
#19
Feb411, 06:04 PM

P: 41

So I know that differentiating both sides of the sin(at) LT will give me the square that I need in the denominator, but then I'm stuck with the d/ds on the sin side. I'm not sure what to do after this.
Thanks 



#20
Feb411, 08:15 PM

HW Helper
P: 1,585

when you differentiate the integral you take in differentiation under the integral sign:
[tex] \frac{d}{ds}\int_{0}^{\infty}e^{st}\sin tdt=\int_{0}^{\infty}\frac{d}{ds}\left( e^{st}\sin t\right) dt [/tex] Differentiating the integrand: [tex] \frac{d}{ds}e^{st}\sin t=te^{st}\sin t [/tex] As now we regard s as variable and t is fixed. So the above can be written as: [tex] (t\sin t)e^{st} [/tex] and so you're taking the Laplace transform of something different now. That something is the answer to your question. Is this making sense now? 



#21
Feb511, 05:25 AM

P: 41

I think I got it now.
After differentiating both sides and dividing by 2: s/(s^2+a^2)^2=L{tsin at/2} Hence, L1{s/(s^2+1)^2}=tsint/2 However, how do we know that "a" isn't 1, since (1)^2=1. 



#22
Feb511, 12:57 PM

HW Helper
P: 1,585

well done, got there in the end.



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