Grigori S. answered • 08/09/13

Certified Physics and Math Teacher G.S.

Let F(s) be the Laplace transform of y(t). Thus

Ly(t) = ∫_{0}^{∞ }y(t) e-^{st}dt (1)

Laplace transforms of derivatives can be found using integration by parts. We have

F_{1}(s) = L y'(t) = ∫_{0}^{∞} y'(t) e^{-st}dt = ∫_{0}^{∞} e^{-st} d y(t) =

y(t) e^{-st} (from 0 to ∞) +s ∫_{0}^{∞} y(t) e^{-st} dt

or

Ly'(t) = sF(s) -1 (2)

after using initial condition for y(t) and the fact that e^{-st} = 0 for t ↔∞.

By following the same steps we can show that fhe Laplace transform for the secpnd derivative

F_{2}(s) = Ly''(t) = sF1(s) + 1 = s[sF(s) - 1] -(-1) (3)

The last term in (3) is just y'(0) = -1. By taking Laplace transforms of all terms in the left side of

your equation and making use of expressions from (1) to (3), we obtain:

F(s) (s2 - s -6) - s + 2 = 0 or F(s) = (s -2)/(s^{2} -s -6) (4)

The denominator can be factored:

s^{2} - s - 6 = (s+2)(s-3)

Thus we have

F(s) = (s-2)/[s+2)(s-3)]= (1/5) (s-2)[ 1/(s-3) - 1/(s+2)] (5)

If you notice that

(s-2)/(s-3) = (s-2-1+1)/(s-3) = 1 + 1/(s-3)

and

(s-2)/(s+2) = (s-2 +4 -4)/(s+2) = 1 - 4/(s+2)

then you will come up with the following expression for F(s):

F(s) = (1/5) [1/(s-3) + 4/(s+2)] (6)

From previous tutoring sections we already know that the Laplace transform of e ^{at} is 1/(s-a)

where a = 3 and a = -2 in in this problem. Thus, taking inverse transforms we will come up with

the final answer:

y(t) = (1/5)( e^{3t} + 4e^{-2t})