We have,
xdy - (y + 2x²)dx = 0
\(\Rightarrow \quad x \frac{d y}{d x}-y-2 x^2=0\)
\(\Rightarrow \quad \frac{d y}{d x}-\frac{1}{x} y=2 x\)          ...(i)
Eq. (i) is a linear differential equation.
On comparing Eq. (i) with \(\frac{d y}{d x}+P y=Q\), we get
\(\begin{gathered} P=-\frac{1}{x} \text { and } Q=2 x \end{gathered}\)
\(\begin{gathered} \therefore \mathrm{FF}=e^{\int P d x}=e^{\int-\frac{1}{x} d x}=e^{-\log x}=e^{\log (x)^{-1}}=(x)^{-1}=\frac{1}{x} \end{gathered}\)
The solution is y.IF = \(\int Q(\mathrm{IF}) d x+C\)
\(\begin{array}{ll} y \times \frac{1}{x}=\int\left(2 x \times \frac{1}{x}\right) d x+C \end{array}\)
\(\begin{array}{ll} \Rightarrow & \frac{y}{x}=\int 2 d x+C \end{array}\)
\(\begin{array}{ll} \Rightarrow & \frac{y}{x}=2 x+C \end{array}\)
\(\begin{array}{ll} \Rightarrow & y=2 x^2+C x \end{array}\)

Consider equation  \( e^{x} \tan y d x+\left(2-e^{x}\right) \sec ^{2} y d y=0\)
\(\Rightarrow\left(2-e^{r}\right) \sec ^{2} y d y=-e^{x} \tan y d x \)
\(\Rightarrow \frac{\sec ^{2} y}{\tan y} d y=\frac{e^{x}}{e^{x}-2} d x \)
Integrating both sides, we get
\(\int \frac{\sec ^{2} y}{\tan y} d y=\int \frac{e^{x}}{e^{x}-2} d x\)
\(\Rightarrow \log |\tan y| =\log \left|e^{x}-2\right|+\log C \)
\(=\log \left|C\left(e^{x}-2\right)\right| \)
\(\Rightarrow \tan y =C\left(e^{x}-2\right) \)
\(\text { Given } y=\frac{\pi}{4}, \text { when } x=0 \)
\(\Rightarrow\tan \frac{\pi}{4}=C\left(e^{0}-2\right) \)
\(\Rightarrow 1=-C \Rightarrow C=-1 \)
Substituting in (i), we get
\(\tan y=-\left(e^{x}-2\right)\)
\(or \tan y=2-e^{x}\text{ is particular solution.}\)

Given equation can be written as
\(\Rightarrow cos\left( \frac { y }{ x } \right) \frac { dy }{ dx } =\left( \frac { y }{ x } \right) cos\left( \frac { y }{ x } \right) +1\)
\(\frac { dy }{ dx } =\frac { 1+\left( \frac { y }{ x } \right) cos\left( \frac { y }{ x } \right) }{ cos\left( \frac { y }{ x } \right) } \)
y/x = v
\(\therefore v+x\frac { dv }{ dx } =\frac { 1+vcosv }{ cosv } \)
\(\Rightarrow x\frac { dv }{ dx } =\frac { 1+vcosv }{ cosv } -v=\frac { 1 }{ cosv } \)
\(\Rightarrow \int { cosv.dv } =\int { \frac { dx }{ x } } \)
sin v = log |x| + C
or sin(y/x) = log |x| + C

We have: \(\left( { tan }^{ 2 }x+2\quad tan\quad x+5 \right) \frac { dy }{ dx } \)
\(=2\left( 1+tan\quad x \right) { sec }^{ 2 }\quad x\)
\(\Rightarrow \frac { dy }{ dx } =\frac { 2\left( 1+tan\quad x \right) { sec }^{ 2 }\quad x }{ { tan }^{ 2 }x+2\quad tan\quad x+5 } .\)

