Recent content by 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!

How will sin and cos go away?

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

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.

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

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?

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...

I think I understand now. Thank you so much!:smile:

Hi! I reallyreallyreally need ur help with this. ex.1 To find the vertical and horizontal components of the force mg, I need to use the given angle \beta. The problem is I don't know how to use it. I have tried to use the corresponding angles postulate. But I can't make it to work. How should...