The highest derivative is of third order and its power is 1.
Order is 3 and degree is 1.

The auxiliary equation is m²-2km+k² = 0
⇒ (m - k)² = 0
⇒ m = k.k
The roots are real and equal
∴ Complimentary function CF is (Ax + B)e^kx
The general solution is y= (Ax + B)e^kx

The auxiliary equation is m² - 4m + 4 = 0
⇒ (m - 2)² = 0
⇒ m 2,2
The roots are real and equal
∴ Complementary function CF is (Ax + B)e^2x
∴ The general solution is y = (Ax + B)e^2x

Given  \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -4\frac { dy }{ dx } +5y\) = 0
(D²−4D+5)y = 0
The auxiliary equation is m²−4m + 5 = 0
⇒ (m−2)²−4 + 5 = 0
(m− 2)² = –1
m - 2 = 土\(\sqrt{-1}\)
m = 2 土 i , it is if the form α 土 iβ
∴ C.F = e^2x[A cos x + B sin x]
The general solution is y = e^2x[A cos x + B sin x]

Given m²\(\frac { dc }{ dm } \)+2mc = 2
Dividing by m², we get,
\(\frac { dC }{ dm } +\frac { 2c }{ m } =\frac { 2 }{ { m }^{ 2 } } \)
This is of form \(\frac { dy }{ dx } \)+Py = Q
where P = \(\frac { 2 }{ m } \) and Q = \(\frac { 2 }{ m } \)
\(\int { p } dm=\int { \frac { 2 }{ m } } \)dm = 2 log m = log m²
∴ Integrating Factor (LF.) = \(e^{ \int { P } dm }=e^{ logm^{ 2 } }\)
= m²
∴ The solution is
\(ce^{ \int { P } dm }=\int { Q } e^{ \int { P } dm }m+c.K\)
cm²=\(\int { \frac { 2 }{ { m }^{ 2 } } { m }^{ 2 } } dm+K=\int { 2 } dm+cK\)
cm² = 2m+K
Given that c = 4, when m = 2
4(2²) = 2(2) + K ⇒ 16 - 4 = K
⇒ K=12
∴ (1) becomes
cm² = 2m+ 12
⇒ cm² = 2 (m + 6)