The given equation can be written as:
\(\sqrt { \left( 1+{ x }^{ 2 } \right) \left( 1+{ y }^{ 2 } \right) } +xy\quad \frac { dy }{ dx } =0.\)
\(\Rightarrow \frac { \sqrt { 1+x^{ 2 } } }{ x } dx+\frac { y }{ \sqrt { 1+{ y }^{ 2 } } } dy=0.\)        
|Variables Separable
Integrating   \(\int { \frac { \sqrt { 1+{ x }^{ 2 } } }{ x } dx+ } \int { \frac { y }{ \sqrt { 1+{ y }^{ 2 } } } } dy=c\)   ...(1)
Now \({ I }_{ 1 }=\int { \frac { \sqrt { 1+{ x }^{ 2 } } }{ x } } dx=\int { \frac { \sqrt { 1+{ x }^{ 2 } } }{ { x }^{ 2 } } } .x\quad dx.\) 
Put   \(1+{ x }^{ 2 }={ u }^{ 2 }\) 
so that \(2\quad x\quad dx=2u\quad i.e\quad x\quad dx=\quad u\quad du.\)
\(\therefore\)   \({ I }_{ 1 }=\int { \frac { u }{ { u }^{ 2 }-1 } } u\ du= \int { \frac { { u }^{ 2 } }{ { u }^{ 2 }-1 } } du\)
\(=\int { \frac { \left( { u }^{ 2 }-1 \right) +1 }{ { u }^{ 2 }-1 } } du\)|
\(=\int { 1.du\quad +\int { \frac { 1 }{ { u }^{ 2 }-{ 1 }^{ 2 } } } } du\\ \)
| "From:  \(\int { \frac { dx }{ { x }^{ 2 }-{ a }^{ 2 } } " } \)
\(=u+\frac { 1 }{ 2\left( 1 \right) } log|\frac { u-1 }{ u+1 } |\)
\(=\sqrt { 1+{ x }^{ 2 } } \frac { 1 }{ 2 } log|\frac { \sqrt { 1+{ x }^{ 2 } } -1 }{ \sqrt { 1+{ x }^{ 2 } } +1 } |.\)   
And \({ I }_{ 2 }=\int { \frac { y }{ \sqrt { 1+{ y }^{ 2 } } } } dy.\)
Put  \({ y }^{ 2 }=v\)     
so that  \(2y\ dy =dv\ i.e.\ y\ dy = \frac { 1 }{ 2 } dv.\)
\(\therefore\)  \({ I }_{ 2 }=\int { \frac { \frac { 1 }{ 2 } dv }{ \sqrt { 1+{ v } } } } =\frac { 1 }{ 2 } { \left( 1+v \right) }^{ -\frac { 1 }{ 2 } }dv\)
\(=\frac { 1 }{ 2 } \frac { { \left( 1+v \right) }^{ -\frac { 1 }{ 2 } } }{ \frac { 1 }{ 2 } } =\sqrt { 1+v } =\quad \sqrt { 1+{ y }^{ 2 } } .\)
\(\sqrt { 1+{ x }^{ 2 } } +\frac { 1 }{ 2 } log|\frac { \sqrt { 1+{ x }^{ 2 } } -1 }{ \sqrt { 1+{ x }^{ 2 } } +1 } |+\sqrt { 1+{ y }^{ 2 } } =c,\)
which is the required solution

