Non-homogeneous laplace

  • Thread starter reece
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  • #1
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Hi, I can solve homogeneous laplace fine, but with RHS i get stuck half way through.

Q:
y' +3y = 8e[tex]^{t}[/tex]
y(0) = 2

Working as if it was homogeneous..

sY(s) - 2 + 3Y(s) = 8 . [tex]\frac{1}{s-1}[/tex]
Y(s) (s+3) - 2 = 8 . [tex]\frac{1}{s-1}[/tex]

I think the next step is
Y(s) = [tex]\frac{2}{s+3}[/tex] + [tex]\frac{8}{s-1}[/tex]
and then do partial fractions but i dont think it leads me to where I need to be. I think i need to make it into a heaviside ??

Any help would be great. thanks
 

Answers and Replies

  • #2
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Hello reece,
It took me a few minutes to figure out what was going on. Your method is OK, except that you should rewrite the final line as:
[tex](s+3)Y(s)=\frac{8}{s-1}+2=2\cdot \frac{s+3}{s-1}[/tex]
From which:
[tex]Y(s)=\frac{2}{s-1}[/tex]
And thus the final solution is:
[tex]y(t)=2e^t[/tex]

I expected two exponentials and therefore it is interesting to solve it without using Laplace. The solution is then
[tex]y(t)=A e^{-3t}+2e^t[/tex]
After applying the boundary condition you get A=0.

It is a strange equation because the exponential is unbounded for large t.
 

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