Integrating, \(y=\int { \frac { 2\left( 1+tan\quad x \right) { sec }^{ 2 }\quad x }{ { tan }^{ 2 }x+2\quad tan\quad x+5 } } dx+c\) ...(1)
Let   \(y=\int { \frac { 2\left( 1+tan\quad x \right) { sec }^{ 2 }\quad x }{ { tan }^{ 2 }x+2\quad tan\quad x+5 } } dx+c\)
Put \(x=t\) so that \({ sec }^{ 2 }\)\(xdx=dt\)
\(\therefore\)  \(I=\int { \frac { 2\left( 1+t \right) }{ { t }^{ 2 }+2t+5 } } dt=\int { \frac { 2t+2 }{ { t }^{ 2 }+2t+5 } } dt\)
Put \({ t }^{ 2 }+2t+5=z\)    so that \(\left( 2t+2 \right) dt=dz.\)
\(\therefore\)   From(2),  \(I=\int { \frac { dz }{ z } =log|z| } =log|{ t }^{ 2 }+2t+5|\)
\(=log|{ tan }^{ 2 }x+2\quad tan\quad x+5|.\)
\(\therefore\)   From(1),\(y=log|{ tan }^{ 2 }x+2\quad tan\quad x+5|+c,\)
Which is the required general solution

Further, let t = 1, (1 + 1)y = -1 \(\Rightarrow \) y(1) = -\(\frac{1}{2}\).

[ t + log|cos t + sin t|] = 2x + c  \(\Rightarrow\)  y- x + log |cos(x + y) + sin(x + y)| = c is the required solution

\({ tan }^{ -1 }\left| \frac { x+y }{ 1-xy } \right| +\frac { 1 }{ 2 } log|(1+{ x }^{ 2 })(1+{ y }^{ 2 })|=c\) is the required solution

\(\int e^{y} d y=\int\left(e^{x}+x^{3}\right) d x\)
\(\Rightarrow { e }^{ y }={ e }^{ x }+\frac { { x }^{ 4 } }{ 4 } +c\)

tan ^-1y = -tan^-1 ex + \(\frac {\pi}{2}\) \(\Rightarrow \) tan^-1y + tan^-1 e^x = \(\frac {\pi}{2}\)  is the required solution.

\(\Rightarrow \) y = 3x² - 2x is the required solution.

Given, equation is \(\cos \left(\frac{d y}{d x}\right)=a\)
which can be rewritten as \(\frac{d y}{d x}=\cos ^{-1} a\)
\(\Rightarrow\)  dy = cos^-1 a dx
On integrating both sides, we get
\(\begin{aligned} \Rightarrow \quad \int d y=\int \cos ^{-1} a d x \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad y=\cos ^{-1} a \cdot x+C \end{aligned}\)
which is the required solution.

Given diflierential equation is \(\sin \left(\frac{d y}{d x}\right)=a \Rightarrow \frac{d y}{d x}=\sin ^{-1} a\) 
On separating the variables, we get
\(d y=\sin ^{-1} a d x\) 
On integrating both sides, we get
\(\int d y=\int \sin ^{-1} a d x \Rightarrow y=x \sin ^{-1} a+C\) 
which is the required solution

Given differential equation is
\(\frac{d y}{d x}-\frac{x}{x^{2}+1}=0  \)
\(\Rightarrow \frac{d y}{d x}=\frac{x}{x^{2}+1} \)
On separating the variables, we get
\(d y=\frac{x}{x^{2}+1} d x \)
On integrating both sides, we get
\(\int d y=\int \frac{x}{x^{2}+1} d x \Rightarrow \int d y=\frac{1}{2} \int \frac{2 x}{x^{2}+1} d x  \)
\(\text { put } x^{2}+1=t \Rightarrow 2 x d x=d t \)
\(\left[\begin{array}{l} ​​​​​​​\left.\therefore \int \frac{2 x}{x^{2}+1} d x=\int_{t}^{1} d t=\log |t|+C=\log \left|x^{2}+1\right|+C\right] \end{array}\right.  \)
\(\therefore \ y=\frac{1}{2} \log \left|x^{2}+1\right|+C \)
which is the required solution.