The equation is: \(\frac { dy }{ dx } +x\quad cot\quad y=2y+{ y }^{ 2 }cot\quad y\left( y\neq 0 \right) \) ...(1)
| Linear Equation
Comparing with \(\frac { dx }{ dy } +px=Q,\) 
we have:
\('P'=\quad cot\quad y\)
\('Q'=\quad 2y+{ y }^{ 2 }cot\quad y\)
\(\therefore \)   \(I.F={ e }^{ \int { P\quad dy } }={ e }^{ \int { cot\quad y\quad dy } }\)
\(={ e }^{ log|sin\quad y| }=sin\quad y.\)
Multiplying (1) by sin y, we get:
\(sin\quad y.\frac { dx }{ dy } +x\quad cos\quad y=2y\quad sin\quad y+{ y }^{ 2 }cot\quad y,\)
\(\Rightarrow \) \(\frac { d }{ dy } \left( x\quad sin\quad y \right) =2y\quad sin\quad y+{ y }^{ 2 }cot\quad y\)
Integrating, \(x\quad sin\quad y=\int { \left( { y }^{ 2 }\quad cos\quad y+2y\quad sin\quad y \right) dy+c } \)
\(\Rightarrow \)   \(x\quad sin\quad y=\int { { y }^{ 2 } } cos\quad y\quad dy+\int { 2y\quad sin\quad y\quad dy+c } \)
\(\Rightarrow \)\(x\quad sin\quad y=\quad { y }^{ 2 }\quad sin\quad y-\int { 2y\quad sin\quad y\quad dy+\int { 2y\quad sin\quad y\quad dy+c } } \)
\(\Rightarrow \)    \(x\quad sin\quad y=\quad { y }^{ 2 }\quad sin\quad y+c\)...(2)
When  \(x=0,y=\frac { \pi }{ 2 } ,\quad \therefore 0=\frac { { \pi }^{ 2 } }{ 4 } sin\frac { \pi }{ 2 } +c\)
\(\Rightarrow \) \(0=\frac { { \pi }^{ 2 } }{ 4 } \left( 1 \right) +c=c=-\frac { { \pi }^{ 2 } }{ 4 } \)
Putting in(2), \(x= sin\ y= { y }^{ 2 }sin y-\frac { { \pi }^{ 2 } }{ 4 } \)
\(\Rightarrow \)  \(x= { y }^{ 2 }-\frac { { \pi }^{ 2 } }{ 4 } cosec\quad y,\)
which is the required solution

The given equation is \(\frac { dy }{ dx } =\frac { xy }{ { x }^{ 2 }+{ y }^{ 2 } } \) ...(1)
| Homogeneous 
Put \(y=vx\) so that \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } .\)
\(\therefore \)   (1) becomes :\(v+x\frac { dv }{ dx } =\frac { x\left( vx \right) }{ { x }^{ 2 }+{ v }^{ 2 }{ x }^{ 2 } } \)
\(\Rightarrow \)  \(v+x\frac { dv }{ dx } =\frac { v }{ 1+{ v }^{ 2 } } \)
\(\Rightarrow \)  \(x\frac { dv }{ dx } =\frac { v }{ 1+{ v }^{ 2 } } -v=\frac { v-v{ -v }^{ 3 } }{ 1+{ v }^{ 2 } } \)
\(\Rightarrow \)  \( x\frac { dv }{ dx } =\frac { -{ v }^{ 3 } }{ 1+{ v }^{ 2 } } \)
\(\Rightarrow \) \(\left( \frac { 1+{ v }^{ 2 } }{ { v }^{ 3 } } \right) dv=\frac { -dx }{ x } \)
|Variables Separable
Integrating, \(\int { \left( \frac { 1 }{ { v }^{ 3 } } +\frac { 1 }{ v } \right) } dv=-\int { \frac { dx }{ x } } +c\)
\(\Rightarrow \) \(\frac { { v }^{ -2 } }{ -2 } +log|v|=-log|x|+c\)
\(\Rightarrow \) \(-\frac { 1 }{ 2{ v }^{ 2 } } +log|v|=-logx|+c\)
\(\Rightarrow \)  \(-\frac { { x }^{ 2 } }{ 2y^{ 2 } } +log|v|=-log|x|+c\)
\(\Rightarrow \) \(-\frac { { x }^{ 2 } }{ 2y^{ 2 } } +log|y|-log|x|=-log|x|+c\)
\(\Rightarrow \) \(-\frac { { x }^{ 2 } }{ 2y^{ 2 } } +log|y|=c\)  ...(2)
When \(x-0,y=1,\)
\(\therefore \ -\frac { 1 }{ 2 } (0)+log|1|=c\)
\(\Rightarrow -0+0=c \Rightarrow c=0\)
Putting in(2),
\(\Rightarrow \)  \(log|y|=\frac { { x }^{ 2 } }{ 2{ y }^{ 2 } } \)
which is the required particular solution

