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 \(\left( y+3{ x }^{ 2 } \right) \frac { dx }{ dy } =x.\)
\(\Rightarrow \)  \(x\frac { dy }{ dx } y+3{ x }^{ 2 }\)
\(\Rightarrow \)  \(\frac { dy }{ dx } -\frac { y }{ x } =3x\) ...(1) 
| Linear Equation 
Comparing with \(\frac { dy }{ dx } +Py=Q\) 
we have:
\('P'=-\frac { 1 }{ x } \) and \('Q'=3x.\)
\(\therefore\)  \(I.F={ e }^{ \int { P\quad dx } }={ e }^{ \int { -\frac { 1 }{ x } dx } }={ e }^{ -logx| }\)
\(={ e }^{ log\frac { 1 }{ x } }=\frac { 1 }{ x } .\)
Multiplying (1) by \(\frac { 1 }{ x } ,\) we get:
\(\frac { 1 }{ x } \frac { dy }{ dx } -\frac { y }{ { x }^{ 2 } } =3\)
\(\Rightarrow \)  \(\frac { d }{ dx } \left( y.\frac { 1 }{ x } \right) =3.\)
Integrating, \(\frac { y }{ x } =\int { 3.dx+c } \)
\(\Rightarrow \) \(\frac { y }{ x } =3x+c\)
\(\Rightarrow \)  \(y=3{ x }^{ 2 }+cx,\)
which is the required solution.

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

The given equation is  \(\frac { dy }{ dx } =1+{ x }^{ 2 }+{ y }^{ 2 }+{ x }^{ 2 }{ y }^{ 2 }\)
\(\Rightarrow \)  \(\frac { dy }{ dx } =\left( 1+{ x }^{ 2 } \right) +{ y }^{ 2 }\left( 1+{ y }^{ 2 } \right) \)
\(\Rightarrow \)  \(\frac { dy }{ dx } =\left( 1+{ x }^{ 2 } \right) +{ y }^{ 2 }\left( 1+{ y }^{ 2 } \right) \)
\(\Rightarrow \)  \(\frac { dy }{ 1+{ y }^{ 2 } } =\left( 1+{ x }^{ 2 } \right) dx\)
|Variables Separable 
Integrating,\(\int { \frac { dy }{ 1+{ y }^{ 2 } } = } \int { \left( 1+{ x }^{ 2 } \right) dx+c } \)
\(\Rightarrow\) \({ tan }^{ -1 }y=x+\frac { { x }^{ 3 } }{ 3 } +c\)         
Now \(y=1\) when \(x=0\)
\(\therefore \)  \({ tan }^{ -1 }(1)=0+0+c\)
\(\Rightarrow\) \(c=\frac { \mu }{ 4 } .\)
Putting in (1),  \({ tan }^{ -1 }y=x+\frac { 1 }{ 3 } { x }^{ 3 }+\frac { \mu }{ 4 } \)
Which is the required particular solution

The given equation is \(\frac { dy }{ dx } =1+x+y+xy\)
\(\Rightarrow \)  \(\frac { dy }{ dx } =\left( 1+x \right) \left( 1+y \right) \)
\(\Rightarrow \)  \(\frac { dy }{ 1+y } =\left( 1+x \right) dx.\)
|Variables Separable 
Integrating, \(\int { \frac { dy }{ 1+y } } =+\left( 1+x \right) dx+c\)
\(\Rightarrow \) \(log|1+y|=x+\frac { { x }^{ 2 } }{ 2 } +c\)
Which is the required solution.
When \(x=1,y=0\) then:
\(log|1+0|=1+\frac { 1 }{ 2 } { \left( 1 \right) }^{ 2 }+c\)
\(\Rightarrow \) \(log \ 1=1+\frac { 1 }{ 2 } +c \Rightarrow 0=\frac { 3 }{ 2 } \)
\(\Rightarrow \) \(c=-\frac { 3 }{ 2 } \)
Putting (1), \(log|1+y|=x+\frac { 1 }{ 2 } { x }^{ 2 }-\frac { 3 }{ 2 } ,\)
which is the required particular solution

y'' + 2y′ + sin y = 0
The highest order derivative present in the differential equation is y''. Therefore, its order is two.
This is a polynomial equation in y'' and y' and the highest power raised to y' is one. Hence, its degree is one.

