Initial Value Problems with Laplace Transform

Question

Solve the following IVP using Laplace Transform

2y′′ + 10y = 3u₁₂(t) − 5δ(t − 4), y(0) = −1 y′(0) = −2

Answer 1

Reading u₁₂(t) as a step funtion that goes from 0 to 1 at t = 12:

Taking the L.T. of the differential equation results in

LT(3u₁₂(t)) = 3e^-12s/s which you can get from a table or doing the LT integral on a unit step function

LT(-5δ(t-4)) =-5e^-4s

LT(2y'' + 10y) = 2(s²Y-sy(0) -y'(0)) +10Y where Y is LT(y)

LT of LHS is (2s²+10)Y + 2s + 4

Full equation is ((2s²+10)Y + 2s + 4 = 3e^-12s/s -5e^-4s

You now have to solve for Y(s) in recognizable (usually simplest) form so that the inverse LT can be worked out:

Y = (-2s- 4 + 3e^-12s/s -5e^-4s)/(2s²+10) (Solve for each of the 4 terms separately as LT^-1 are linear.

-2s/(2s²+10) looks like bs/(s²+a²) divide top and bottom by 2:

-1(s/(s²+5)) a = sqrt(5) The LT^-1 is -cos(sqrt(5)x) (If you take derivative twice, you get 5cos(sqrt(5)x)) which is a homogeneous solution.

-4/(2s²+10) goes to -2/sqrt(5)*(sqrt(5)/(s²+5)) which has a LT^-1 of -2sin(sqrt(5)t)/sqrt(5)

3e^-12s/(2s²+10) goes to (3/2)/(s²+5) * e^-12s and a LT^-1 of 3u₁₂(t)sin(sqrt(5)(t-12))/2sqrt(5)

-5e^-4s has a LT^-1 of -5δ(t-4)

Lots of room for errors, so you need to check it all. Hope that helps.