The given differential equation is:
(x e^y/x+y)dx = x dy
\(\Rightarrow \frac { dy }{ dx } ={ e }^{ y/x }+\left( \frac { y }{ x } \right) \)...(i)
Let f(x, y) = e y/x+y/x
On putting, \(x=\lambda x, y=\lambda y\)
then \(f(\lambda x,\lambda y)=e^{ \lambda y/\lambda x }+\frac { \lambda y }{ \lambda x } \)
\(={ e }^{ y/x }+\frac { y }{ x } =f(x,y)\)
So this equation is homogeneous.
Now put y = vx in equation (i)
On differentiating both sides w.r.t. x we get
\(\frac { dv }{ dx } =v+x\frac { dv }{ dx } \)
From (i), \(v+x\frac { dv }{ dx } =e^{ vx/x }+\frac { vx }{ x } \)
\(\Rightarrow\) \(v+x\frac { dv }{ dx } ={ e }^{ v }+v\)
\(\Rightarrow\) \(\int { { e }^{ -v } } dv=\int { \frac { dx }{ x } } \)
\(\Rightarrow\) \(\frac { { e }^{ -v } }{ -1 } =log|x|+C\)
\(\Rightarrow\) C = -e^-1
\(\Rightarrow\)e^-1-e^-y/x = log |x|
Hence required solution is:
e^-1-e^-y/x = log |x|

\(xcos\left( \frac { y }{ x } \right) (ydx+xdy)\)
\(=ysin\left( \frac { y }{ x } \right) (xdy-ydx)\)
\(\Rightarrow \left( y+x\frac { dy }{ dx } \right) xcos\left( \frac { y }{ x } \right) \)
\(=\left( x\frac { dy }{ dx } -y \right) ysin\left( \frac { y }{ x } \right) \)
\(\Rightarrow \frac { dy }{ dx } =\frac { y\left\{ xcos\left( \frac { y }{ x } \right) +ysin\left( \frac { y }{ x } \right) \right\} }{ x\left\{ ysin\left( \frac { y }{ x } \right) -xcos\left( \frac { y }{ x } \right) \right\} } \)
Eqn (i) is homogeneous So, putting y = vx and \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
\(v+x\frac { dv }{ dx } =\frac { vx[xcosv+vxsinv] }{ x[vxsinv-xcosv] } \)
\(\Rightarrow v+x\frac { dv }{ dx } =\frac { v[cosv+vsinv] }{ vsinv-cosv } \)
\(=\frac { 2vcosv }{ vsinv-cosv } \)
\(\Rightarrow \frac { 1-cosv-vsinv }{ vcosv } dv=-\frac { 2dx }{ x } \)
\(\Rightarrow \int { \frac { 1.cosv-vsinv }{ vcosv } dv=-2\int { \frac { dx }{ x } } } \)
\(\Rightarrow\) log |v cosv| = -2log x + log C
\(\Rightarrow\) x²v cosv = C
\(\Rightarrow\) xy cos(y/x) = C[as y = vx]

The given D.E. is:
\(xcos\left( \frac { y }{ x } \right) \frac { dy }{ dx } =ycos\left( \frac { y }{ x } \right) +x\) ..(i)
When, \(x=1, y=\frac { \pi }{ 4 } \)
\(cos\left( \frac { y }{ x } \right) \frac { dy }{ dx } =\frac { y }{ x } cos\left( \frac { y }{ x } \right) +1\)....(ii)
Put, y/x = v
\(\Rightarrow\) \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
From (ii), \(cosv\left( v+x\frac { dv }{ dx } \right) =v\quad cosv+1\)
\(\Rightarrow\) v cosv+x cosv dv/dx = v cos v + 1
\(\Rightarrow\) \(\\ xcos\frac { dv }{ dx } =1\)
\(\Rightarrow\) \(\int { cosv\quad dv=\int { \frac { dx }{ x } } } \)
Put \(y=\frac { \pi }{ 4 } \) , x = 1
\(\Rightarrow\) \(\frac { 1 }{ \sqrt { 2 } } =C\)
The particular solution is
\(sin\left( \frac { y }{ x } \right) =logx+\frac { 1 }{ \sqrt { 2 } } \)

