inverse Laplace transform: y''-y = 2t, y(0) = 0, y'(0) = -1

mathstresser

Junior Member
Joined
Jan 28, 2006
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134
Use the Laplace transform to solve the second-order initial value problem.

\(\displaystyle \L\\ y''-y = 2t\)

\(\displaystyle \L\\ y(0)= 0, y'(0)= -1\)

I use:

\(\displaystyle \L\\ L(y')(s)= sY(s)-y(0)\)

and

\(\displaystyle \L\\ L(y'')(s)= s^2Y(s)-sy(0)-y'(0)\)

I get

\(\displaystyle \L\\ Y(s)= {2}/{(s^2)(s^2\, -\, s)}\)

I factored it to be

\(\displaystyle \L\\ {2}/{(s^3)(s\, -\, 1)}\)

So, then I do partial fraction decomposition and get

\(\displaystyle \L\\ {2}/{(s^3)(s\, -\, 1)} = {A}/{s^3} + {B}/{s^2} + {C}/{s} + {D}/{(s-1)}\)

Now… is where I get lost. (That is assuming I did all of that right…)
 
Please check my calculations, but:

\(\displaystyle \L\\y''-y=2t, \;\ y(0)=0, \;\ y'(0)=1\)

\(\displaystyle \L\\p^{2}Y-py(0)-y'(0)=\frac{2}{p^{2}}\)


\(\displaystyle \L\\\frac{-1}{2(p+1)}+\frac{1}{2(p-1)}-\frac{2}{p^{2}}\)

\(\displaystyle \L\\y=\frac{e^{t}}{2}-\frac{1}{2}e^{-t}-2t\)

I put this back into the equation and it all checks out.
 
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