# Initial value problem (Laplace)

1. Jul 11, 2008

### math_04

1. The problem statement, all variables and given/known data

Use the Laplace transform method to solve the differential equation

y" -4y' +4y = e^2t subject to initial conditions y(0)=1, y'(0)=0

2. Relevant equations

3. The attempt at a solution

s^2Y(s) -sy(0) - y'(0) - 4[ sY(s) - y(0)] + 4Y(s)= L (e^2t)

Y(s) [s^2 -4s +4] - s -4 = 1/(s-2)

Y(s) [ (s-2)^2 ] = 1/(s-2) + s +4

Y(s) = 1/(s-2)^3 + s/(s-2)^2 + 4/(s-2)^2

Is this right, how do I find out the inverse laplace of Y(s) like for example 1/(s-2)^3 seems impossible!

Thanks.

2. Jul 11, 2008

### HallsofIvy

Staff Emeritus
??? Pretty nearly every inverse lapalce transform is done by looking it up in a table of transforms. But, as I pointed out in response to your other post, generally, you find the Laplace transform (and almost always the inverse transform) by looking it up in a table. Here is a good one:
http://www.vibrationdata.com/Laplace.htm

In particular, the inverse transform of 1/(s-2)3 is (1/2)t2e-2t.

Last edited: Jul 11, 2008
3. Jul 11, 2008

### math_04

how about s/(s-2)^2. It isnt a cosine or sine laplace thing?

Thanks.

4. Jul 11, 2008

### jeffreydk

You can find $$\frac{1}{(s-2)^2}$$ and s in that table he linked.

5. Jul 12, 2008

### HallsofIvy

Staff Emeritus
$$\frac{1}{(s-2)^3}$$
and
$$\frac{1}{(s-2)^2}$$
are given by number 2.11 in that table.

By partial fractions,
$$\frac{s}{(s-2)^2}= \frac{1}{s-2}+ \frac{2}{(s-2)^2}$$