We have: sin x cos y dx+ cos x sin y dy=0.
\(\Rightarrow\) \(\frac { sin\quad x }{ cos\quad x } dx+\frac { sin\quad y }{ cos\quad y } dy=0\) 
| Variables Separable
Integrating,  - \(\int { \frac { -sin\quad x }{ cos\quad x } dx-\int { \frac { -sin\quad x }{ cos\quad y } dy } } \)
= Constant
\(\Rightarrow\) \(-log|cos x|-log|cos\ y| =-\quad log|C|\)
\(\Rightarrow\)  \(-log|cos x\ cos\ y|=-\quad log|C|\)
\(\Rightarrow\) \(cos x\cos \ y=C\) ...(1)
Since the curve passes thro'\(\left( 0,\frac { \pi }{ 4 } \right) \),
\(\therefore\)   \(cos\ 0\ cos\frac { \pi }{ 4 } =C\)
\(\Rightarrow\left( 1 \right) \left( \frac { 1 }{ \sqrt { 2 } } \right) =C\Rightarrow C=\frac { 1 }{ \sqrt { 2 } } .\)
Putting in (1), cos x cos y = \(\frac{1}{\sqrt2}\)
\(\Rightarrow\) \(cos\quad y=\frac { sec\quad x }{ \sqrt { 2 } } ,\)
Which is the reqd. equation of the curve.

Since \(1+y^2 \neq 0\) therefore separating the variables, the given differential equation can be written as
\(\frac { dy }{ dx } =\frac { 1+{ y }^{ 2 } }{ 1+{ x }^{ 2 } } \)....(1)
Integrating both sides of equation (1), we get
\(\int \frac{d y}{1+y^2}=\int \frac{d x}{1+x^2}\)
or \(\tan ^{-1} y=\tan ^{-1} x+C\)
which is the general solution of equation (1).

Given function is y = e^{– 3x}. Differentiating both sides of equation with respect to x , we get
\(\frac{d y}{d x}=-3 e^{-3 x}\)
Now, differentiating (1) with respect to x, we have
\(\frac{d^{2} y}{d x^{2}}=9 e^{-3 x}\)
Substituting the values of \(\frac{d^{2} y}{d x^{2}} \frac{d y}{d x^{2}}\) and y in the given differential equation, we get
L.H.S. = 9 e^{– 3x} + (–3e^{– 3x}) – 6.e^{– 3x} = 9 e^{– 3x} – 9 e^{– 3x} = 0 = R.H.S..
Therefore, the given function is a solution of the given differential equation.

(i) The highest order derivative present in the differential equation is \(\frac { dy }{ dx } \) so its order is one. It is a polynomial equation in y′ and the highest power raised to \(\frac { dy }{ dx } \) is one, so its degree is one.
(ii) The highest order derivative present in the given differential equation is \(\frac{d^{2} y}{d x^{2}}\) so its order is two. It is a polynomial equation in \(\frac{d^{2} y}{d x^{2}} \text { and } \frac{d y}{d x}\)and the highest power raised to \(\frac{d^{2} y}{d x^{2}}\) is one, so its degree is one.
(iii) The highest order derivative present in the differential equation is y′′′ , so its order is three. The given differential equation is not a polynomial equation in its derivatives and so its degree is not defined.

