# Differential Eq: Inverse Laplace Transforms

1. May 10, 2010

### Gogeta007

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

I have a couple of problems that im stuck on. The following:

y'' - 6y' + 9y = t
y(0) = 0
y'(0) = 1

and

y'' - 6y' + 13y = 0
y(0) = 0
y'(0) = -3

2. Relevant equations

y'' = s2Y(s) - sf(0) - f'(0)
y' = sY(s) - f(0)
y = Y(s)

3. The attempt at a solution

===================================
for the first one:

once I substitute into the original equation, I can move things around and I came out with the following:
Y(s){s2-6s+9} = (1 +s2)/s2

So I get Y(s) = (1+s2)/s2(s-3)2

IIRC for partial fractions it should be the following: A/(s-3) + B/(s-3)2 + (Cs+D)/s2

I dont know if this is where i messed up but I got:
A=-2/27
B=-1/9
C=2/27
D=1/9

as a final result I get:
-2/27e^3t - 1/92e^3t and the other part (Cs + D)/s^2 . . .I cant find any way to transform that

the back of the book says:
1/9t + 2/27 - 2/27e^3t + 10/9te^3t

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The second one:

Starting by substitution, plugging in the values and solving for Y(s)

Y(s) = -3/(s2-6s + 13)

and well. . Im lost from this point onwards. . .I dont remember how to do partial fractions if you cant factorize that denominator. And the quadratic formula (A=1 B = 6 C=13) gives imaginary numbers

=============================

Thank you!

Last edited: May 10, 2010
2. May 10, 2010

### vela

Staff Emeritus
You just have to separate it into individual terms.

$$\frac{Cs+D}{s^2} = \frac{C}{s} + \frac{D}{s^2}$$

Also, according to Mathematica, you should have B=10/9. The other coefficients you found are correct.

Last edited: May 10, 2010
3. May 10, 2010

### vela

Staff Emeritus
Complete the square in the bottom so you get something that looks like

$$Y(s) = -\frac{3}{(s-a)^2+b^2}$$

You should be able to find the inverse of that using the tables and the properties of Laplace transforms.

4. May 10, 2010

### Gogeta007

wow. . .I cant believe I missed that. . .how does really old math comes back and haunt you huh?. . .thanks a lot