m + n = 4
\(\because \ m=2\)  (second order derivative)
\(\therefore \ n=2\)
(degree of the highest order derivative)

Degree of the given differential equation = 3
Order of the given differential equation = 2
Hence, the sum of order and degree = 2 + 3 = 5

Given, differential equation is \(\frac{d y}{d x}+y \sec x=\tan x\)
which is a linear differential equation of first order and is of the form
\(\frac{d y}{d x}+P y=Q\)       ...(i)
Here,   P = sec x and Q = tan x
\(\begin{aligned} \therefore \quad \text { IF }= & e^{\int P d x}=e^{\int \sec x d x}=e^{\log |\sec x+\tan x|} \end{aligned}\)
\(\begin{aligned} {\left[\because \int \sec x d x=\log |\sec x+\tan x|+C\right] } \end{aligned}\)
\(\Rightarrow\)   IF = sec x + tan x
The general solution is \(y \times \mathrm{IF}=\int(Q \times \mathrm{IF}) d x+C\)
\(\begin{aligned} & \Rightarrow y(\sec x+\tan x)=\int \tan x \cdot(\sec x+\tan x) d x \end{aligned}\)
\(\begin{aligned} & \Rightarrow y(\sec x+\tan x)=\int \sec x \tan x d x+\int \tan ^2 x d x \end{aligned}\)
\(\begin{aligned} & \Rightarrow y(\sec x+\tan x)=\sec x+\int\left(\sec ^2 x-1\right) d x \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow y(\sec x+\tan x)=\sec x+\tan x-x+C \end{aligned}\)
                                                \(\left[\because \int \sec ^2 x d x=\tan x+C\right]\)
On dividing both sides by (sec x + tan x), we get
\(y=1-\frac{x}{\sec x+\tan x}+\frac{C}{\sec x+\tan x}\)

Given, differential equation is a linear differential equation of the form \(\frac{d y}{d x}+P y=Q\) and its solution is given by \(y \cdot(\mathrm{IF})=\int Q \cdot(\mathrm{IF})+C\),
where \(\mathrm{IF}=e^{\int P d x}\)
\(y e^{\tan ^{-1} x}=\frac{e^{2 \tan ^{-1} x}}{2}+C\)

We have, \(\frac{d y}{d x}-y=\sin x\), which is a linear differential equation of the form
\(\frac{d y}{d x}+P y=Q\), here P = -1 and Q = sin x
\(\therefore \quad \mathrm{IF}=e^{\int P d x}=e^{\int(-1) d x}=e^{-x}\)
Now, the general solution of given differential equation is given by \(\begin{aligned} y \cdot(\mathrm{IF}) & =\int(\mathrm{IF}) \cdot Q d x+C \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad y \cdot e^{-x} & =\int e^{-x} \sin x d x+C \end{aligned}\)     ...(i)
Let \(I=\int_{II}^{e^{-x}} \)\(\underset{1}{sin}\) x.dx                                   ...(ii)
By using the method of integration by parts, we get
\(\begin{aligned} I & =\sin x \frac{e^{-x}}{(-1)}-\int \cos x \frac{e^{-x}}{(-1)} d x \end{aligned}\)
= -sin x e^-x + \(\int e_{\mathrm{II}}^{-x}\) \(\underset{I}{cos}\) x dx
Again, by using integration by parts, we get
\(\begin{aligned} I & =-\sin x e^{-x}+\cos x \frac{e^{-x}}{(-1)}-\int(-\sin x) \frac{e^{-x}}{(-1)} d x \end{aligned}\)
\(\begin{aligned} =-\sin x e^{-x}-\cos x e^{-x}-\int e^{-x} \sin x d x \end{aligned}\)
= -sin x e^-x - cos x e^-x - I    [from Eq. (ii)]
\(\Rightarrow\)  2I = -e^-x (sin x + cos x)
\(\Rightarrow \quad I=-\frac{e^{-x}}{2}(\sin x+\cos x)\)
Then, from Eq. (1), we get
\(\begin{aligned} y \cdot e^{-x} & =-\frac{e^{-x}}{2}(\sin x+\cos x)+C \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad y & =-\frac{1}{2}(\sin x+\cos x)+C e^x \end{aligned}\)

