inverse laplace transform

[hardyaa1]

I'm looking over an old exam to study for tomorrow's final, and I'm trying to figure out where I went wrong in this problem:

Solve using Laplace Transform: y" - 4y' + 4y = e[sup:nvsar3b8]2t[/sup:nvsar3b8] , y(0) = y'(0) = 0

It doesn't look like I lost points up to Y(s) = 1 / (s-2)[sup:nvsar3b8]3[/sup:nvsar3b8]
From there, I did partial fractions: 1 = (A+B+C)(s-2)(s-2), from which I gather that A=B=C=1/3
So here's where I lost points...

Y = 3 (A/(s-2)) where A= 1/3
So,
y(t) = 3 * (1/3) * (L[sup:nvsar3b8]-1[/sup:nvsar3b8] {1/(s-2}) = e[sup:nvsar3b8]2t[/sup:nvsar3b8]

these last 2 steps simply had a big "X" through them on my graded exam and our teacher doesn't go over / give us solutions,
so if anyone could help me figure this one out before 9am tomorrow (wednesday) I'd greatly appreciate it.

-Aaron

 

[DrMike]

I'm looking over an old exam to study for tomorrow's final, and I'm trying to figure out where I went wrong in this problem:

Solve using Laplace Transform: y" - 4y' + 4y = e[sup:y9lcvmp3]2t[/sup:y9lcvmp3] , y(0) = y'(0) = 0

It doesn't look like I lost points up to Y(s) = 1 / (s-2)[sup:y9lcvmp3]3[/sup:y9lcvmp3]
From there, I did partial fractions: 1 = (A+B+C)(s-2)(s-2), from which I gather that A=B=C=1/3

This doesn't look right.

Since you already know Y(s) = 1 / (s-2)[sup:y9lcvmp3]3[/sup:y9lcvmp3] can't you just invert that directly?

In the more general case, if (say) you had Y(s) = (1+3s+2s[sup:y9lcvmp3]2[/sup:y9lcvmp3])/(s-2)[sup:y9lcvmp3]3[/sup:y9lcvmp3], the correct way to do partial fractiosn would be to write

\(\displaystyle Y(s) = \frac{1+3s+2s^2}{(s-2)^3} = \frac{A}{s-2}+\frac{B}{(s-2)^2}+\frac{C}{(s-2)^3}\)

and so on.