The given equation is  \({ e }^{ x }\sqrt { 1-{ y }^{ 2 } } dx+\frac { y }{ x } =0,\)
\(\Rightarrow \) \(x{ e }^{ x }\quad dx+\frac { y }{ \sqrt { 1-{ y }^{ 2 } } } dy=0\)
|Variables Separable 
Integrating, \(\int { x\quad { e }^{ x }\quad dx+ } \int { \frac { y }{ \sqrt { 1-{ y }^{ 2 } } } dy= } c\)
\(\Rightarrow \) \(x{ e }^{ x }-\int { \left( 1 \right) } { e }^{ x }\quad dx-\frac { 1 }{ 2 } \int { { \left( 1-{ y }^{ 2 } \right) }^{ -\frac { 1 }{ 2 } } } \left( -2\quad y \right) dy=c\)
\(\Rightarrow \) \(x{ e }^{ x }- { e }^{ x }-\frac { 1 }{ 2 } \frac { { \left( 1-{ y }^{ 2 } \right) }^{ -\frac { 1 }{ 2 } } }{ \frac { 1 }{ 2 } } =c\)
\(\Rightarrow \) \(\left( x-1 \right) { e }^{ x }-\sqrt { 1-{ y }^{ 2 } } =c\) ...(1)
When \(x=0,y=1\)
\(\therefore \)    \(\left( -1 \right) \left( 1 \right) -0=c\Rightarrow c=-1\)
Putting in (1), \(\left( x-1 \right) { e }^{ x }-\sqrt { 1-{ y }^{ 2 } } =-1\)
\(\Rightarrow \)  \(\sqrt { 1-{ y }^{ 2 } } =\left( x-1 \right) { e }^{ x }+1,\)
which is the required solution

We have:  \(cosec\ x\ log\ y\ dy+{ x }^{ 2 }{ y }^{ 2 }dx=0\)
\(\Rightarrow\)  \(\frac { log\quad y }{ { y }^{ 2 } } dy+{ x }^{ 2 }\quad sin\quad x\quad dx=0\)
| Variables Separable 
Intregrating,  \(\int { log\quad y.\frac { 1 }{ { y }^{ 2 } } } dy+\int { { x }^{ 2 }\quad sin\quad x\quad dx=c } \)
\(\Rightarrow\) \(log\quad y.\frac { { y }^{ -1 } }{ -1 } -\int { \frac { 1 }{ y } } .\frac { { y }^{ -1 } }{ -1 } dy+[{ x }^{ 2 }\left( -cos\quad x \right) \)
\(-\int { 2x } \left( -cos\quad x \right) dx=c\)   
[Integrating by Parts]
\(\Rightarrow\)  \(\frac { -log\quad y }{ { y } } +\int { { y }^{ -2 } } dy-{ x }^{ 2 }cos x+2\int { xcos x\ dx=c } \)
\(\Rightarrow\)  \(\frac { -log y }{ { y } } +\frac { { y }^{ -1 } }{ -1 } -{ x }^{ 2 }cos x+2\left[ x\ sin\ x-\int { 1.sin x\ dx } \right] =c\)
[Again integrating by Parts]
\(\Rightarrow \frac { -log\ y }{ { y } } -\frac { 1 }{ y } -{ x }^{ 2 }cos\ x+2\left[ x\ sin\ x+cos x \right] =c,\)
which is the required solution

\(\mathrm{e}^{\mathrm{x} / \mathrm{y}}+\frac{x}{y}=\frac{c}{y} \Rightarrow \mathrm{ye}^{\mathrm{x} / \mathrm{y}}+\mathrm{x}=\mathrm{c}\)is the required solution.