Given, differential equation is \(\left(x^2+1\right) \frac{d y}{d x}+2 x y=\frac{1}{x^2+1}\)
On dividing both sides by (x² + 1),  we get
\(\frac{d y}{d x}+\frac{2 x}{x^2+1} y=\frac{1}{\left(x^2+1\right)^2}\)
which is a linear differential equation of the form \(\frac{d y}{d x}+P y=Q, \text { here } P=\frac{2 x}{x^2+1} \text { and } Q=\frac{1}{\left(x^2+1\right)^2}\)
Now, integrating factor, IF \(=e^{\int P d x}=e^{\int-\frac{2 x}{x^2+1} d x}\)
\(=e^{\log \left|x^2+1\right|}=x^2+1\)
         \(\left[\begin{array}{l} \text { put } x^2+1=t \Rightarrow 2 x d x=d t, \text { then } \\ \int \frac{2 x}{x^2+1} d x=\int \frac{1}{t} d t=\log |t|=\log \left|x^2+1\right| \end{array}\right]\)
So, the required general solution is
\(\begin{aligned} y \times \mathrm{IF}=\int(Q \times \mathrm{IF}) d x+C \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad y\left(x^2+1\right) & =\int \frac{1}{\left(x^2+1\right)^2} \times\left(x^2+1\right) d x+C \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad y\left(x^2+1\right) & =\int \frac{1}{x^2+1} d x+C \end{aligned}\)
\(\Rightarrow\)   y(x² + 1) = tan^-1 x + C                   ...(i)
when x = 1, then y = 0
\(\therefore 0=\tan ^{-1} 1+C \Rightarrow C=\frac{-\pi}{4}\)
Now, \(y\left(x^2+1\right)=\tan ^{-1} x-\frac{\pi}{4}\)  [from Eq. (i)]
which is the required differential equation.

Given, \(\frac{d v}{d t}=g \cos \alpha-k v\) 
\(\Rightarrow \frac{d v}{d t}+k v=g \cos 0\) 
which is a linear differential equation of the form
\(\frac{d v}{d t}+P v=Q\) 
Here, \(P=k \text { and } Q=g \cos \alpha\) 
Now, \(\mathrm{IF}=e^{\int P d t}=e^{\int k d t}=e^{k t}\) 
and the solution of the differential equation is given by
\(v \cdot e^{k t}=\int e^{k t} \cdot g \cos \alpha d t+C\) 
\(\Rightarrow v \cdot e^{k t}=g \cos \alpha \int e^{k t} d t+C\) 
\(\Rightarrow v \cdot e^{k t}=\frac{g \cos \alpha e^{k t}}{k}+C\) 
It is given that v = 0, when t = 0.
\(\therefore\) From Eq. (i), we get
\(0=\frac{g \cos \alpha}{k}+C\) 
\(\Rightarrow C=\frac{-g \cos \alpha}{k}\) 
On putting the value of C in Eq. (i), we get
\(v \cdot e^{k t}=\frac{g \cos \alpha e^{k t}}{k}-\frac{g \cos \alpha}{k}\) 
\(\Rightarrow v \cdot e^{k t}=\frac{g \cos \alpha}{k}\left(e^{k t}-1\right)\)
\(\Rightarrow v=\frac{g \cos \alpha}{k}\left(1-e^{-k t}\right)\) 
which is the required expression.

