Laplace Transform of this Differential Equation

[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!
Thanks in advance!

 

[tylerc1991]

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!
Thanks in advance!

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

 

[s3a]

I do have to use the method of Laplace Transforms.

 

[vela]

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.

 

[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.

 

[vela]

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.

 

[s3a]

Yes! I did it! Thank you!