\(\int { \frac { 1-y }{ { y }^{ 2 } } } dy=\int { \frac { -{ y }^{ 2 }(1+{ x }^{ 2 }) }{ { x }^{ 2 }-{ x }^{ 2 }y } } dx\)  
\(\Rightarrow \)  \(-\frac { 1 }{ y } -log\left| y \right| =\frac { 1 }{ x } -x+c\) 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

Given, differential equation is \(\left(1+y^2\right)+(2 x y-\cot y) \frac{d y}{d x}=0 .\)
The above equation can be rewritten as \(\begin{aligned} (\cot y-2 x y) \frac{d y}{d x}=1+y^2 \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad & \frac{\cot y-2 x y}{\left(1+y^2\right)}=\frac{d x}{d y} \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{d x}{d y}=\frac{\cot y}{1+y^2}-\frac{2 x y}{1+y^2} \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{d x}{d y}+\frac{2 y}{1+y^2} \cdot x=\frac{\cot y}{1+y^2} \end{aligned}\)
which is a linear differential equation of the form
\(\frac{d x}{d y}+P x=Q, \text { here } P=\frac{2 y}{1+y^2} \text { and } Q=\frac{\cot y}{1+y^2}\)
Now, integrating factor = \(e^{\int P d y}=e^{\int \frac{2 y}{1+y^2} d y}\)
On putting 1 + y² = t \(\Rightarrow\) 2ydy = dt
\(\therefore \quad \mathrm{IF}=e^{\int \frac{d t}{t}}=e^{\log |t|}=t=1+y^2\)

Given differential equation is \(\log \left(\frac{d y}{d x}\right)=3 x+4 y\) 
\(\Rightarrow \frac{d y}{d x}=e^{3 x+4 y} \Rightarrow \frac{d y}{d x}=e^{3 x} e^{4 y}\) 
On separating the variables, we get
\(\frac{1}{e^{4 y}} d y=e^{3 x} d x\) 
On integrating both sides, we get
\(\int e^{-4 y} d y=\int e^{3 x} d x \Rightarrow \frac{e^{-4 y}}{-4}=\frac{e^{3 x}}{3}+C\) ...(i)
which is the required general solution of given differential equation.

Writing the given equation as
\(\frac { dy }{ dx } +\frac { 1 }{ \sqrt { x } } y=\frac { { e }^{ -2\sqrt { x } } }{ \sqrt { x } } \)
Hence, \(P=\frac { 1 }{ \sqrt { x } } \)
\(\therefore I.F.={ e }^{ \int { Pdx } }\)
\(={ e }^{ \int { \frac { 1 }{ \sqrt { x } } dx } }\)
\(I.F.={ e }^{ 2\sqrt { x } }\)

As equation cannot be represented as a polynomial of derivatives.

As differential equation is
1+ \(3\frac { dy }{ dx } +3{ \left( \frac { dy }{ dx } \right) }^{ 2 }+{ \left( \frac { dy }{ dx } \right) }^{ 3 }={ \left( \frac { dy }{ dx } \right) }^{ 2 }\)
Exponent of highest order derivative is 3.

Given, \(\frac{d y}{d x}-\frac{2 x}{1+x^2} y=x^2+2\)       ...(i)
This is a linear differential equation with
\(\begin{aligned} P & =\frac{-2 x}{1+x^2} \text { and } Q=x^2+2 \end{aligned}\)
\(\begin{aligned} \therefore \quad \mathrm{IF} =e^{\int P d x}=e^{\int \frac{-2 x}{x^2+1} d x} \end{aligned}\)
\(\begin{aligned} =e^{-\int \frac{2 x}{x^2+1} d x}=e^{-\log \left(x^2+1\right)}=\frac{1}{x^2+1} \end{aligned}\)
\(\therefore \quad y \cdot \frac{1}{\left(x^2+1\right)}=\int\left(x^2+2\right) \cdot \frac{1}{\left(x^2+1\right)} d x+C\)
                                [using y . (IF) = \(\left.\int Q \cdot(\mathrm{IF}) d x+C\right]\)
\(\begin{aligned} \Rightarrow \quad \frac{y}{x^2+1}=\int \frac{\left(x^2+1\right)+1}{x^2+1} d x+C \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{y}{x^2+1}=\int 1 d x+\int \frac{1}{x^2+1} d x+C \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{y}{x^2+1}=x+\tan ^{-1} x+C \end{aligned}\)
\(\begin{aligned} \Rightarrow y=x\left(x^2+1\right)+\left(\tan ^{-1} x\right)\left(x^2+1\right)+C^{\prime} \end{aligned}\)
\(\begin{aligned} {\left[\because C^{\prime}=C \cdot\left(x^2+1\right)\right] } \end{aligned}\)

