y''-5y'-6y=e^-2t where y'(0)=1, y(0) = -1

Question: Solve the following equation using Laplace transform.

y''-5y'-6y=e^-2t where y(0)=-1, y’(0) = 1

Laplace transforms for the 1st and 2nd differentials.
L{y’(t)} = s.Y(s) – y(0)
L{y’’(t)} = s^2.Y(s) – (s.y(0) + y’(0))

Laplace transform for the exponential.
L{e^(-2t)} = 1/(s+2)

Substituting for their Laplace transforms into the DE,
s^2.Y(s) – (s.y(0) + y’(0)) – 5(s.Y(s) – y(0)) – 6.Y(s) = 1/(s+2)

Putting in the initial condition values,
s^2.Y(s) – (-s + 1) – 5(s.Y(s) + 1) – 6.Y(s) = 1/(s+2)
(s^2 – 5s – 6)Y(s) + s – 1 – 5 = 1/(s+2)
(s – 6)(s + 1)Y(s) = 1/(s+2) – (s – 6) = (1 – (s+2)(s-6))/(s+2) = (1 – (s^2 – 4s – 12))/(s+2)
Y(s) = -(s^2 – 4s – 13)/(s+1)(s+2)(s-6)

Converting to partial fractions, this becomes
Y(s) = (1/56)*(1/(s-6)) – (8/7)*(1/(s+1)) + (1/8)*(1/(s+2))

Taking the inverse transforms,
y(t) = L^(-1){Y(s)} = (1/56)*e^(6t) – (8/7)*e^(-t) + (1/8)*e^(-2t)

Answer: y(t) = (1/56)*e^(6t) – (8/7)*e^(-t) + (1/8)*e^(-2t)

by Level 11 User (81.5k points)