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# Laplace Transform: e to the power of -2t

DIFFEQ-7XB4JE

If $X(t)$ and $y(t)$ are the solutions to the initial value problem:

\begin{align} x'+2x-3y &= e^{-2t}, & x(0)=0 \\\ y'-2x+3y &=0, &y(0)=0 \end{align}

...their Laplace transforms are given by:

A

$X(s)=\cfrac{s+3}{(s+2)(s^{2}+5s+2)},Y(S)=\cfrac{2}{(s+2)(s^{2}+5s+2)}$

B

$X(s)=\cfrac{s-3}{(s+2)(s^{2}-5s+2)},Y(S)=\cfrac{-2}{(s+2)(s^{2}-5s+2)}$

C

$X(s)=\cfrac{s-2}{(s+2)(s^{2}-5s+2)},Y(S)=\cfrac{s+3}{(s+2)(s^{2}-5s+2)}$

D

None of the above