Question

Solve the following differential equation:

`(4"D"^2 + 16"D" + 15)y = 4"e"^((-3)/2x)`

Answer 1

The auxiliary equation is 4m² + 16m + 15 = 0

4m² + 16m + 10m + 15 = 0

2m(2m + 3) + 5(2m + 3) = 0

(2m + 3)(2m + 5) = 0

2m = – 3, – 5

∴ m = `(-3)/2, (-5)/2`

Roots are real and different

C.F = (Ax + B) e^m₁x + Be^m₂x

C.F = `"Ae"^((-3)/2 x) + "Be"^((-5)/2 x)`

P.I = `1/((4"D"^2 + 16"D" + 15)) 4"e"^((-3)/2 x)`

Replace D by `(-3)/2`

4D² + 16D + 15 = 0

When D = `(-3)/2`

P.I = `x * 4/(4(2"D") + 16(1)) 4"e"^((-3)/2x)`

= `x * 1/(8"D" + 16) 4"e"^((-3)/2x)`

= `x * 1/(8((-3)/2) + 16) 4"e"^((-3)/2 x)`

= `x * 1/(-12 + 16) 4"e"^((-3)/2 x)`

= `x * 1/4 4"e"^((-3)/2 x)`

P.I = `x"e"^((-3)/2 x)`

The general solution is y = C.F + P.I

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