# Laplace Transform of this Differential Equation

1. Jan 20, 2012

### s3a

I tried doing this problem soooo many times for several days and keep failing and then I'm so worn out I can't keep thinking straight and I can't move on because of this! I'm attaching my work and it would be great if I can get back and forth feedback if necessary to find my mistake(s).

The question is:
"Solve the initial value problem: y'' + 18y' + 90y = -6e^(-9t) cos(2t); y(0) = 3, y'(0) = -1"

Any help would be GREATLY appreciated!

#### Attached Files:

• ###### P5.jpg
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2. Jan 20, 2012

### tylerc1991

Do you have to use the Laplace transform? What about the method of undetermined coefficients?

3. Jan 21, 2012

### s3a

I do have to use the method of Laplace Transforms.

4. Jan 21, 2012

### vela

Staff Emeritus
You seem to have made a mistake right before you started in on the partial fractions. You should get
$$Y(s) = \frac{3(s+9)}{(s+9)^2+9} + \frac{26}{3}\frac{3}{(s+9)^2+9} -\frac{6(s+9)}{[(s+9)^2+4][(s+9)^2+9]}$$When you deal with that last term, I'd pull the overall factor of -6 out and replace s+9 with s to make the algebra simpler. In other words, use partial fractions on
$$Y_3(s) = \frac{s}{(s^2+4)(s^2+9)}$$I think you'll find that much more pleasant to deal with. Then take into account the effect of replacing s by s+9 and multiply by the overall factor of -6.

Last edited: Jan 21, 2012
5. Jan 24, 2012

### s3a

Sorry for the late response. I have sooooo much work to do otherwise, I'd respond immediately. I'm attaching my latest failure.

The answer is supposed to be:
21/5 cos(3t) e^(-9t) - 6/5 e^(-9t) cos(2t) + 26/3 e^(-9t) sin(3t)

and based on the last step I did, I already have mistakes.

#### Attached Files:

• ###### laplace_redone.pdf
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40
6. Jan 24, 2012

### vela

Staff Emeritus
When you collected the Y(s) terms on the lefthand side, you should have gotten $Y(s)(s^2+18s+90) - 3s - 53$. You accidentally included the 3s term.

Last edited: Jan 24, 2012
7. Jan 25, 2012

### s3a

Yes! I did it! Thank you!!!!!!