Given differential equation is
\(\left(1+x^{2}\right) d y+2 x y d x=\cot x d x \quad[\because x \neq 0]\) 
Above equation can be rewritten as,
\(\left(1+x^{2}\right) d y+(2 x y-\cot x) d x=0\) 
\(\Rightarrow\left(1+x^{2}\right) d y=(\cot x-2 x y) d x\) 
On dividing both sides by \(1+x^{2}\) ,we get
\(d y=\frac{\cot x-2 x y}{1+x^{2}} d x\) 
\(\Rightarrow \frac{d y}{d x}=\frac{\cot x}{1+x^{2}}-\frac{2 x y}{1+x^{2}}\) 
\(\Rightarrow \frac{d y}{d x}+\frac{2 x}{1+x^{2}} y=\frac{\cot x}{1+x^{2}}\) 
which is a linear differential equation of the form of
\(\frac{d y}{d x}+P y=Q\) 
Here, \(P=\frac{2 x}{1+x^{2}} \text { and } Q=\frac{\cot x}{1+x^{2}}\) 
Now,\(\mathrm{IF}=e^{\int P d x}=e^{\frac{1}{1+x^{2}} d x}=e^{\log \left|1+x^{2}\right|}=1+x^{2}\) 
\(\left[\because I_{1}=\int \frac{2 x}{1+x^{2}} d x,\right. \text { put } 1+x^{2}=t \Rightarrow 2 x d x=d t\) 
\(\left.\Rightarrow I_{1}=\int \frac{d t}{t}=\log |t|=\log \left|1+x^{2}\right|\right]\)
and the solution of linear differential equation is given by
\(y \times I F=\int(Q \times I F) d x+C\) 
\(\Rightarrow y\left(1+x^{2}\right)=\int \frac{\cot x}{1+x^{2}} \times\left(1+x^{2}\right) d x+C\) 
\(\Rightarrow y\left(1+x^{2}\right)=\int \cot x d x+C\) 
\(\Rightarrow y\left(1+x^{2}\right)=\log |\sin x|+C\) 
\(\Rightarrow y=\frac{\log |\sin x|}{1+x^{2}}+\frac{C}{1+x^{2}}\) 
which is the required solution.

Given differential equation can be rewritten as
\(F(x, y)=\frac{d y}{d x}=\frac{y}{2 x-x \log \left(\frac{y}{x}\right)}\) 
Verify \(F(\lambda x, \lambda y)=F(x, y)\) 
On putting \(y=v x \text { and } \frac{d y}{d x}=v+x \frac{d v}{d x}\) ,then given equation becomes
\(v+x \frac{d v}{d x}=\frac{v x}{2 x-x \log \left(\frac{v x}{x}\right)}\) 
\(\Rightarrow x \frac{d v}{d x}=\frac{v}{2-\log v}-v\) 
\(\Rightarrow \int \frac{2-\log v}{v(\log v-1)} d v=\int \frac{d x}{x}\) 
On putting log v = t and \(\frac{1}{v} d v=d t, \text { we get }\) 
\(\int \frac{2-t}{t-1} d t=\log |x|+C\) 
\(\Rightarrow \int\left(\frac{1}{t-1}-1\right) d t=\log |x|+C\) 
\(=\log \left|\frac{\log \frac{y}{x}-1}{y}\right|=C\) 

Given, \({ x }^{ 2 }dy+\left( xy+{ y }^{ 2 } \right) dx=0\)
\(\therefore \ \frac { dy }{ dx } =\frac { -\left( xy+{ y }^{ 2 } \right) }{ { x }^{ 2 } } \)
Put  y = vx
\(\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
\(\therefore \) The differential equation becomes
\(v+x\frac { dv }{ dx } =-\left( v+{ v }^{ 2 } \right) \)
\(\Rightarrow \frac { dv }{ { v }^{ 2 }+2v } =\frac { dx }{ x } \)
\(\Rightarrow \int { \frac { dv }{ \left( v+1 \right) ^{ 2 }-{ 1 }^{ 2 } } } =-\int { \frac { dx }{ x } } \)
\(\Rightarrow \frac { 1 }{ 2 } \log { \frac { v }{ v+2 } } =-\log { x } +\log { C } \)
\(\Rightarrow \frac { C }{ x } =\sqrt { \frac { y }{ y+2x } } \)
If x = 1, y = 1 then \(c=\frac { 1 }{ \sqrt { 3 } } \)
\(\Rightarrow \frac { 1 }{ \sqrt { 3 } x } =\sqrt { \frac { y }{ y+2x } } \)