Question

Solve the following differential equation:

(D² + D – 2)y = e^3x + e^–3x 

Answer 1

The auxiliary equation is

m² + m – 6 = 0

(m + 3)(m – 2) = 0

Roots are real and different

The complementary function is

C.F = Ae^m₁x + Be^m₂x

C.F = Ae^–3x + Be^2x 

P.I = `1/("d"^2 + "D" - 6)  "e"^(3x)`

= `1/((3)^2 + 3 - 6) "e"^(3x)`

P.I = `"e"^(3x)/(9 + 3 - 6)`

P.I₍₁₎ = `"e"^(3x)/6`

P.I₍₂₎ = `1/("D"^2 + "D" - 6) "e"^(-3x)`

Replace D by – 3

D² + D – 6 = 0

When D = – 3

∴ P.I₂ = `x * 1/(2"D" + 1) "e"^(-3x)`

= `x * 1/(2(-3) + 1) "e"^(-3x)`

= `x * 1/(-5) "e"^(-3x)`

P.I₂ = `(-x)/5 "e"^(-3x)`

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

y = `"Ae"^(-3x) + "Be"^(2x) + "e"^(3x)/6 - x/5 "e"^(-3x)`