Initial value ODE with shifting forcing function

  • #1
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0

Main Question or Discussion Point

Use laplace Transform to solve this ode:
upload_2016-7-21_21-49-59.png


So I got:

sV(s)-V(0)-12V(s)=U(s+5)
V(s)(s-12)=U(s+5)+1
V(s)=[U(s+5)+1]/(s-12)

Now to go back to time domain with Inverse Laplace Transform...My question is, how to transform U(s+5)/(s-12)?

Any help?

Thanks guys
 

Answers and Replies

  • #2
Twigg
Gold Member
196
35
I think it may help you to know that u(t) is the standard symbol for the step function, which has a known Laplace transform. Check your tables.
 
  • #3
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Thanks
 
  • #4
9
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@Twigg , can you shed some lights on Laplace Transform of e^(at)*u(t)?

Thanks
 
  • #5
Twigg
Gold Member
196
35
I'm pretty sure you just apply the shifting property to the Laplace transform of the Heaviside step function. The Laplace transform of ##u(t)## is ##\frac{1}{s}##, so the Laplace transform of ##e^{-5t} u(t)## is ##\frac{1}{s + 5}##. Just like you did in your first post.
 
  • #6
9
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Thanks
 

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