GGRRR Laplace transform

I have the following equation

d^2y/dt^2 + 3dy/dt -4y = -8e^-3t

subject to y = 3, dy/dt = 0 at t = 0

If i transfrom both sides using laplace tranfroms i get...

(s^2 + 3s -4)Y -3s -9 = -8/(s+3)

then..

(s^2 + 3s -4)Y = (3s^2 + 18s + 19)/ (s+3)

I know i'm goin wrong as the (s^2 + 3s -4) should cancel out, I've gone though it a couple of times and keep getting the same answer!!!

If some could help it would save me from topping myself :smile:

Thanks

Reply 1

Jonny W

The LHS here is correct, but the RHS should be -8/(s + 3)^2.

(s^2 + 3s -4)Y -3s -9 = -8/(s+3)

After doing partial fractions,

Y(s) = (11/5)[1/(4 + s)] - (3/2)[1/(3 + s)] + (23/10)[1/(-1 + s)] + 2/(3 + s)^2

y(t) = (11/5)e^(-4t) - (3/2)e^(-3t) + (23/10)e^t + 2t e^(-3t)

Reply 2

RichE

The RHS should be -8/(s+3).

The Laplace Transform of exp(at) is 1/(s-a).

And then use partial fractions.

Reply 3

stu_724

partial fractions to which part??

Reply 4

RichE

From your last equation solve for Y in terms of S.

The denominator factorises nicely.

Use partial fractions to write it a the sum of functions of the form A/(s-a).