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

\(x\frac { dy }{ dx } +y={ e }^{ 2 }\)
Divide by x both the sides,
\(\frac { dy }{ dx } +\frac { y }{ x } =x\)
\(\frac { dy }{ dx } +Py =Q\),
Where \(P=\frac{1}{x},Q=x\)
\(I.F.={ e }^{ \int { Pdx } }={ e }^{ \int { \frac { dx }{ x } } }={ e }^{ logx }=x\)
\(y\left( I.F. \right) =\int { Q.I.F.dx } \)
\(xy=\int { x.xdx } =\int { { x }^{ 2 }dx } \)
\(\Rightarrow yx=\frac { { x }^{ 3 } }{ 3 } +C\)
\(\Rightarrow y=\frac { 1 }{ 3 } { x }^{ 2 }+\frac { C }{ x } \)

\(\frac { dy }{ dx } +y={ e }^{ -x }\)
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,
\(\int { dy } =\int { { e }^{ x+1 }dx } \)
\(\Rightarrow y={ e }^{ x+1 }+C\)

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

\(\frac { dy }{ dx } =\frac { 1+\cos { 2y } }{ 1+\cos { 2x } } \)
\(\Rightarrow \frac { dy }{ 1+\cos { 2y } } =\frac { dx }{ 1+\cos { 2x } } \)
as \(\cos { 2x } =2\cos ^{ 2 }{ x } -1\)
\(\cos { 2y } =2\cos ^{ 2 }{ y } -1\)
Integrating both the sides,
\(\int { \frac { dy }{ 2\cos ^{ 2 }{ y } } } =\int { \frac { dx }{ 2\cos ^{ 2 }{ x } } } \)
\(\Rightarrow \frac { 1 }{ 2 } \int { \sec ^{ 2 }{ ydy } } =\frac { 1 }{ 2 } \int { \sec ^{ 2 }{ xdx } } \)
\(\Rightarrow \int { \sec ^{ 2 }{ ydy } } =\int { \sec ^{ 2 }{ xdx } } \)
\(\Rightarrow \tan { y } =\tan { x } +C\)

\(\frac { dy }{ dx } =xy+x+y+1\)
= (y + 1)(x + 1)
\(\frac { dy }{y + 1 } =dx(x+1)\)
Integrating both the sides
\(\int { \frac { dy }{ y+1 } } =\int { (x+1)dx} \)
\(\Rightarrow \log { \left| y+1 \right| } =\frac { { x }^{ 2 } }{ 2 } +x+C\)

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

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

\(\frac { dy }{ dx } =\sqrt { \frac { 1-\cos { x } }{ 1+\cos { x } } } \)
Integrating both the sides,
\(\int { dy } =\int { \sqrt { \frac { 1-\cos { x } }{ 1+\cos { x } } } } dx\)
\(\Rightarrow \int { dy } =\int { \sqrt { \frac { 2\sin ^{ 2 }{ \frac { x }{ 2 } } }{ 2\cos ^{ 2 }{ \frac { x }{ 2 } } } } } dx\)
\(\left\{ \because \quad \cos { x } =2\cos ^{ 2 }{ \frac { x }{ 2 } } -1\quad and\quad \cos { x } =1-2\sin ^{ 2 }{ \frac { x }{ 2 } } \right\} \)
\(\Rightarrow \int { dy } =\int { \tan { \frac { x }{ 2 } } } dx\)
\(\Rightarrow y=-2\log { \cos { \frac { x }{ 2 } } } +c\)
\(\Rightarrow \log { \cos { \frac { x }{ 2 } } } +c\)

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

\({ x }^{ 2 }+\left( y-b \right) ^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }\)
or \({ x }^{ 2 }+{ y }^{ 2 }-2by={ a }^{ 2 }\) ... (i)
\(\Rightarrow 2x+2y\frac { dy }{ dx } -2b\frac { dy }{ dx } =2b\)
\(\Rightarrow 2b=\frac { 2x+2y\frac { dy }{ dx } }{ \frac { dy }{ dx } } \)
Substituting the value in eqn. (i),
\(\left( { x }^{ 2 }-{ y }^{ 2 }-{ a }^{ 2 } \right) \frac { dy }{ dx } -2xy=0\)

Given, differential equation is 
\(\frac{d^2 y}{d x^2}+\sqrt[3]{\frac{d y}{d x}}+(1+x)=0 \Rightarrow \frac{d^2 y}{d x^2}+(1+x)=-\sqrt[3]{\frac{d y}{d x}}\)
On cubing both sides, we get
\(\left\{\frac{d^2 y}{d x^2}+(1+x)\right\}^3=-\frac{d y}{d x}\)
Hence, the order is 2 and degree is 3. So, the sum is 5.

y = a cos2x + b sin2x
\(\Rightarrow \frac { dy }{ dx } =-a\sin { 2x } \times 2+b\cos { 2x } \times 2\)
\(\Rightarrow \frac { dy }{ dx } =2\left[ -a\sin { 2x } +b\cos { 2x } \right] \)
\(\Rightarrow \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =2\left[ -a\cos { 2x } \times 2-b\sin { 2x } \times 2 \right] \)
\(\Rightarrow \frac { { x }^{ 2 }y }{ { dx }^{ 2 } } =-4\left[ a\cos { 2x } +b\sin { 2x } \right] =-4y\)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +4y=0\)

\(y=\left( { 2x }^{ 2 }-1 \right) +c{ e }^{ { -x }^{ 2 } }\)
\(\Rightarrow \frac { dy }{ dx } =2\left( 2x \right) +c{ e }^{ { -x }^{ 2 } }\times -2x\)
\(\Rightarrow \frac { dy }{ dx } =4x-2xc{ e }^{ { -x }^{ 2 } }\) ...(i)              
\(y=2\left( { x }^{ 2 }-1 \right) +c{ e }^{ { -x }^{ 2 } }\)
\(\Rightarrow c{ e }^{ { -x }^{ 2 } }=y-{ 2x }^{ 2 }+2\)...(ii)
Putting the value of \(c{ e }^{ { -x }^{ 2 } }\) from eqn. (ii) to eqn.(i),
\(\frac { dy }{ dx } =4x-2x\left( y-{ 2x }^{ 2 }+2 \right) \)
\(\Rightarrow \frac { dy }{ dx } =4x-2xy+{ 4x }^{ 3 }-4x\)
\(\therefore \ \frac { dy }{ dx } +2xy+{ 4x }^{ 3 }=0\)