Laplacetransform an Integral, problem with the inverstransform

[Muffin]

Homework Statement

Hey! I have tried to solve this problem but I get stuck when it comes to the inverstransforming. Anyway here is the problem and my attempt to a solution:

Solve $$f(t)=2\int_{0}^{t}sin(9u)f'(t-u)du+sin9t,t\geq 0$$ for f(t)

The Attempt at a Solution

Laplacetransforming:

$$F(s)=\frac{18}{s^{2}+9^{2}}sF(s) + \frac{9}{s^{2}+9^{2}}$$

Solving F(s)
$$F(s)=\frac{9}{s^{2}+9^{2}-18s}$$

I don't know what to do now, with the inverstransform? And the F(s) feels wrong..Is my Laplacetransforming correct, I have tried to use the convolution theorem..but I don't know if I used it correct..

Thanks!

 

[HallsofIvy]

Did you notice that $s^2- 18s+ 9^2= (s- 9)^2$?

 

[Muffin]

Yes, But its something wrong bcs in the end f(t) should be sin9t.

And this answer will give me 9te^(9t).

Do you know what am I doing wrong?

 

[vela]

Your answer is correct; 9te^9t satisfies the original equation. Did you copy the problem correctly?

 

[Muffin]

Your answer is correct; 9te^9t satisfies the original equation. Did you copy the problem correctly?

Really?? :D hmm, yes but I tried the problem in wolframalpha.com and the answer was sin9t. I maybe did write the problem wrong in wolfram..

Thank you for the answer! :D

 

[Muffin]

Sorry but.. How does 9te^(9t) satisfies the answer.
Im trying to derivate it f'(t-u) and put that into the original equation. But I get a strange answer.

 

[vela]

Show us your work. When I do what you describe, it works out.

 

[Muffin]

Okey :)

$$f'(t-u)= 9e^{t-u}(9(t-u)+1)$$

$$f(t)=2\int_{0}^{t}(sin9u)(9e^{t-u}(9(t-u)+1)du + sin9t$$

$$f(t)=\frac{2}{9}(cos9t-1)+ 2(e^{9t}(t-\frac{1}{9})+\frac{1}{9})+sin9t$$

 

[vela]

You dropped the factor of 9 in the exponent.

 

[Muffin]

How will sin and cos go away?

 

[vela]

They end up canceling out if you integrate everything correctly. If you're just doing this to check the answer, why don't you just evaluate the integrals using Wolfram Alpha?

 

[Muffin]

No, I wanted to try it by my self. And now I did :)
Thank you so much for the help, I really appreciate it!

 

[vela]

Good work! I totally understand wanting to crank it out by hand. There's nothing better than practice to learn how to calculate quickly and how to avoid making mistakes.