( \frac{dz}{dx}=2x u^2(2u^2-1)=2xz(2z-1) \).

Put \(v=\frac{1}{z}\), then \( \frac{dv}{dx}=-\frac{1}{z^2}\frac{dz}{dx} \).

Hence \( \frac{dv}{dx}=-\frac{2xz(2z-1)}{z^2}=-2x\left(2-\frac{1}{z}\right) \).

So \( \frac{dv}{dx}=2xv-4x \).

Therefore \( \frac{dv}{dx}-2xv=-4x \).

Here integrating factor \(=\mathrm{e}^{\int -2x\,dx}=\mathrm{e}^{-x^2}\).

Multiplying by \( \mathrm{e}^{-x^2} \), we get

\( \frac{d}{dx}\left(v\mathrm{e}^{-x^2}\right)=-4x\mathrm{e}^{-x^2} \).

Integrating, \( v\mathrm{e}^{-x^2}=2\mathrm{e}^{-x^2}+C \).

Thus \(v=2+C\mathrm{e}^{x^2}\).

Since \(y(0)=1\), we get \(u(0)=0+y(0)=1\).

So \(z(0)=1\) and \(v(0)=1\).

Therefore \(1=2+C\), hence \(C=-1\).

So \(v=2-\mathrm{e}^{x^2}\).

Since \(v=\frac{1}{z}\) and \(z=u^2\), we get \(u^2=\frac{1}{2-\mathrm{e}^{x^2}}\).

Now \(u=x+y\).

Therefore \( \left(x+y(x)\right)^2=\frac{1}{2-\mathrm{e}^{x^2}} \).

At \(x=\frac{1}{\sqrt{2}}\), \(x^2=\frac{1}{2}\).

Hence \( \left(\frac{1}{\sqrt{2}}+y\left(\frac{1}{\sqrt{2}}\right)\right)^2=\frac{1}{2-\mathrm{e}^{\frac{1}{2}}}=\frac{1}{2-\sqrt{\mathrm{e}}} \).

Equation:

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

Note $3\tan^2x+3=3(\tan^2x+1)=3\sec^2x$, so

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

Integrating factor:

$$\mu(x)=e^{\int 3\sec^2x\,dx}=e^{3\tan x}$$

Thus

$$\frac{d}{dx}\left(y e^{3\tan x}\right)=\sec^2x\,e^{3\tan x}$$

Let $t=\tan x$, $dt=\sec^2x\,dx$:

$$y e^{3t}=\int e^{3t}dt + C = \frac13 e^{3t}+C$$ $$\Rightarrow y=\frac13 + Ce^{-3\tan x}$$

Use $y(0)=\frac13+e^3$ and $\tan 0=0$:

$$\frac13 + C = \frac13 + e^3 \Rightarrow C=e^3$$

So

$$y(x) = \frac13 + e^3 e^{-3\tan x}$$

At $x=\frac{\pi}{4}$, $\tan\frac{\pi}{4}=1$:

$$y\left(\frac{\pi }{4}\right)=\frac{1}{3}+e^3{e}^{-3}=\frac{1}{3}+1=\frac{4}{3}$$

\(\begin{aligned}
& \frac{d y}{d x}+\left(\frac{x}{1+x^2}\right) y=\frac{\sqrt{x}}{\sqrt{1+x^2}}; \quad y(0)=0 \\
& \text { I.f} =e^{\int \frac{x}{1+x^2} d x} e^{\frac{1}{2} \ln \left(1+x^2\right)} \\
& I.f=\sqrt{1+x^2} \\
& y \sqrt{1+x^2}=\int \sqrt{x} d x \\
& y \sqrt{1+x^2}=\frac{2}{3} x^{\frac{3}{2}}+C \\
& x=0, \quad y=0 \\
& c=0 \\
& y \sqrt{1+x^2}=\frac{2}{3} x^{\frac{3}{2}} \\
& x=1 \\
& y \cdot \sqrt{2}=\frac{2}{3} \\
& y=\frac{2}{3 \sqrt{2}}
\end{aligned}\)

Let the centre of the circle be \((a,a)\), since the centre lies on the line \(y=x\).