We have: \(\frac { dx }{ dy } =\frac { 2x{ e }^{ \frac { x }{ y } }-y }{ 2y{ e }^{ \frac { x }{ y } } } \) .(1)
Here \(f\left( x,y \right) =\frac { 2x{ e }^{ \frac { x }{ y } }-y }{ 2y{ e }^{ \frac { x }{ y } } } .\)
\(=\frac { 2x{ e }^{ \frac { x }{ y } }-y }{ 2y{ e }^{ \frac { x }{ y } } } ={ \lambda }^{ 0 }f\left( x,y \right) .\)
Thus \(f(x,y)\) is homogeneous function of degree 0.
To solve:
Put  \(\frac { x }{ y } =v\quad i.e\quad x=vy\)
So that   \(\frac { dx }{ dy } =v+y\frac { dy }{ dy } .\)     
\(\therefore\) (1) becomes : \(v+y\frac { dy }{ dx } =\frac { 2vy\quad { e }^{ v }-y }{ 2y\quad { e }^{ v } } \)
\(\Rightarrow \)  \(y\frac { dv }{ dy } =\frac { 2\quad v\quad { e }^{ v }-1 }{ 2\quad { e }^{ v } } -v\)
\(\Rightarrow\)  \(y\frac { dv }{ dy } =\frac { 2\quad v\quad { e }^{ v }-1 }{ 2\quad { e }^{ v } } -v\)
\(\Rightarrow \)  \(y\frac { dv }{ dy } =\frac { -1 }{ 2{ e }^{ v } } \)
\(\Rightarrow \)  \(2{ e }^{ v }dv=-\frac { dy }{ y } \)
Integrating,\(2\int { { e }^{ v }\ dv=-log|y|+c } \)
\(\Rightarrow \) \(2{ e }^{ v }=-log|y|+c\)
\(\Rightarrow \)\(2{ e }^{ \frac { x }{ y } }=-log|y|+c\) ...(2)
When \(x=0,y=1\)
\(\therefore\)  \(2\left( 1 \right) =-log|1|+c\)
\(\Rightarrow \) \(2=-0+c\Rightarrow =2.\)
Putting in (2),
\(2{ e }^{ \frac { x }{ y } }+log|y|=2,\)
Which is the required solution.

Let the equation of parabola be:\({ x }^{ 2 }=4ay\)  ...(1)
Diff. w.r.t. x,\(2x = 4ay'\)
\(\Rightarrow\)  \(x= 2ay'\Rightarrow 2a\frac { x }{ { y }^{ ' } } .\)
Putting in (1),\({ x }^{ 2 }=\frac { 2x }{ { y }^{ ' } } y\)
\(\Rightarrow\) \(xy'= 2y\)​​​
\(\Rightarrow\) \(x{ y }^{ ' }-2y=0,\)
Which is the required solution.

Given differential equation can be written as
\({dy\over dx}={x+2y\over x-y}\)
\(\Rightarrow c+x{dv\over dx}={1+2v\over 1-v},\) where y=vx
\(\Rightarrow {v-1\over v^2+v+1}dv={1\over x}dx\)
Integrating both sides, we get
\({1\over 2}\int {2v+1\over v^2+v+1}dv-{3\over 2}\int{1\over v^2+v+1}dv\)
\(=-log|x|\)
\(\Rightarrow{1\over 2}log |v^2+v+1|-\sqrt3 tan^{-1}\left(2v+1\over \sqrt3\right)\)
= - log |x| + c
\(\Rightarrow{1\over 2}log\left|{y^2\over xz^2}+{y\over x}+1\right|-\sqrt3tan^{-1}\left(2y+x\over \sqrt3x\right)\)
= - log |x| + c
\(\Rightarrow\ log|x^2+xy+y^2|=2\sqrt3tan^{-1}\left(x+2y\over \sqrt3x\right)+C_1\)
where C₁=2C

\(\frac { dy }{ dx } +y={ e }^{ -x }\)
Let \(\frac { dy }{ dx } +Py=Q\)
Where P = 1, Q = e^-x
\(y\left( I.F. \right) =\int { I.F.\times Qdx } \)
\(\Rightarrow I.F.={ e }^{ \int { Pdx } }={ e }^{ \int { Pdx } }={ e }^{ x }\)
\(\Rightarrow y{ e }^{ x }=\int { { e }^{ x }.{ e }^{ -x } } dx=\int { dx } \)
\(\Rightarrow y{ e }^{ x }=x+C\)

\(\log { \left( \frac { dy }{ dx } \right) } =x+1\)
\(\Rightarrow \frac { dy }{ dx } ={ e }^{ x+1 }\)
\(\Rightarrow dy={ e }^{ x+1 }dx\)
Integrating both the sides,
\(\int { dy } =\int { { e }^{ x+1 }dx } \)
\(\Rightarrow y={ e }^{ x+1 }+C\)