(Inverse) Laplace Transforms

  • Thread starter Knissp
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Homework Statement


[tex]y'' + 2y' + 2y = t^2 + 4t [/tex]
[tex]y(0) = 0[/tex]
[tex]y'(0) = -1[/tex]

Homework Equations


[tex]L(y) = Y [/tex]
[tex]L(y') = sY - y(0) [/tex]
[tex]L(y'') = s^2 Y - s y(0) - y'(0) [/tex]
[tex]L(t^n) = \frac {n!}{s^{n+1}}[/tex]
[tex]L(a*b) = AB[/tex]
[tex]a*b = L^{-1}(AB)[/tex]

The Attempt at a Solution



Take the Laplace transform of both sides.

[tex]L(y'' + 2y' + 2y) = L(t^2 + 4t) [/tex]

By linearity:

[tex]L(y'') + 2L(y') + 2L(y) = L(t^2) + 4L(t) [/tex]

Substitute:

[tex](s^2 Y - s y(0) - y'(0)) + 2(sY - y(0)) + 2Y = \frac{2}{s^3} + \frac{4}{s^2} [/tex]

Plug in initial conditions:

[tex](s^2 Y + 1) + 2(sY) + 2Y = \frac{2}{s^3} + \frac{4}{s^2} [/tex]

Solve for Y:

[tex] Y (s^2 + 2s + 2) -1 = \frac{2}{s^3} + \frac{4}{s^2} [/tex]

[tex] Y = \frac {\frac{2}{s^3} + \frac{4}{s^2} + 1}{s^2 + 2s + 2} [/tex]

Now I must take the inverse Laplace transform. This is where I get confused.

Rewrite:


[tex] Y = \frac{1}{s^2 + 2s + 2} (\frac{2}{s^3} + \frac{4}{s^2} + 1)[/tex]

Distribute:

[tex] Y = \frac{1}{s^2 + 2s + 2} \frac{2}{s^3} + \frac{1}{s^2 + 2s + 2} \frac{4}{s^2} + \frac{1}{s^2 + 2s + 2} [/tex]

Completing the square:

[tex] Y = \frac{1}{(s+1)^2 + 1} \frac{2}{s^3} + \frac{1}{(s+1)^2 + 1} \frac{4}{s^2} + \frac{1}{(s+1)^2 + 1} [/tex]

Now to take the inverse Laplace transform of both sides:

[tex] y = L^{-1}(\frac{1}{(s+1)^2 + 1} \frac{2}{s^3}) + L^{-1}(\frac{1}{(s+1)^2 + 1} \frac{4}{s^2}) + L^{-1}(\frac{1}{(s+1)^2 + 1}) [/tex]

[tex] y = L^{-1}(\frac{1}{(s+1)^2 + 1} \frac{2}{s^3}) + L^{-1}(\frac{1}{(s+1)^2 + 1} \frac{4}{s^2}) + e^{-t}sin(t) [/tex]

So now I need to do the separate inverse transforms (via convolutions):

[tex]L^{-1}(\frac{1}{(s+1)^2 + 1} \frac{2}{s^3})[/tex]

[tex]= e^{-t}sin(t) * t^2[/tex]

[tex]= \int_0^t (e^{-(t-v)}sin(t-v))(v^2)) dv [/tex]

[tex]L^{-1}(\frac{1}{(s+1)^2 + 1} \frac{4}{s^2})[/tex

[tex]= e^{-t}sin(t) * 4t [/tex]

[tex]= \int_0^t (e^{-(t-v)}sin(t-v))(4v) dv [/tex]

I don't know how to figure those convolution integrals. Any ideas?
Alternatively, is there a better way to approach the problem?
 

Answers and Replies

  • #2
1,341
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Perhaps you might have better luck using partial fractions, instead of completing the square... I don't know for sure since I didn't do it, but it's a suggestion you might want to look into.
 

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