Given \(\frac { a+{ x }^{ 2 } }{ 1+y } =xy\frac { dy }{ dx } \)
Separating the variables we get,
\(\frac { (1+{ x }^{ 2 })dx }{ x } \) = y(1+y)dy
⇒ \(\left( \frac { 1 }{ x } +\frac { { x }^{ 2 } }{ x } \right) \)dx = (y+y²)dy
⇒ \(\left( \frac { 1 }{ x } +x \right) \) = (y+y²)dy
Integrating both sides we get,
\(\int { \left( \frac { 1 }{ x } +x \right) dx } =\int { (y+y^{ 2 })dy } \)
⇒ logx + \(\frac { { x }^{ 2 } }{ 2 } =\frac { { y }^{ 2 } }{ 2 } +\frac { y^{ 3 } }{ 3 } \) + C.

Let the equation of straight lines passing through the origin be
y = mx ..(1)
where m is the arbitrary constant
Differentiating w.r.t 'x' we get,
\(\frac { dy }{ dx } \)= m(1) ⇒ \(\frac { dy }{ dx } \) = m  .....(2)
Substituting (2) in (1) we get,
\(y=x\frac { dy }{ dx } \).

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

The auxiliary equation is m² - 3m + 2 = 0
⇒ (m -2) (m -1) = 0
⇒ m = 1, 2
The roots are real and equal.
∴ Complementary function CF is Ae^x + Be^2x
PI = \(\frac { 1 }{ \phi D } f(x)=\frac { 1 }{ (D-2)(D-1) } \)
= \(\frac { e^{ 4x } }{ (4-2)(4-1) } =\frac { e^{ 4x } }{ 2(3) } =\frac { e^{ 4x } }{ 6 } \)
∴ y = CF+PI
y = Ae^x+Be^2x+\(\frac { { e }^{ 4x } }{ 6 } \)   ..(1)
Given that when x = 0, y = 0
⇒ 0 = Ae⁰+Be⁰+\(\frac { { e }^{ 0 } }{ 6 } \) ⇒ A+B+\(\frac { 1 }{ 6 } \)
⇒ A + B = -\(\frac { 1 }{ 6 } \) ...(2)
Als, when x = 1, y = 0
⇒ 0 = Ae¹+ Be² +\(\frac { { e }^{ 4 } }{ 6 } \) ⇒ Ae + Be² = -\(\frac { { e }^{ 4 } }{ 6 } \)
(2) xe ⟶ Ae + Be = -\(\frac { { e }^{ 4 } }{ 6 } \)

⇒ \(-\frac { 1 }{ 6 } \)(e²+e+1)
Substituting the value of A and B in (2) we get,
A-\(\frac { 1 }{ 6 } \)(e²+e+1) = -\(\frac { 1 }{ 6 } \)

⇒ A = \(\frac { 1 }{ 6 } \)(e² + e)
Substituting the value of A and B in (1) we get,
y = \(\frac { 1 }{ 6 } \)(e²+e)e^x-\(\frac { 1 }{ 6 } \)(e²+e+1)e^2x+\(\frac { { e }^{ 4x } }{ 6 } \)
⇒ 6y = (e²+e)e^x-(e²+e+1)e^2x+e^4x

The auxiliary equation is m² - 10m + 25 = 0
(m - 5) = 0
⇒ m = 5, 5
The roots are real and equal.
∴ Complementary function CF is (Ax + B) e_5x
Particular Integral PI1 =\(\frac { 1 }{ \phi (D) } \)f₁(x)
= \(\frac { 1 }{ (D-5)^{ 2 } } \).4 x e^5x = 4.e^5x.\(\frac { x^{ 2 } }{ 2 } \)
= 2x² e^5x [∵ (D-5)² =0 when D=5]
PI₂ = \(\frac { 1 }{ \phi (D) } \)f₂(x)
= \(\frac { 1 }{ (D-5)^{ 2 } } 5=\frac { 5.{ e }^{ 0x } }{ (D-5)^{ 2 } } =\frac { 5.{ e }^{ 0x } }{ (0-5)^{ 2 } } \)
= \(\frac { 5 }{ 25 } =\frac { 1 }{ 5 } \)
∴ y = CF + PI₁ + PI₂
∴ The general solution is
y = (Ax + B) e^5x + 2x² e^5x + 5