Since the circle passes through the origin \((0,0)\),

its radius is the distance between \((a,a)\) and \((0,0)\).

So, \(r^2=a^2+a^2=2a^2\).

Therefore, the equation of the circle is \((x-a)^2+(y-a)^2=2a^2\).

\(x^2+y^2-2a(x+y)=0\).

\(2x+2y\frac{dy}{dx}-2a\left(1+\frac{dy}{dx}\right)=0\)

Let \(\frac{dy}{dx}=y'\).

So, \(2x+2yy'-2a(1+y')=0\).

Hence, \(a=\frac{x+yy'}{1+y'}\).

From the circle equation,

\(x^2+y^2=2a(x+y)\).

Substitute \(a=\frac{x+yy'}{1+y'}\).

\(x^2+y^2=2(x+y)\frac{x+yy'}{1+y'}\)

\((x^2+y^2)(1+y')=2(x+y)(x+yy')\).

\(x^2+y^2+x^2y'+y^2y'=2x^2+2xy+2xyy'+2y^2y'\)

So, \((x^2-2xy-y^2)y'+y^2-x^2-2xy=0\).

Therefore, \((x^2-2xy-y^2)\frac{dy}{dx}=x^2+2xy-y^2\).

$$\begin{gathered} (1+y^2)+(x-2{\tan }^{-1}y)\frac{dy}{dx}=0 \\ (1+y^2)=-(x-2{\tan }^{-1}y)\frac{dy}{dx} \\ \frac{dx}{dy}=-\frac{-x}{1+y^2}+\frac{2{\tan }^{-1}y}{1+y^2} \\ \frac{dx}{dy}+\frac{x}{1+y^2}=\frac{2{\tan }^{-1}y}{1+y^2} \\ On comparing with \frac{dx}{dy}+Px=Q \\ P = \frac{1}{1+y^2}, Q = \frac{2{\tan }^{-1}y}{1+y^2} \\ I.F. = e^{\int \frac{1}{1+y^2}dy}=e^{{\tan }^{-1}y} \\ Now general solution is \\ x. (I.F) = \int Q. (I.F) dy + c \\ x.e^{{\tan }^{-1}y}=\int \frac{2{\tan }^{-1}y}{1+y^2}.e^{{\tan }^{-1}y} dx \\ let {\tan }^{-1}y = t \\ \frac{1}{1+y^2}dy= dt \\ x.e^{{\tan }^{-1}y}=\int 2t.e^{t}dt \\ x.e^{{\tan }^{-1}y}=2\left(t\int e^{t}dt-\int \left\{\frac{dt}{dt}\int e^{t}dt\right\}dt\right) \\ x.e^{{\tan }^{-1}y}=2\left(t.e^t-e^t\right)+c \\ x.e^{{\tan }^{-1}y}=2e^{{\tan }^{-1}y}({\tan }^{-1}y-1)+c \\ x =2\left({\tan }^{-1}y-1\right)+c{e}^{-{\tan }^{-1}y} \end{gathered}$$

\(
\begin{gathered}
\begin{array}{l}
y=0, x=1 \\
c=3 \\
x e^{\tan ^{-1} y}=2 \tan ^{-1} y \cdot e^{-tan ^{-1} y } -2 e^{\tan ^{-1} y}+3 \\
f(y)=x=2 \tan ^{-1} y-2+3 e^{-tan ^{-1} y } \\
f\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{3}-2+3 e^{-\pi / 6}
\end{array}
\end{gathered}
\)