\({ x }^{ 2 }\frac { dy }{ dx } -xy=1+\cos { \left( \frac { y }{ x } \right) } ,x\neq 0,when\quad x=1,y=\frac { \pi }{ 2 } \)
\({ x }^{ 2 }\frac { dy }{ dx } -xy=1+\cos { \left( \frac { y }{ x } \right) } ,x\neq 0\)
When x = 1, \(y=\frac { \pi }{ 2 } \)
as \(\left( 1+\cos { x } =2\cos ^{ 2 }{ \frac { x }{ 2 } } \right) \)
\(\frac { { x }^{ 2 }\frac { dy }{ dx } -xy }{ 2\cos ^{ 2 }{ \left( \frac { y }{ 2x } \right) } } =1\)
\(\frac { \sec ^{ 2 }{ \left( \frac { y }{ 2x } \right) } }{ 2 } \left[ { x }^{ 2 }\frac { dy }{ dx } -xy \right] =1\)
Dividing both sides by x³
\(\Rightarrow \frac { \sec ^{ 2 }{ \left( \frac { y }{ 2x } \right) } }{ 2 } \left[ \frac { 1 }{ x } .\frac { dy }{ dx } -\frac { y }{ { x }^{ 2 } } \right] =\frac { 1 }{ { x }^{ 3 } } \)
\(\Rightarrow \frac { d }{ dx } \left[ \tan { \left( \frac { y }{ 2x } \right) } \right] ={ x }^{ -3 }\)
Integrating both sides w.r.t.x, we get
\(\left[ \tan { \left( \frac { y }{ 2x } \right) } \right] =\int { { x }^{ -3 } } dx\)
\(\Rightarrow \tan { \left( \frac { y }{ 2x } \right) } =\frac { { x }^{ -3+1 } }{ -3+1 } +C\)
\(\Rightarrow \tan { \left( \frac { y }{ 2x } \right) } =-\frac { 1 }{ 2 } { x }^{ -2 }+C\)
\(\Rightarrow \tan { \left( \frac { y }{ 2x } \right) } =-\frac { 1 }{ 2{ x }^{ -2 } } +C\)
For particular solution when x = 1, \(y=\frac { \pi }{ 2 } \), we have
\(\tan { \left( \frac { \pi }{ 4 } \right) } =-\frac { 1 }{ 2 } +C\quad \)
\(\Rightarrow 1+\frac { 1 }{ 2 } =C\)
\(\Rightarrow C=\frac { 3 }{ 2 } \)
Hence particular solution is :
\(\tan { \left( \frac { y }{ 2x } \right) } =-\frac { 1 }{ 2{ x }^{ 2 } } +\frac { 3 }{ 2 } \)

The given D.E. is x²dy + y(x + y)dx = 0... (i)
when x = 1, y = 1
From (i), \(\frac { dy }{ dx } =-\frac { xy+{ y }^{ 2 } }{ { x }^{ 2 } } \) ... (ii)
Put,   y = vx
\(\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
From (ii), \(v+x\frac { dv }{ dx } =\frac { xvx+{ v }^{ 2 }{ x }^{ 2 } }{ { x }^{ 2 } } \)
\(\Rightarrow x\frac { dv }{ dx } =\frac { -v{ x }^{ 2 }-{ v }^{ 2 }{ x }^{ 2 }-v{ x }^{ 2 } }{ { x }^{ 2 } } \)
\(\Rightarrow x\frac { dv }{ dx } =-2v-{ v }^{ 2 }\)
\(\Rightarrow \frac { dv }{ v\left( v+2 \right) } =-\frac { dx }{ x } \)
\(\Rightarrow \frac { 1 }{ 2 } \left[ \frac { 1 }{ v } -\frac { 1 }{ v+2 } \right] dv+\frac { dx }{ x } =0\)
Integrating it we get
\(\frac { 1 }{ 2 } \left[ \log { v } -\log { \left| v+2 \right| } \right] +\log { x } ={ C }_{ 1 }\)
\(\Rightarrow \log { \frac { v }{ \left| v+2 \right| } } +\log { { x }^{ 2 } } ={ 2C }_{ 1 }\)
\(\Rightarrow \log { \left[ \frac { y }{ y+2x } { x }^{ 2 } \right] } ={ C }_{ 1 }\)
\(\Rightarrow \log { \left[ \frac { y{ x }^{ 2 } }{ y+2x } \right] } =C\)
\(\Rightarrow \log { \left[ \frac { 1 }{ 1+2 } \right] } =C\Rightarrow C=\log { \frac { 1 }{ 3 } } \)
\(\therefore \log { \left[ \frac { y{ x }^{ 2 } }{ y+2x } \right] } =\log { \frac { 1 }{ 3 } } \)
\(\Rightarrow \)  3yx² = y + 2x
3yx² - y = 2x
y(3x² - 1) = 2x
\(y=\frac { 2x }{ 3{ x }^{ 2 }-1 } \)