The auxiliary equation is m² + m - 6 = 0
⇒ (m + 3) (m - 2) = 0
⇒ m = -3, 2
The roots are real and different.
∴ The complementary function CF is Ae^-3x + B^ex
Particular Integral
PI₁=\(\frac { 1 }{ \phi (D) } \)f₁(x)
=\(\frac { 1 }{ ({ D }^{ 2 }+D-6) } \)e^3x
PI₁=\(\frac { { e }^{ 3x } }{ (D+3)(D-2) } =\frac { { e }^{ 3x } }{ (3+3)(3-2) } \)
= \(\frac { { e }^{ 3x } }{ 6(1) } =\frac { e^{ 3x } }{ 6 } \)
PI₂ = \(\frac { 1 }{ \phi (D0 } { f }_{ 2 }(x)=\frac { e^{ -3x } }{ (D+3)(D-2) } \)
= x\(\frac { { e }^{ -3x } }{ (-3-2) } \) [∵ when D = - 3, D + 3 = 0]
PI₂ = \(-\frac { x }{ 5 } \)e^-3x
∴ y = CF+PI₁+PI₂
∴ The general solution is
y = Ae^-3x+Be^2x+\(\frac { { e }^{ 3x } }{ 6 } -\frac { x }{ 5 } \)e^-3x

The auxiliary equation is 3m² - 5m + 2 = 0
⇒ (m - 1)(3m - 2) = 0
⇒ m = 1, \(\frac { 2 }{ 3 } \)
The roots are real and different
∴ Complementary function CF is Ae^x + \({ Be }^{ \frac { 2 }{ 3 } x }\)
∴ The general solution is y = Ae^x + \({ Be }^{ \frac { 2 }{ 3 } x }\).

Given \(\frac { dC }{ dq } =a-\frac { c }{ q } \)
⇒ \(\frac { dC }{ dq } +\frac { C }{ q } \) = a
The given differential equation is of the form
\(\frac { dC }{ dq } \)+PC = Q where
P = \(\frac { 1 }{ q } \) and Q = a
\(\int { p } dq=\int { \frac { 1 }{ q } } \)
Integrating factor I.F = e^{log q} =q
∴ The solution is C y\(e^{ \int { P } dq }=\int { Q } e^{ \int { P } dq }\)+C
⇒ C(q) = \(\int { a } .qdq+C\)
⇒ C.q = a\(\left( \frac { { q }^{ 2 } }{ 2 } \right) \)+C
⇒ 2Cq = aq² + K  where K = 2C.

x dy = -y dx
Separating the variables we get
\(\frac { dy }{ y } =-\frac { dx }{ x } \)
Integrating, \(\int { \frac { dy }{ y } } =-\int { \frac { dx }{ x } } \)
⇒ log y = -log x + log C
⇒ log y = log\(\left( \frac { C }{ x } \right) \Rightarrow y=\frac { C }{ x } \) ⇒ xy = C.

Given y = mx + \(\frac { a }{ m } \) ...(1)
Differentiating w.r.t. 'x' we get,
\(\frac { dy }{ dx } \) = m(1)+0 ⇒ m = \(\frac { dy }{ dx } \) ...(2)
Substituting (2) in (1) we get,
y = \(\left( \frac { dy }{ dx } \right) x+\frac { a }{ \frac { dy }{ dx } } \Rightarrow y=\frac { \left( \frac { dy }{ dx } \right) ^{ 2 }x+a }{ \left( \frac { dy }{ dx } \right) } \)
⇒ y\(\left( \frac { dy }{ dx } \right) =x\left( \frac { dy }{ dx } \right) ^{ 2 }\)+ a which is the required differential equation.

Taking power 3 both sides, we get
\(\left[ 1+\left( \frac { dy }{ dx } \right) ^{ 2 } \right] ^{ \frac { 2 }{ 3 } \times 3 }=\left( \frac { d^{ 2 }y }{ { dx }^{ 2 } } \right) ^{ 3 }\)
⇒ \(\left[ 1+\left( \frac { dy }{ dx } \right) ^{ 2 } \right] ^{ 2 }=\left( \frac { d^{ 2 }y }{ { dx }^{ 2 } } \right) ^{ 3 }\)
The highest derivative is of order 2 and its power is 3
∴ order is 2 and degree is 3.

The highest derivative if of order 3 and its power is 1
∴ order is 3 and degree is 1.

⇒ \(\frac { dy }{ dx } \) = -e^x(1+y)
Separating the variables we get,
⇒ \(\frac { dy }{ 1+y } \) = -e^x dx
⇒ log(1+y) = -e^x+c

\(\left( \frac { dy }{ dx } \right) ^{ 3 }+y=x-\frac { 1 }{ \left( \frac { dy }{ dx } \right) } \)
⇒ \(\left( \frac { dy }{ dx } \right) ^{ 3 }+y=\frac { x\left( \frac { dy }{ dx } \right) -1 }{ \left( \frac { dy }{ dx } \right) } \)
⇒ \(\left( \frac { dy }{ dx } \right) ^{ 4 }+y\left( \frac { dy }{ dx } \right) =x\left( \frac { dy }{ dx } \right) -1\)
The highest derivative is of first order and its power is 3.
∴ Order is 1 and degree is 4.

Squaring both Sides, we get
\(\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) ^{ 2 }=y-\frac { dy }{ dx } \)
The highest derivative is second order and its degree is 2.