\(\begin{aligned}
& y=\left(x-y \frac{d x}{d y}\right) \sin \left(\frac{x}{y}\right) \\
& 1=\left(\frac{x}{y}-\frac{d x}{d y}\right) \sin \left(\frac{x}{y}\right) \\
& \frac{x}{y}=v \\
& \frac{d x}{d y}=v+y \frac{d v}{d y} \\
& 1=\left(v-\left(v+y \frac{d v}{d y}\right)\right) \sin v \\
& 1=v-v-y \frac{d v}{d y} \cdot \sin v \\
& 1=-y \frac{d v}{d y} \cdot \sin v \\
& \frac{d y}{y}=-\sin v d v \\
& \ln y=\cos v+c \\
& \ln y=\cos \frac{x}{y}+c \\
& 0=0+c \\
& c=0 \\
& \ln y=\cos \frac{x}{y} \\
& \ln 2=\cos \left(\frac{x}{2}\right) \\
\end{aligned}\)

\(\begin{aligned}
& \cos x=2 \cos ^2 \frac{x}{2}-1 \\
& =2 \ln ^2 2-1
\end{aligned}\)

$\frac{dy}{dx}-y{\log }_{e}2=2^{\sin x}(\cos x-1){\log }_{e}2$

This is linear differential equation

$\text{ I.F. }=e^{-{\log }_{e}2\int dx}=e^{-x{\log }_{e}2}=2^{-x}$

Solution is

$$\begin{gathered} y{2}^{-x}=\int 2^{-x}{2}^{\sin x}(\cos x-1){\log }_{e}2dx \\ \text{ put }\sin x-x=t\Rightarrow (\cos x-1)dx=dt \\ \therefore y{2}^{-x}={\log }_{e}2\int 2^{t}dt \\ \therefore y{2}^{-x}=2^t+c \\ \therefore y=2^{x+t}+c{2}^x \\ \therefore y=2^{\sin x}+c{2}^x \end{gathered}$$

$$\begin{gathered} \sqrt{4-x^2}\frac{dy}{dx}=\left[{\left({\sin }^{-1}\left(\frac{x}{2}\right)\right)}^2-2\right]{\sin }^{-1}\frac{x}{2} \\ \Rightarrow \int dy=\int \frac{\left[{\left({\sin }^{-1}\left(\frac{x}{2}\right)\right)}^2-2\right]{\sin }^{-1}\frac{x}{2}}{\sqrt{4-x^2}}dx \\ Let, t ={\sin }^{-1}\frac{x}{2} \Rightarrow \frac{1}{\sqrt{1-\frac{x^2}{4}}}\times \frac{1}{2}dx=dt \\ \Rightarrow dt=\frac{dx}{\sqrt{4-x^2}} \\ \int dy=\int t^3-2t dt \\ y = \frac{t^4}{4}-t^2+c \\ y=\frac{1}{4} {\left({\sin }^{-1}x\right)}^4-{\left({\sin }^{-1}x\right)}^2+c \\ \mathrm{Now} y\left(0\right) = -6 \mathrm{then} c=-6. \\ y\left(2\right)=\frac{1}{4}{\left({\sin }^{-1}1\right)}^2-{\left({\sin }^{-1}1\right)}^2-6 \\ =\frac{{\mathrm{\pi }}^4}{64}-\frac{{\mathrm{\pi }}^2}{4}-6 \end{gathered}$$

\(\sin \frac{\mathrm{y}}{\mathrm{x}} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{\mathrm{x}} \sin \frac{\mathrm{y}}{\mathrm{x}}-1 \)
Put, \( y=t x \Rightarrow \frac{d y}{d x}=t+x \frac{d t}{d x} \)

\(\sin t\left(t+x \frac{d t}{d x}\right)=t \sin t-1 \)
\(\mathrm{x} \sin \mathrm{t} \frac{\mathrm{dt}}{\mathrm{dx}}+1=0 \)
\(\sin t d t+\frac{d x}{x}=0 \)
\(\Rightarrow-\cos t+\ell \mathrm{nx}=\mathrm{C} \)
\(\Rightarrow-\cos \left(\frac{\mathrm{y}}{\mathrm{x}}\right)+\ell \mathrm{nx}=\mathrm{C}\)
\(\mathrm{y}(1)=\frac{\pi}{2} \Rightarrow \mathrm{C}=0 \)
\(\Rightarrow \cos \left(\frac{\mathrm{y}}{\mathrm{x}}\right)=\ln \mathrm{x} \)
\(\operatorname{cos}\left(\frac{\mathrm{y}\left(\mathrm{e}^{12}\right)}{\mathrm{e}^{12}}\right)=12\)

which is not possible.