We have, (x² + xy) dy = (x² + y²)dx
\(\Rightarrow \quad \frac{d y}{d x}=\left(\frac{x^2+y^2}{x^2+x y}\right)\)     ...(i)
This is a homogeneous differential equation.
On putting \(y=v x \Rightarrow \frac{d y}{d x}=v \cdot 1+x \frac{d v}{d x}\) in Eq. (i),
we get
\(\begin{aligned} v+x \frac{d v}{d x} & =\left(\frac{x^2+v^2 x^2}{x^2+x \cdot x v}\right) \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad x \frac{d v}{d x} & =\frac{1+v^2}{1+v}-v=\frac{1+v^2-v-v^2}{1+v}=\frac{1-v}{1+v} \end{aligned}\)
\(\therefore\left(\frac{1+v}{1-v}\right) d v=\frac{1}{x} d x\)
On integrating both sides, we get
\(\begin{aligned} & \int\left(\frac{1+v}{1-v}\right) d v=\int \frac{1}{x} d x \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad \int\left[-1+\frac{2}{1-v}\right] d v=\log |x|+\log C \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad-v-2 \log (1-v)=\log |x|+\log C \end{aligned}\)
\(\begin{aligned} & \Rightarrow-v=2 \log (1-v)+\log |x|+\log C \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad-v=\log (1-v)^2+\log \{C|x|\} \end{aligned}\)
                                  [\(\because\) log m + log n = log mn]
\(\Rightarrow\)    -v = log{C |x| (1 - v)²}
\(\Rightarrow\)    C|x|(1 - v)² = e^-v
\(\Rightarrow C|x|\left(1-\frac{y}{x}\right)^2=e^{-y / x} \quad\left[\because v=\frac{y}{x}\right]\)   ...(ii)
On puting x = 1 and y = 0 in Eq. (ii), we get
C.1(1 - 0) = e⁰ \(\Rightarrow\) C = 1
Thus, the required solution is
\(\begin{aligned} |x|\left(1-\frac{y}{x}\right)^2 & =e^{-y / x} \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad(x-y)^2 & =|x| e^{-y / x} \end{aligned}\)
which is the required particular solutions.

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

Given differential equation is
\(\frac { dy }{ dx } =\frac { { y }/{ x } }{ 1+\left( { y }/{ x } \right) ^{ 2 } } =f\left( \frac { y }{ x } \right) \)
Hence, homogeneous.
Put y = vx
\(\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
\(\therefore \)  The differential equation becomes
\(v+x\frac { dv }{ dx } =\frac { v }{ 1+{ v }^{ 2 } } \)
\(x\frac { dv }{ dx } =\frac { v }{ 1+{ v }^{ 2 } } -v\)
\(=-\frac { { v }^{ 3 } }{ 1+{ v }^{ 2 } } \)
\(\Rightarrow \log { \left| v \right| } -\frac { 1 }{ 2{ v }^{ 2 } } =-\log { \left| x \right| } +C\)
\(\Rightarrow \log { v } +\log { x } -\frac { 1 }{ 2{ v }^{ 2 } } =C\)
\(\Rightarrow \log { \left( vx \right) } -\frac { 1 }{ 2{ v }^{ 2 } } =C\)
\(\therefore \log { y } -\frac { { x }^{ 2 } }{ 2{ y }^{ 2 } } =C\)
For particular solution
x = 0, y = 1  \(\Rightarrow \)  c = 0
\(\therefore \log { y } -\frac { { x }^{ 2 } }{ 2{ y }^{ 2 } } =0\)

