Question

Solve the following differential equation:

(D² – 10D + 25) y = 4e^5x + 5

Answer 1

The auxiliary equation is

m² – 10m + 25 = 0

(m – 5)(m – 5) = 0

m = 5, 5

Roots are real and equal

C.F = (Ax + B) e^mx

C.F = (Ax + B) e^5x 

P.I₍₁₎ = `x * 4/(2"D" - 10) "e"^(5x)`

Replace D by 5

2D – 10 = 0

when D = 5

P.I₍₁₎ = `x^2 * 4/2 "e"^(5x)`

P.I₁ = `2x^2 "e"^(5x)`

P.I₂ = `5/("D"^2 - 10"D" + 25)`

= `(5"e"^(0_x))/("D"^2 - 10"D" + 25)`

= `(5"e"^(0_x))/(0 - 10(0) + 25)`

== `5/25`

= `1/5`

P.I₂ - `1/5`

The general solution is y = C.F + P.I₁ + P.I₂

∴ y = `("A"x + "B")"e"^(5x) + 2x^2"e"^(5x) + 1/5`