Its a bonus Question

$$\begin{gathered} \text{Since, it is a Linear Differential equation:} \\ \text{ I.F.= }{\mathrm{e}}^{\int \tan xdx}={\mathrm{e}}^{ln\left(\sec x\right)}=\sec x \\ \Rightarrow y\left(I.F.\right)=\int \left(I.F.\right)Q\left(x\right) dx + C \\ \therefore y·\sec x=\int \left\{\frac{2+\sec x}{(1+2\sec x{)}^2}\right\}\sec xdx \\ =\int \frac{2\cos x+1}{(\cos x+2{)}^2}dx \\ \text{ Let }\cos x=\frac{1-t^2}{1+t^2} \\ =\int \frac{2\left(\frac{1-t^2}{1+t^2}\right)+1}{{\left(\frac{1-t^2}{1+t^2}+2\right)}^2}2\mathrm{dt} \\ =\int \frac{2-2t^2+1+t^2}{{\left(1-t^2+2+2t^2\right)}^2}\times 2dt \\ =2\int \frac{3-t^2}{{\left(t^2+3\right)}^2}dt \\ \text{ Let }t+\frac{3}{t}=z \\ \left(1-\frac{3}{t^2}\right)\mathrm{dt}=dz \\ =-2\int \frac{dz}{z^2} \\ y·\left(\sec x\right)=\frac{2}{z}+c \\ y·\sec x=\frac{2}{t+\frac{3}{t}}+c ... \left(1\right) \\ \text{ At }\mathrm{x}=\frac{\pi }{3},\mathrm{t}=\tan \frac{\mathrm{x}}{2}=\frac{1}{\sqrt{3}} \\ 2·\frac{\sqrt{3}}{10}=\frac{2}{\frac{1}{\sqrt{3}}+3\sqrt{3}}+\mathrm{c} \\ \text{ 2. }\frac{\sqrt{3}}{10}=\frac{2\sqrt{3}}{10}+\mathrm{c} \\ \Rightarrow \mathrm{C}=0 \\ \text{ At }\mathrm{x}=\frac{\mathrm{\pi }}{4},\mathrm{t}=\tan \frac{\mathrm{x}}{2}=\sqrt{2}-1 \\ \therefore \mathrm{y}·\sqrt{2}=\frac{2}{\sqrt{2}-1+\frac{3}{\sqrt{2}-1}} \\ \mathrm{y}·\sqrt{2}=\frac{2(\sqrt{2}-1)}{6-2\sqrt{2}} \\ \mathrm{y}=\frac{\sqrt{2}(\sqrt{2}-1)}{2(3-\sqrt{2})}=\frac{1}{\sqrt{2}}\times \frac{2\sqrt{2}-1}{7} \\ =\frac{4-\sqrt{2}}{14} \end{gathered}$$

Divide the entire equation by $(x^2 + 1)$ to get the standard linear form

$y' - \frac{2x}{x^2 + 1}y = \frac{(x^4 + 2x^2 + 1)\cos x}{x^2 + 1}$

$y' - \frac{2x}{x^2 + 1}y = (x^2 + 1)\cos x$

The integrating factor is $e^{\int P(x)dx}$.