Given \(\left( \tan ^{ -1 }{ y } -x \right) dy=\left( 1+{ y }^{ 2 } \right) dx\)
\(\Rightarrow \frac { dx }{ dy } =\frac { \tan ^{ -1 }{ y } -x }{ 1+{ y }^{ 2 } } \)
\(\Rightarrow \frac { dx }{ dy } +\frac { x }{ 1+{ y }^{ 2 } } =\frac { \tan ^{ -1 }{ y } }{ 1+{ y }^{ 2 } } \)  ...(i)
This is a linear differential equation with :
\(P=\frac { 1 }{ 1+{ y }^{ 2 } } \) and \(Q=\frac { \tan ^{ -1 }{ y } }{ 1+{ y }^{ 2 } } \)
\(\Rightarrow I.F.={ e }^{ \int { \frac { 1 }{ 1+{ y }^{ 2 } } } }dy={ e }^{ \tan ^{ -1 }{ y } }\)
Multiplying both sides of eqn. (i) by
\(I.F.={ e }^{ \tan ^{ -1 }{ y } },\) we get
\(x.I.F.=\int { Q.I.F.dy } \)
\(x{ e }^{ \tan ^{ -1 }{ y } }=\int { \frac { \tan ^{ -1 }{ y } }{ 1+{ y }^{ 2 } } } { e }^{ \tan ^{ -1 }{ y } }dy+C\)
\(\Rightarrow x{ e }^{ \tan ^{ -1 }{ y } }=\int { t{ e }^{ t }dt+C } \), where \(t=\tan ^{ -1 }{ y } \)
\(\Rightarrow x{ e }^{ \tan ^{ -1 }{ y } }={ e }^{ t }\left( t-1 \right) +C\)
\(\Rightarrow x{ e }^{ \tan ^{ -1 }{ y } }={ e }^{ \tan ^{ -1 }{ y } }\left( \tan ^{ -1 }{ y } -1 \right) +c\) .. (ii)
It is given that y(0) = 0 i.e., y = 0 when x = 0
Putting x = 0, y = 0 in eqn. (ii), we get
\(0={ e }^{ 0 }\left( 0-1 \right) +c\Rightarrow c=1\)
Putting c = 1 in eq (ii), we get
\(x{ e }^{ \tan ^{ -1 }{ y } }={ e }^{ \tan ^{ -1 }{ y } }\left( { e }^{ \tan ^{ -1 }{ y } }-1 \right) +1\)
\(\Rightarrow \left( x-\tan ^{ -1 }{ y+1 } \right) { e }^{ \tan ^{ -1 }{ y } }=1\)

\(\left( x-a \right) ^{ 2 }+\left( y-b \right) ^{ 2 }={ r }^{ 2 }\) ... (i)
\(\Rightarrow 2\left( x-a \right) +2\left( y-b \right) \frac { dy }{ dx } =0\).. (ii)
\(\Rightarrow 1+\left( y-b \right) \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +\left( \frac { dy }{ dx } \right) ^{ 2 }=0\). (iii)
\(\therefore \left( y-b \right) =-\frac { \left( 1+{ y }_{ 1 }^{ 2 } \right) }{ { y }_{ 2 } } \)
From (ii), \(\left( x-a \right) =\frac { { y }_{ 1 }\left( 1+{ y }_{ 1 }^{ 2 } \right) }{ { y }_{ 2 } } \)
Putting these values in eqn. (i),
\(\frac { { y }_{ 1 }^{ 2 }\left( 1+{ y }_{ 1 }^{ 2 } \right) ^{ 2 } }{ { y }_{ 2 }^{ 2 } } +\frac { \left( 1+{ y }_{ 1 }^{ 2 } \right) ^{ 2 } }{ { y }_{ 2 }^{ 2 } } ={ r }^{ 2 }\)
or \(\left[ 1+\left( \frac { dy }{ dx } \right) ^{ 2 } \right] ^{ 3 }={ r }^{ 2 }\left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) ^{ 3 }\)