Initial value ODE with shifting forcing function

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SUMMARY

The discussion focuses on solving an initial value ordinary differential equation (ODE) using the Laplace Transform. The equation presented is sV(s) - V(0) - 12V(s) = U(s+5), leading to V(s) = [U(s+5) + 1] / (s - 12). Participants clarify the application of the shifting property in Laplace Transforms, specifically how to transform U(s+5)/(s-12) and the Laplace Transform of e^(at) * u(t), where u(t) represents the Heaviside step function.

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  • Understanding of Laplace Transforms
  • Familiarity with ordinary differential equations (ODEs)
  • Knowledge of the Heaviside step function
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  • Study the properties of the Laplace Transform, focusing on the shifting property
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Mathematicians, engineers, and students studying differential equations, particularly those interested in the application of Laplace Transforms in solving initial value problems.

Houeto
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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
 
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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.
 
Thanks
 
@Twigg , can you shed some lights on Laplace Transform of e^(at)*u(t)?

Thanks
 
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.
 
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