$\text{IF} = e^{\int -\frac{2x}{x^2 + 1}dx} = e^{-\ln(x^2 + 1)} = \frac{1}{x^2 + 1}$

$\frac{1}{x^2 + 1}y' - \frac{2x}{(x^2 + 1)^2}y = \cos x$

$\frac{y}{x^2 + 1} = \sin x + C$ $y(x) = (x^2 + 1)(\sin x + C)$

Given $y(0) = 1$:

$1=(0^2+1)(\sin 0+C)$$⟹1=1(0+C)⟹C=1$

So the solution is:

$y\left(x\right)=(x^2+1)(\sin x+1)$$=(x^2+1)\sin x+(x^2+1)$

$I = \int_{-3}^{3} \left[ (x^2 + 1)\sin x + (x^2 + 1) \right] dx$

$I = \int_{-3}^{3} (x^2 + 1)\sin x \, dx + \int_{-3}^{3} (x^2 + 1) \, dx$

$2 \int_{0}^{3} (x^2 + 1) dx = 2 \left[ \frac{x^3}{3} + x \right]_{0}^{3}$

$=2((\frac{27}{3}+3)-0)=2(9+3)=2(12)=24$

Given \( y\frac{dx}{dy}=x\left(\log_e x-\log_e y+1\right) \).

Since \( \log_e x-\log_e y=\log_e\frac{x}{y} \), we get \( y\frac{dx}{dy}=x\left(\log_e\frac{x}{y}+1\right) \).

Put \(v=\frac{x}{y}\), so \(x=vy\).

Then \( \frac{dx}{dy}=v+y\frac{dv}{dy} \).

Substituting, \( y\left(v+y\frac{dv}{dy}\right)=vy(\log_e v+1) \).

\( v+y\frac{dv}{dy}=v\log_e v+v \).

So \( y\frac{dv}{dy}=v\log_e v \).

Therefore \( \frac{dv}{v\log_e v}=\frac{dy}{y} \).

Integrating, \( \log_e|\log_e v|=\log_e y+C \).

Using the point \( (e,1) \), we get \(v=\frac{x}{y}=e\) and \(y=1\).

So \( \log_e|\log_e e|=\log_e 1+C \), hence \(C=0\).

Therefore \( \log_e|\log_e v|=\log_e y \).

Thus \( |\log_e v|=y \).

Since \(v=\frac{x}{y}\), the solution curve is

\( \left|\log_e\frac{x}{y}\right|=y \).

Given differential equation is

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

Dividing above equation by (1+x)

$\frac{dy}{dx}-\frac{1}{(1+x)}y=e^{3x}(1+x)\dots \text{ eq(1) }$

Equation (1) is of the form

$$\begin{gathered} \frac{dy}{dx}+Py=Q \\ \text{ Where, }P=\frac{-1}{(1+x)}\text{ and }Q=e^{3x}(1+x) \end{gathered}$$

Therefore, the integrating factor is

$$\begin{gathered} \text{ I.F. }=e^{\int Pdx} ={\mathrm{e}}^{-\int \frac{1}{(1+\mathrm{x})}\mathrm{dx}} \\ ={\mathrm{e}}^{-\log (1+\mathrm{x})} \\ ={\mathrm{e}}^{\log \frac{1}{(1+\mathrm{x})}} (\hspace{0.17em}\mathrm{using} \log \mathrm{properties}) \\ =\frac{1}{(1+\mathrm{x})} \end{gathered}$$

General solution is

$$\begin{gathered} y.\left(\text{ I.F. }\right)=\int \text{ Q. (I. F.) }dx+c \\ \therefore y·\left(\frac{1}{(1+x)}\right)=\int {\mathrm{e}}^{3\mathrm{x}}(1+\mathrm{x})\left(\frac{1}{(1+\mathrm{x})}\right)\mathrm{dx}+\mathrm{c} \\ \therefore \mathrm{y}·\left(\frac{1}{(1+\mathrm{x})}\right)=\int {\mathrm{e}}^{3\mathrm{x}}\mathrm{dx}+\mathrm{c} \\ \therefore \mathrm{y}·\left(\frac{1}{(1+\mathrm{x})}\right)=\frac{1}{3}{\mathrm{e}}^{3\mathrm{x}}+\mathrm{c}\dots \dots \dots \left(\because \int {\mathrm{e}}^{\mathrm{kx}}\mathrm{dx}=\frac{1}{\mathrm{k}}{\mathrm{e}}^{\mathrm{kx}}\right) \\ \mathrm{Multiplying} \mathrm{above} \mathrm{equation} \mathrm{by} (1+\mathrm{x}) \end{gathered}$$

