Given \(\frac { dy }{ dx } \) = e^x−y + x²e^−y = e^−ye^x + e^−yx²
= e^−y(e^x + x²)
Separating the variables, we get e^ydy=(e^x + x²)dx
Integrating, we get ഽe^ydy = ഽ(e^x+x²)dx
e^y = e^x + \(\frac { x^{ 3 } }{ 3 } \) + c

Given  \(y=\frac { a }{ x } +b\)
Differentiating w.r.t x , we get
\(\frac { dy }{ dx } =\frac { -a }{ { x }^{ 2 } } \)
\({ x }^{ 2 }\frac { dy }{ dx } =-a\)
Again differentiating w.r.t x we get
\({ x }^{ 2 }\frac { dy }{ dx } +2x\frac { dy }{ dx } =0\)
⇒ \({ x }^{ 2 }\frac { dy }{ dx } +2x\frac { dy }{ dx } =0\) which is the required differential equation

Equation of family of parabolas with foci at the origin and axis along the x-axis is
y² = 4a(x+a) ...(1)
[ ∵ the focus is at the origin its vertex will be (-a, 0) and latus rectum is 4a]
Differentiating w.r.t. 'x' we get,
2y\(\left( \frac { dy }{ dx } \right) \) = 4a(1) ....(1)
⇒ 2y\(\left( \frac { dy }{ dx } \right) \) = 4a ...(2)
Also \(\frac { 2y }{ 4 } \left( \frac { dy }{ dx } \right) \) = a
⇒ \(\frac { y }{ 2 } \left( \frac { dy }{ dx } \right) \) = a...(3)
Substituting (2) and (3) in (1) we get,
y² = 2y\(\left( \frac { dy }{ dx } \right) \left[ x+\frac { y }{ 2 } \left( \frac { dy }{ dx } \right) \right] \)
⇒ y² = 2xy\(\left( \frac { dy }{ dx } \right) +{ y }^{ 2 }\left( \frac { dy }{ dx } \right) ^{ 2 }\)
Dividing by Y we get,
\(y=2x\frac { dy }{ dx } { +y\left( \frac { dy }{ dx } \right) }^{ 2 }\).

Equation of all circles passing through the origin and having their centres on the y-axis.
x² + (y - k)² = k² ....(1)
[where (0, k) is the centre of the cirde which lies on the y-axis and radius is k].
Differentiating w.r.t. 'x
⇒ 2x + 2(y-k)\(\frac { dy }{ dx } \) = 0
⇒ (y-k) =\(\frac { -x }{ \frac { dy }{ dx } } \) ....(2)
Also, y+\(\frac { x }{ \frac { dy }{ dx } } \) = k ....(3)
Substituting (2) & (3) in (1) we get
\({ x }^{ 2 }+\left( \frac { -x }{ \frac { dy }{ dx } } \right) ^{ 2 }=\left( y+\frac { x }{ \frac { dy }{ dx } } \right) ^{ 2 }\)

⇒ x² = y²+\(\\ \frac { 2xy }{ \frac { dy }{ dx } } \)
⇒ \(y^{2}-x^{2}-2 x y \frac{d y}{d x}=0\)

Given equation is (x - α)² + (y - β)² = r²
Differentiating w.r.t. 'x' we get,
2 (x - α)² + (y - β)\(\frac { dy }{ dx } \) = r²
⇒ (x - α) + (y - β)\(\frac { dy }{ dx } \) = 0....(2)
Differentiating again w.r.t. 'x' we get
1+(y-β) \(\frac { d^{ 2 }y }{ dx^{ 2 } } +\frac { dy }{ dx } .\frac { dy }{ dx } \) = 0
⇒ 1+(y-β) \(\frac { d^{ 2 }y }{ dx^{ 2 } } +\left( \frac { dy }{ dx } \right) ^{ 2 }\)= 0
⇒ y-β = \(\frac { -1\left( 1+\left( \frac { dy }{ dx } \right) ^{ 2 } \right) }{ \frac { d^{ 2 }y }{ dx^{ 2 } } } \) .....(3)
Substituting this value in (2) we get
(x-α) = \(\left( \frac { 1+\left( \frac { dy }{ dx } \right) ^{ 2 }\frac { dy }{ dx } }{ \frac { d^{ 2 }y }{ dx^{ 2 } } } \right) \) ......(4)
Substituting (3) and (4) in (1) we get
\(\frac { \left\{ 1+\left( \frac { dy }{ dx } \right) ^{ 2 }\left( \frac { dy }{ dx } \right) ^{ 2 } \right\} }{ \left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) ^{ 2 } } +\frac { \left\{ 1+\left( \frac { dy }{ dx } \right) ^{ 2 } \right\} ^{ 2 } }{ \left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) ^{ 2 } } \)
=\(\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) ^{ 2 }\)
⇒ \(\left[ 1+\left( \frac { dy }{ dx } \right) ^{ 2 } \right] \left[ \left( \frac { dy }{ dx } \right) ^{ 2 }+1 \right] ={ r }^{ 2 }\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) \)
⇒ \(\left[ 1+\left( \frac { dy }{ dx } \right) ^{ 2 } \right] ^{ 3 }\)
= \({ r }^{ 2 }\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) ^{ 2 }\)