Let us radius = a, Centre is (0,a)
Equation is (x-0)² + (y-a)² = a²
x²+y²-2ay=0... (1)
Differentiating both sides W.r.t x, we get
\( 2 x+2 y y^{\prime}-2 a y^{\prime}=0 \)
\(\Rightarrow 2 a=\frac{2 x+2 y y^{\prime}}{y^{\prime}} \)
\(\text { Substituting in }(i) \text { we get }\)
\(x^{2}+y^{2}-y\left[\frac{2 x+2 y y^{\prime}}{y^{\prime}}\right]=0\)
\(\Rightarrow\left(x^{2}+y^{2}\right) y^{\prime}-2 x y-2 y^{2} y^{\prime}=0\)
\(\Rightarrow\left(x^{2}-y^{2}\right) y^{\prime}-2 x y=0 \text { is required equation } \)

\(\frac{d y}{d x}=A-\frac{B}{x^{2}} \)
\(\Rightarrow x \frac{d y}{d x}=A x-\frac{B}{x} \)
\(\Rightarrow y+x \frac{d y}{d x}=A x+\frac{B}{x}+A x-\frac{B}{x}=2 A x \)
Again differentiating, we get
\(\frac{d y}{d x}+x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=2 A \Rightarrow x^{2} \frac{d^{2} y}{d x^{2}}+2 x \frac{d y}{d x}=2 A x\)
\(\Rightarrow x^{2} \frac{d^{2} y}{d x^{2}}+2 x \frac{d y}{d x}=y+x \frac{d y}{d x} \quad \text { [from (i)]}\)
\(\Rightarrow x^{2} \frac{d^{2} y}{d r^{2}}+x \frac{d y}{d x}-y=0 \)
As differential equation corresponding to function
\(y=A x+\frac{B}{x} x \text { is } x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}-y=0 . \text { Hence }\)
\(y=A x+\frac{B}{x} \text { is solution of } x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}-y=0 \text { . } \)

Substiuting in (i), we get, \(\frac {1}{2}\) log |1 + y²|  = tan^-1 x - \(\frac {\pi}{4}\) is the required solution.

\(\frac{d y}{d x} =e^{a x+b y} \Rightarrow \frac{d y}{e^{b y}}=e^{a x} \cdot d x \)
\(\Rightarrow \int e^{-b y} d y =\int e^{a x} d x \)
\(\Rightarrow - \frac {1}{b} {e}^{-by}\) c is solution.

[ t + log|cos t + sin t|] = 2x + c  \(\Rightarrow\)  y- x + log |cos(x + y) + sin(x + y)| = c is the required solution

Solution is y sin x =  \(\int\) sin 2x dx  \(\Rightarrow\)  y sin x  = \(\frac {-cos2x}{2}\) + c is the required solution.

When x = 1, y = \(\frac {\pi}{4}\)  \(\Rightarrow \) c = 2. Substituting in (i), we get (1 + x²) tan y = 2

y(1 + ax)(1 - a²) = x(1 + a²)(1 - ay) is the required solution.

(x + c) e^x+y + 1 = 0 is the required solution

\(xcos\left( \frac { y }{ x } \right) (ydx+xdy)\)
\(=ysin\left( \frac { y }{ x } \right) (xdy-ydx)\)
\(\Rightarrow \left( y+x\frac { dy }{ dx } \right) xcos\left( \frac { y }{ x } \right) \)
\(=\left( x\frac { dy }{ dx } -y \right) ysin\left( \frac { y }{ x } \right) \)
\(\Rightarrow \frac { dy }{ dx } =\frac { y\left\{ xcos\left( \frac { y }{ x } \right) +ysin\left( \frac { y }{ x } \right) \right\} }{ x\left\{ ysin\left( \frac { y }{ x } \right) -xcos\left( \frac { y }{ x } \right) \right\} } \)
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
\(sin\left( \frac { y }{ x } \right) =logx+\frac { 1 }{ \sqrt { 2 } } \)

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

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
\(\therefore \log { y } -\frac { { x }^{ 2 } }{ 2{ y }^{ 2 } } =0\)

The given differential equation can be written as :
\(\frac { dx }{ dy } +\left( \cot { y } \right) x=\cos { y } \)
This is linear differential equation.
\(\Rightarrow I.F.={ e }^{ \int { \cot { y } dy } }={ e }^{ \log { \sin { y } } }=\sin { y } \)
\(\therefore\)  The solution is :
\(x\sin { y } =\int { \sin { y } \cos { y } dy } +C\)
\(=\frac { 1 }{ 2 } \int { \sin { 2y } } dy+C\)
or \(x\sin { y } =\frac { -1 }{ 4 } \cos { 2y } +C\)
It is given that y = 0, when x = 0
\(c-\frac { 1 }{ 4 } =0\Rightarrow c=\frac { 1 }{ 4 } \)
\(\therefore x\sin { y } =\frac { 1 }{ 4 } \left( 1-\cos { 2y } \right) =\frac { 1 }{ 2 } \sin ^{ 2 }{ y } \)
\(\Rightarrow 2x=\sin { y } \)
which is the required solution.

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 \(\left[ 1+\left( \frac { dy }{ dx } \right) ^{ 2 } \right] ^{ 3 }={ r }^{ 2 }\left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) ^{ 3 }\)