$\therefore y=\frac{1}{3}(1+x)e^{3x}+c(1+x)$

$\Rightarrow \frac{3y}{x+1}=e^{3x}+C$

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

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

$$\begin{array}{l} \text {I . F .} = e ^ {\int \frac {e ^ {x (x - 2)}}{x (x ^ {2} + e ^ {x})} d x} \\ = e ^ {\int \frac {x . e ^ {x} - 2 e ^ {x}}{x (x ^ {2} + e ^ {x})}} \\ \end{array}$$

$$\begin{array}{l} = e \int \frac {e ^ {x} + 2 x ^ {2} - 2 e ^ {x} - 2 x ^ {2}}{x \left(x ^ {2} + e ^ {x}\right)} \\ = e ^ {\int \frac {e ^ {x} + 2 x ^ {2}}{e ^ {x} + x ^ {2}} - \frac {2}{x}} \\ = e ^ {\ln \left(e ^ {x} + x ^ {2}\right) - \frac {2}{x}} \\ = e ^ {\ln \frac {e ^ {x} + x ^ {2}}{x ^ {2}}} = \frac {e ^ {x} + x ^ {2}}{x ^ {2}} \\ \end{array}$$

Solution

$$y \left(\frac {e ^ {x} + x ^ {2}}{x ^ {2}}\right) = \int \frac {x ^ {3}}{x \left(x ^ {2} + e ^ {x}\right)} \cdot \frac {e ^ {x} + x}{x ^ {2}}$$ $$y \left(\frac {e ^ {x} + x ^ {2}}{x ^ {2}}\right) = x + c$$

Use $x = 1, y = 0 \Rightarrow c = -1$

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

Put $x = 2$

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

$$\begin{gathered} \sqrt{4-x^2}\frac{dy}{dx}=\left[{\left({\sin }^{-1}\left(\frac{x}{2}\right)\right)}^2-2\right]{\sin }^{-1}\frac{x}{2} \\ \Rightarrow \int dy=\int \frac{\left[{\left({\sin }^{-1}\left(\frac{x}{2}\right)\right)}^2-2\right]{\sin }^{-1}\frac{x}{2}}{\sqrt{4-x^2}}dx \\ Let, t ={\sin }^{-1}\frac{x}{2} \Rightarrow \frac{1}{\sqrt{1-\frac{x^2}{4}}}\times \frac{1}{2}dx=dt \\ \Rightarrow dt=\frac{dx}{\sqrt{4-x^2}} \\ \int dy=\int t^3-2t dt \\ y = \frac{t^4}{4}-t^2+c \\ y=\frac{1}{4} {\left({\sin }^{-1}x\right)}^4-{\left({\sin }^{-1}x\right)}^2+c \\ \mathrm{Now} y\left(0\right) = -6 \mathrm{then} c=-6. \\ y\left(2\right)=\frac{1}{4}{\left({\sin }^{-1}1\right)}^2-{\left({\sin }^{-1}1\right)}^2-6 \\ =\frac{{\mathrm{\pi }}^4}{64}-\frac{{\mathrm{\pi }}^2}{4}-6 \end{gathered}$$

