## anonymous 5 years ago how would I do the laplace transform of sin(2t)e^(5t)?

1. anonymous

L(sin(2t)e^5t)=2/((s-5)^2+2^2) $2 \over (s-5)^2+4$

2. anonymous

would you show me the steps, because I got the answer from a website, but I can't actually get there on my own with integration by parts. I keep on ending up with sin and cos on the top.

3. anonymous

I see.. I actually just apply it to the formula, you can find it in any laplace table.. I will try the integration now

4. anonymous

thanks

5. anonymous

any luck?

6. anonymous

I guess.. just gimme a minute

7. anonymous

thank you so much for your help

8. anonymous

you need to do integration by parts twice for the integration $I=\int\limits_{0}^{\infty}e ^{(5-s)t}\sin(2t)dt$

9. anonymous

$\int\limits_{}^{}udv=uv-\int\limits_{}^{}vdu$ u=e^(5-s)t , dv=sin(2t) ... I think you know how to proceed. you will get another integration with cos this time, do integration by parts once more with the substitution for u and proceed.

10. anonymous

then you can substitute the new result in the previous one to get the final integration.

11. anonymous

I got all of that, but I still end up with wonky sin and cos stuff on the top, even when using the indicator function with integration by parts twice.

12. anonymous

you're right.. the final step is the trick, if I may say, you will notice that the integral you got at last is the same one you have in the beginning, right?

13. anonymous

let me write, directly the result of the two integrations by parts: (1) $\int\limits_{0}^{\infty}e ^{(5-s)t}\sin(2t)dt=-1/2e ^{(5-t)s}\cos(2t)-1/2\int\limits_{0}^{\infty}e ^{(5-s)t}\cos(2t)dt$

14. anonymous

yep, that's what I got as well

15. anonymous

I think that I also got the next part, but when I do the algebra for the last step and get I on its own again, I end up with those wonky sin and cos bits in the numerator.

16. anonymous

now integrate the new integral: (2) $\int\limits_{0}^{\infty}e ^{(5-s)t}\cos(2t)dt=1/2e ^{(5-s)t}\sin(2t)+\int\limits_{0}^{\infty}e ^{(5-s)t}\sin(2t)dt$

17. anonymous

now just substitute (2) in (1) to, and take the part with integration in one side..

18. anonymous

try it out and see if it works with you

19. anonymous

are you talking about the substitution in the end of infinity and zero?