How do I solve y''+5y'+6y=e^(-t) 𝛿(t-4); y(0) =2, y'(0) =-5 given this symbolic initial value problem?

Question

How do I solve y''+5y'+6y=e^(-t) 𝛿(t-4); y(0) =2, y'(0) =-5 given this symbolic initial value problem? I am trying to solve for y(t).

Answer 1

This problem can be solved with Laplace transforms. The left hand side of the equation transforms easily using any table of Laplace transform pairs to give

F(s) [s² + 5s + 6] - 2s - 5.

The right hand side of the equation can be found using the definition of the Laplace transform, namely

L[f(t)] = F(s) = ∫ f(t) e^-st dt.

The bounds of integration are t∈(0,∞). Plugging in, we find for the right hand side of the equation

∫e^-t δ(t-4) e^-st dt = ∫ e^-t(1+s) δ(t-4) dt = e^-4(1+s).

Now we can solve for F(s) to give

F(s) = (2s + 5 + e^-4(1+s)) / (s² + 5s + 6).

Using the method of partial fractions, we find that the first two terms can be found to be

1/(s+2) + 1/(s+3).

The second two terms, also found from partial fractions, are

e^-4(1+s) [ 1/(s+2) - 1/(s+3) ].

Now it is a simple matter of inverse Laplace transforming to find the answer. From a table of Laplace transforms, we find

y(t) = e^-2t + e^-3t + [ e^4-2t - e^8-3t ] h(t-4),

where h(t) is the Heaviside step function.