$$\begin{gathered} \text{Given,} \\ \left(xy-5x^2\sqrt{1+x^2}\right)dx+\left(1+x^2\right)dy=0 \\ \left(1+x^2\right)\frac{dy}{dx}+xy=5x^2\sqrt{1+x^2} \\ \text{Divide }\left(1+x^2\right)\text{ on both sides, we get :} \\ \frac{dy}{dx}+\frac{xy}{1+x^2}=\frac{5x^2}{\sqrt{1+x^2}} \\ \text{On Comparing with standard L.D.E} \\ \frac{dy}{dx}\text{+Py=Q, we get} \\ P =\frac{x}{1+x^2} \& Q =\frac{5x^2}{\sqrt{1+x^2}} \\ Now, I.F. =e^{\int P dx} \\ =e^{\int \frac{x}{1+x^2}dx} \\ =e^{\frac{\ln \left(1+x^2\right)}{2}} \\ =\sqrt{1+x^2} \\ \text{Now, general solution,} \\ y. IF = \int \mathrm{Q}. \mathrm{IF} \mathrm{dx}+\mathrm{C} \\ \therefore \mathrm{y}\sqrt{1+{\mathrm{x}}^2}=\int \frac{5{\mathrm{x}}^2}{\sqrt{1+{\mathrm{x}}^2}}·\sqrt{1+{\mathrm{x}}^2}\mathrm{dx} +C \\ \mathrm{y}\sqrt{1+{\mathrm{x}}^2}=\frac{5{\mathrm{x}}^3}{3}+\mathrm{C} \\ \mathrm{Now}, \mathrm{x}=0, \mathrm{y}=0 \\ \Rightarrow 0=0+\mathrm{C} \\ \Rightarrow \mathrm{C}=0 \\ \therefore \mathrm{y}=\frac{5{\mathrm{x}}^3}{3\sqrt{1+{\mathrm{x}}^2}} \\ \mathrm{Put}, \mathrm{x} =\sqrt{3}, \mathrm{we} \mathrm{get} \\ \mathrm{y}=\frac{15\sqrt{3}}{3\times 2}=\frac{5\sqrt{3}}{2} \end{gathered}$$

\(\begin{aligned}
& y=\left(x-y \frac{d x}{d y}\right) \sin \left(\frac{x}{y}\right) \\
& 1=\left(\frac{x}{y}-\frac{d x}{d y}\right) \sin \left(\frac{x}{y}\right) \\
& \frac{x}{y}=v \\
& \frac{d x}{d y}=v+y \frac{d v}{d y} \\
& 1=\left(v-\left(v+y \frac{d v}{d y}\right)\right) \sin v \\
& 1=v-v-y \frac{d v}{d y} \cdot \sin v \\
& 1=-y \frac{d v}{d y} \cdot \sin v \\
& \frac{d y}{y}=-\sin v d v \\
& \ln y=\cos v+c \\
& \ln y=\cos \frac{x}{y}+c \\
& 0=0+c \\
& c=0 \\
& \ln y=\cos \frac{x}{y} \\
& \ln 2=\cos \left(\frac{x}{2}\right) \\
\end{aligned}\)

\(\begin{aligned}
& \cos x=2 \cos ^2 \frac{x}{2}-1 \\
& =2 \ln ^2 2-1
\end{aligned}\)

\(\begin{aligned}
\frac{d y}{d x} & =\frac{6-2 e^{2 x} y}{1+e^{2 x}} \\
\frac{d y}{d x} & +\frac{2 e^{2 x}}{1+e^{2 x}} y=\frac{6}{1+e^{2 x}} \\
I F & =e^{\int \frac{2 e^{2 x}}{1+e^{2 x}} d t} \\
& =e^{\ln \left(1+e^{2 x}\right)} \\
& =1+e^{2 x}
\end{aligned}\)

\(\begin{aligned}
& y\left(1+e^{2 x}\right)=\int \frac{6}{1+e^{2 x}} \cdot\left(1+e^{2 x}\right) d x \\
& y\left(1+e^{2 x}\right)=\int 6 d x \\
& y\left(1+e^{2 x}\right)=6 x+c \\
& c=0 \\
& y\left(1+e^{2 x}\right)=6x \\
& k\left(1+e^{2 \ln 2})=6 \ln 2\right. \\
& k=\frac{6 \ln 2}{5}
\end{aligned}\)