x²dx - aydx = axdy - y²dy
⇒ \(\int { { x }^{ 2 } } dx+\int { { y }^{ 2 } } dy=a\left[ \int { xdy } +\int { y } dx \right] \)
=\(\frac { { x }^{ 3 } }{ 3 } +\frac { { y }^{ 3 } }{ 3 } \) = a(xy)+C [∵ d(xy) = x.dy + y.dy]

The given equation is of the form \(\frac { dy }{ dx } \)+ py = Q
where P = a and Q = e^x
\(\int { P } dx=\int { a } dx\) = ax
Integrating factor (L.F.) = \(e^{ \int { P } dx }\) = eax
∴ The solution is y \(e^{ \int { P } dx }=\int { Q } e^{ \int { P } dx }dx\)+C
⇒ y.e^ax =\(\int { { e }^{ x }.{ e }^{ ax } } dx+C\)
⇒ y.e^ax = \(\int { e^{ (a+1)x }dx+C } \)
⇒ y.e^ax = \(\frac { { e }^{ (a+1)x } }{ a+1 } \)+C

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.

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 + \({ Be }^{ \frac { 2 }{ 3 } x }\).

The auxiliary equation is m² - 6m + 25 = 0
Here a = 1, b = -6, c = 25
∴ m = \(\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } =\frac { 6\pm \sqrt { 36-4(1)(25) } }{ 2 } \)
= \(\frac { 6\pm \sqrt { 36-100 } }{ 2 } =\frac { 6\pm \sqrt { -64 } }{ 2 } =\frac { 6\pm 8i }{ 2 }\)

= 3 ± 4i
∴ α = 3, β = 4
Complementary function CF is e^ax
[A cosβx + B sinβx]
⇒ CF = e^3x[A cos4x + B sn4x]
∴ The general solution is e^3x
[A cos4x + B sin4x]