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

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

\(\frac { dy }{ dx } ={ e }^{ 3x-4y }\)
\(\frac { dy }{ dx } ={ e }^{ 3x }.{ e }^{ -4y }\)
\(\Rightarrow \frac { dy }{ { e }^{ -4y } } ={ e }^{ 3x }dx\)
integrating both the sides
\(\int { { e }^{ 4y }dy } =\int { { e }^{ 3x }dx } \)
\(\Rightarrow \frac { 1 }{ 4 } { e }^{ 4y }=\frac { 1 }{ 3 } { e }^{ 3y }+C\)

\(\frac { dy }{ dx } =xy+2y\)
\(\Rightarrow \frac { dy }{ dx } =y\left( x+2 \right) \)
Integrating both the sides,
\(\int { \frac { dy }{ y } } =\int { \left( x+2 \right) dx } \)
\(\Rightarrow \log { y } =\frac { { x^{ 2 } } }{ 2 } +2x+c\)

\(\frac { dy }{ dx } ={ e }^{ x-y }+x{ e }^{ -y }\)
\(={ e }^{ -y }\left[ { e }^{ x }+x \right] \)
\(\Rightarrow \frac { dy }{ dx } =\left( { e }^{ x }+x \right) { e }^{ -y }\)
\(\Rightarrow \int { { e }^{ y }dy } =\int { \left( { e }^{ x }+x \right) { dx } } \)
\(\Rightarrow { e }^{ y }={ e }^{ x }+\frac { { x }^{ 2 } }{ 2 } +C\)

Given, differential equation is \(\frac{d y}{d x}=x^3 e^{-2 y}\)
On separating the variables, we get
e^2ydy = x³dx
On integrating both sides, we get
\(\begin{aligned} \int e^{2 y} d y & =\int x^3 d x \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad & \frac{e^{2 y}}{2}=\frac{x^4}{4}+C_1=2 e^{2 y}=x^4+4 C_1 \end{aligned}\)
\(\therefore\) 2e^2y = x⁴ + C, where C = 4C₁

y = mx
\(\Rightarrow \frac { dy }{ dx } =m\)
\(\therefore \quad y=\frac { dy }{ dx } .x\) or \(\frac { dy }{ dx } -\frac { y }{ x } =0\)

Generate equation of family of lines passing through origin
y = mx 
\(m=\frac { y }{ x } \)
\(\frac { dy }{ dx } =m\)
\(\frac { dy }{ dx } =\frac { y }{ x } \)
\(x\frac { dy }{ dx } -y=0\)

Given \({ x }^{ 3 }\left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) ^{ 2 }+x\left( \frac { dy }{ dx } \right) ^{ 4 }=0\)
The highest order derivative is \(\left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) \) and its power is 2.
So, the degree of differential equation is 2.

Degree = 1

The given equation is:\(\left( x-y \right) \left( dx+dy \right) =dx-dy\)
\(\Rightarrow \left( x-y-1 \right) dx+\left( x-y+1 \right) dy=0\)
\(\Rightarrow\) \(\frac { dy }{ dx } =-\frac { x-y-1 }{ x-y+1 } \)
Put \(x-y=z\)
so that  \(1-\frac { dy }{ dx } =\frac { dy }{ dx } \)
\(\Rightarrow\)  \(\frac { dy }{ dx } =1-\frac { dz }{ dx } .\)
\(\therefore\) from (1),  \(1-\frac { dz }{ dx } =-\frac { 2z }{ z+1 } \)
\(\Rightarrow \frac { dz }{ dx } =1+\frac { z-1 }{ z+1 } \Rightarrow \frac { dz }{ dx } =\frac { 2z }{ z+1 } \)
\(\Rightarrow\)  \(\frac { z+1 }{ z } dz=2dx\)
|Variables Separable 
Integrating, \(\int { \left( 1+\frac { 1 }{ 2 } \right) } dz=2\int { dx+C } \)
\(\Rightarrow\) \(z+log|z|=2x+C\)
\(\Rightarrow\) \(x-y+log|x-y|=2x+C\) ...(2)
When  \(x=0,y=-1.\) 
\(\therefore\)  \(0+1+log|0+1|2(0)+C\)                   
\(\Rightarrow\) \(1+0=0+C\Rightarrow C=1.\)
Putting in  (2),\(x-y+log|x-y|=2x+1\)
\(\Rightarrow\) \(log|x-y|=x+y+1,\)
Which is required solution.

