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

mathstresser

Junior Member
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Jan 28, 2006
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134
Solve the second-order problem

\(\displaystyle \L\\ y"\, -\, y'\, -\, 2y\, =\, e^{2t}\) (y-double-prime - y-prime...)

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


\(\displaystyle \L\\ Y(s)\, =\, \frac{1}{(s\, -\, 2)^{2}(s\, +\, 1)}\, - \, \frac{s\, +\, 1}{(s\, -\, 2)(s\, +\, 1)}\)

Then with \(\displaystyle \L\\ L^{-1}\) and partial fraction decomposition I get

\(\displaystyle \L\\ (1/3)(te^{2t})\, +\, (-7/3)e^{(2t)}\, +\, e^{t}\)

But the answer is

\(\displaystyle \L\\ (-5/9)e^{(-t)}\, -\, (4/9)e^{(2t)}\, +\, (1/3)te^{(3t)}\)

What am I doing wrong ?
 
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