Given differential equation is
\(\left(\tan ^{-1} y-x\right) d y=\left(1+y^{2}\right) d x\) 
\(\Rightarrow \frac{\tan ^{-1} y-x}{1+y^{2}}=\frac{d x}{d y}\) 
\(\Rightarrow\frac{d x}{d y}=\frac{-x}{1+y^{2}}+\frac{\tan ^{-1} y}{1+y^{2}}\) 
\(\Rightarrow \frac{d x}{d y}+\frac{1}{1+y^{2}} \cdot x=\frac{\tan ^{-1} y}{1+y^{2}}\) 
which is a linear differential equation of first order
On comparing with \(\frac{d x}{d y}+P x=Q\), we get
\(P=\frac{1}{1+y^{2}} \text { and } Q=\frac{\tan ^{-1} y}{1+y^{2}}\) 
Now, \(\mathrm{IF}=e^{\int P d y}=e^{\int \frac{d y}{1+y^{2}}}=e^{\tan ^{-1} y}\) 
and the required solution is given by
\(x \cdot \mathrm{IF}=\int Q \cdot \mathrm{IF} d y+C\) 
\(\Rightarrow x \cdot e^{\tan ^{-1} y}=\int \frac{\tan ^{-1} y}{1+y^{2}} \times e^{\tan ^{-1} y} d y+C\) 
Put \(t=\tan ^{-1} y, \text { then } d t=\frac{1}{1+y^{2}} d y\) 
\(\therefore x \cdot e^{\tan ^{-1} y}=\int t \cdot e^{t} d t+C\) 
\(\Rightarrow x \cdot e^{\tan ^{-1} y}=t \cdot e^{t}-\int 1 \cdot e^{t} d t+C\) [integration by parts]
\(\Rightarrow x \cdot e^{\tan ^{-1} y}=t \cdot e^{t}-e^{t}+C\)
\(\Rightarrow x \cdot e^{\tan ^{-1} y}=\left(\tan ^{-1} y-1\right) e^{\tan ^{-1} y}+C\left[\text { put } t=\tan ^{-1} y\right] \ldots(\mathrm{i})\) 
\(\therefore\) It is given that x = 1, when y = 0,
Therefore, we have
\(1 \cdot e^{0}=(0-1)+C\) 
\(\Rightarrow 1=-1+C\) 
\(\Rightarrow C=2\) 
Hence, \(x \cdot e^{\tan ^{-1} y}=e^{\tan ^{-} y}\left(\tan ^{-1} y-1\right)+2\) 
which is the required particular solution of the differential equation.

Write the given differential equation as
\(\frac{d y}{d x}-\frac{2 x y}{\left(x^{2}+1\right)}=\frac{\left(x^{2}+1\right)^{2}}{\left(x^{2}+1\right)} \cos x\) 
\(\Rightarrow \frac{d y}{d x}-\frac{2 x y}{x^{2}+1}=\left(x^{2}+1\right) \cos x\) 
and \(\mathrm{IF}=e^{\int \frac{-2 x}{1+x^{2}} d x}=e^{-\log \left|x^{2}+1\right|}=\left(x^{2}+1\right)^{-1}\) 
\(y=\left(x^{2}+1\right) \sin x\)