\(\frac { dy }{ dx } +\frac { 2 }{ y } y\) = x³
The given differential equation is of this form [Divided by x]
\(\frac { dy }{ dx } \)+Py = Q where
P =\(\int { \frac { 2 }{ x } } \) and Q = x³
∴ \(\int { P } dx=\int { \frac { 2 }{ x } dx } \) = 2 log x = log x²
∴ Integrating factor (I. F) =\(e^{ \int { p } dx }=e^{ logx^{ 2 } }\)= x²
Hence the solution is
\(ye^{ \int { p } dx }=\int { Q } .e^{ \int { p.dx } }dx+c\)
⇒ y.x²=\(\int { { x }^{ 3 }.{ x }^{ 2 } } dx+c\)
⇒ x²y=\(\\ \int { { x }^{ 5 }dx } +c\)
⇒ x²y = \(\frac { { x }^{ 6 } }{ 6 } \) + c

Given equation is y = ax² + bx....(1)
Differentiating w. r. t. x : we get,
\(\frac { dy }{ dx } \) = 2ax+b...(2)
Differentiating again w. r, t. 'x:we get,
\(\frac { d^{ 2 }y }{ dx^{ 2 } } =2a\Rightarrow \frac { 1 }{ 2 } \left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right) \)=a ...(3)
Substituting (3) in (2) we get,
\(\frac { dy }{ dx } =x\left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right) +b\)
⇒ b=\(\frac { dy }{ dx } -x\left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right)\)
⇒ bx = \(x\left( \frac { dy }{ dx } \right) -x^{ 2 }\left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right) \)  ....(4)
Substituting (3) and (4) in (1) we get
y = \(\frac { 1 }{ 2 } \left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right) x^{ 2 }+x\frac { dy }{ dx } -{ x }^{ 2 }\left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right) \)
y = \(-\frac { 1 }{ 2 } { x }^{ 2 }\left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right) +x\frac { dy }{ dx } \)
Multiplying by 2,
2y = \(-x^{ 2 }\left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right) +2x\frac { dy }{ dx } \)
⇒ \(-x^{ 2 }\left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right) -2x\frac { dy }{ dx } \)+2y = 0

The auxiliary equation is 4m² + 4m - 3 = 0
∴ (2m +3)(2m - 1) = 0
⇒ m = \(\frac { -3 }{ 2 } \) and m =\(\frac { 1 }{ 2 } \)

∴ Complementary function CF is \(Ae^{ \frac { -3x }{ 2 } }+Be^{ \frac { 1x }{ 2 } }\)
Particular Integral PI =\(\frac { 1 }{ \phi (D) } \)
PI=\(\frac { 1 }{ 4{ D }^{ 2 }+4D-3 } \).e^2x
=\(\frac { 1 }{ (2D+3)(2D-1) } \).e^2x
=\(\frac { 1 }{ 4\left( D+\frac { 3 }{ 2 } \right) \left( D-\frac { 1 }{ 2 } \right) } \).e^2x
=\(\frac { e^{ 2x } }{ 4\left( 2+\frac { 3 }{ 2 } \right) \left( 2-\frac { 1 }{ 2 } \right) } \)
=\(\frac { e^{ 2x } }{ 4\left( \frac { 7 }{ 2 } \right) \left( \frac { 5 }{ 2 } \right) } =\frac { { e }^{ 2x } }{ 21 } \)
General solution is y = CF + PI
⇒ y=\({ Ae }^{ \frac { -3x }{ 2 } }+Be^{ \frac { x }{ 2 } }=\frac { { e }^{ 2x } }{ 21 } \).

\(\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } -3\frac { dx }{ dt } +2x=0\)
Given (D²−3D+2) x = 0 where D = \(\frac{d}{dt}\)
A.E is m²− 3m + 2 = 0
(m−1) (m−2) = 0
m = 1, 2
C.F = Ae^t+ Be^2t
The general solution is x = Ae^t+ Be^2t   (1)
Now when t=0, x=0 (given)
(1) ⇒ 0 = A + B   (2)
Differentiating (1) w.r.t ‘t’
\(\frac { dx }{ dt } \)= Ae^t+ 2Be^2t
When t = 0, \(\frac { dx }{ dt } \) = 1
A + 2B = 1
Thus we have A + B = 0 and A + 2B = 1
Solving, we get A = –1, B = 1
∴ (1) ⇒ x = –e^t + e^2t
(i.e.) x = e^2t–e^t
Type II: \(f(x)=e^{a x}(\text { i.e }) \phi(D) y=e^{a x}\)
\(\text { P.I }=\frac{1}{\phi(D)} e^{a x}\)
Replace D by a, provided \(\phi(D) \neq 0\) when D = a
If \(\phi(D)=0\) when D = a, then
\(\text { P.I }=x \frac{1}{\phi^{\prime}(D)} e^{a x}\)
Replace D by a, provided \(\phi(D) \neq 0\) when D = a
If \(\phi^{\prime}(D)=0\) when D = a, then
\(\text { P.I }=x^{2} \frac{1}{\phi^{\prime \prime}(D)} e^{a x}\) and so on 

The highest derivative is third order and its power is one
∴ order : 3,
degree : 1