We have: \(y'={ e }^{ x }sin x\)
\(\Rightarrow\) \(\frac { dy }{ dx } ={ e }^{ x }sin\quad x\)
\(\Rightarrow\) \(dy={ e }^{ x }sin x dx\)
Variables Separable
Integrating,\(\int { dy=\int { { e }^{ x } } sin x dx+C } \)
\(\Rightarrow \)  \(y=\int { { e }^{ x } } sin xdx + C\) ...(1)
Now \(I=\int { { e }^{ x } sin x dx } \)
\(= { e }^{ x }\left( -cos x \right) -\int { { e }^{ x }\left( - cos x \right) dx } \)
[Integrating by Parts]
\(=-{ e }^{ x }\left( -cos x \right) +\int { { e }^{ x }cos xdx } \)
\(=- { e }^{ x }cos x\)\(+\left[ { e }^{ x } cos x-\int { { e }^{ x } sinx dx } \right] \)
[Integrating by Parts]
\(=-{ e }^{ x }cos x+{ e }^{ x }sin x-I\)
\(\Rightarrow\) \(2I={ e }^{ x }\left( sin x- cos x \right) \)
\(\Rightarrow\) \(I=\frac { { e }^{ x } }{ 2 } \left( sin x- cos x \right) .\)  
Putting in (1),
\(y=\frac { { e }^{ x } }{ 2 } \left( sinxcos x \right) +C\)
When \(x=0, y=0,\) then
\(0=\frac { 1 }{ 2 } \left( 0-1 \right) +C\Rightarrow C=\frac { 1 }{ 2 } .\)
Putting in (2),
\(y=\frac { { e }^{ x } }{ 2 } \left( sin x-cos x \right) +\frac { 1 }{ 2 } \)
\(\Rightarrow\)  \(2y-1={ e }^{ x }\left( sinx -cos x \right) ,\)
Which is the required equation of the curve.

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.

The given differential equation can be expressed as
\(d y^*=\left(\frac{2 x^2+1}{x}\right) d x^*\)
or \(d y=\left(2 x+\frac{1}{x}\right) d x\).... (1)
\(\Rightarrow \) \(y=\frac { 2{ x }^{ 2 } }{ 2 } +log|x|+c\)
Integrating both sides of equation (1), we get
\(\int d y=\int\left(2 x+\frac{1}{x}\right) d x\)                   
or  \(y=x^2+\log |x|+C\)... (2)
Equation (2) represents the family of solution curves of the given differential equation but we are interested in finding the equation of a particular member of the family which passes through the point (1, 1). Therefore substituting x = 1, y = 1 in equation (2), we get C = 0.
Now substituting the value of C in equation (2) we get the equation of the required curve as y = x² + log |x|.

The given function is
y = a cos x + b sin x ... (1)
Differentiating both sides of equation (1) with respect to x, successively, we get
\(\frac{d y}{d x}=-a \sin x+b \cos x \)
\(\frac{d^{2} y}{d x^{2}}=-a \cos x-b \sin x \)
Substituting the values of \(\frac{d^{2} y}{d x^{2}}\)and y in the given differential equation, we get
L.H.S. = (– a cos x – b sin x) + (a cos x + b sin x) = 0 = R.H.S..
Therefore, the given function is a solution of the given differential equation.

\(\int { \frac { dy }{ 1+cos2y } } =-\int { \frac { dx }{ 1-cos2x } } \Rightarrow \int { { sec }^{ 2 } } ydy= -\int { { cosec }^{ 2 } } x\quad dx\)  \(\Rightarrow \)tany=cotx+c

Given differential equation is homogeneous.
\(\therefore \) Putting y = vx to get \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
\(\frac { dy }{ dx } =\frac { y\sin { \left( \frac { y }{ x } \right) } -x{ e }^{ \frac { y }{ x } } }{ x\sin { \left( \frac { y }{ x } \right) } } \)
\(\Rightarrow v+x\frac { dv }{ dx } =\frac { v\sin { v } -{ e }^{ v } }{ \sin { v } } \)
\(\Rightarrow v+x\frac { dv }{ dx } =v-\frac { { e }^{ v } }{ \sin { v } } \)
\(\Rightarrow x\frac { dv }{ dx } =-\frac { { e }^{ v } }{ \sin { v } } \)
\(\therefore \int { \sin { v{ e }^{ -v }dv } } =-\int { \frac { dx }{ x } } \)
\(\Rightarrow { I }_{ 1 }=-\log { x } +{ C }_{ 1 }\) ...(i)
\(\Rightarrow { I }_{ 2 }=-\sin { v. } { e }^{ -v }+\int { \cos { v } } { e }^{ -v }dv\)
\(\Rightarrow { I }_{ 2 }=-\sin { v. } { e }^{ -v }-\cos { v } { e }^{ -v }-\int { \sin { v } } { e }^{ -v }dv\)
\(\Rightarrow { I }_{ 2 }=-\frac { 1 }{ 2 } \left( \sin { v } +\cos { v } \right) { e }^{ -v }\)
Putting (i), \(\left( \sin { v } +\cos { v } \right) { e }^{ -v }=\log { { x }^{ 2 } } +{ C }_{ 2 }\)
\(\Rightarrow \left[ \sin { \left( \frac { y }{ x } \right) +\cos { \left( \frac { y }{ x } \right) } } \right] { e }^{ \frac { -y }{ x } }=\log { { x }^{ 2 } } +C\)
For x = 1, y = 0 \(\Rightarrow \)  C = 1
Hence, solution is
\(\left[ \sin { \left( \frac { y }{ x } \right) +\cos { \left( \frac { y }{ x } \right) } } \right] { e }^{ \frac { -y }{ x } }=\log { { x }^{ 2 } } +1\)

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 \(c=\frac { 1 }{ \sqrt { 3 } } \)
\(\Rightarrow \frac { 1 }{ \sqrt { 3 } x } =\sqrt { \frac { y }{ y+2x } } \)