<s>[INST] Laplace differential equation Can somebody help me work out $2y''+y'-y=\mathrm{e}^{3t}$, y(0)=2 and y'(0)=0 with the method of Laplace? I got
\begin{align*}
Y(s)&=\frac{1}{(s-3)(2s^2+s-1)}+ \frac{2+4s}{(s-3)(2s^2+s-1)}\\
&=-\frac{4}{15}\frac{1}{2s-1}+\frac{1}{12}\frac{1}{s+1}+\frac{1}{20}\frac{1}{s-3}-\frac{1}{6}\frac{1}{s+1}-\frac{16}{15}\frac{1}{2s-1}+\frac{7}{10}\frac{1}{s-3}\\
&=-\frac{2}{15}\mathrm{e}^{0.5t}+\frac{1}{12}\mathrm{e}^{-t}+\frac{1}{20}\mathrm{e}^{3t}-\frac{1}{6}\mathrm{e}^{-t}-\frac{8}{15}\mathrm{e}^{0.5t}+\frac{7}{10}\mathrm{e}^{3t}
\end{align*}
Is this correct?
 [/INST] Given:
$$\tag 1 2 y''+ y'- y = e^{3t},~ y(0)=2, ~ y'(0)=0$$
Taking the Laplace Transform of $(1)$, we have:
$$\mathcal{L}(2 y''+ y'- y) = 2(s^2y(s) - sy(0)-y'0) + (sy(s)-y(0)) - y(s) = \dfrac{1}{s-3}$$
Using the initial conditions from $(1)$, this reduces to:
$$y(s)(2s^2 + s - 1) -4s - 2 = \dfrac{1}{s-3}$$
Solving for $y(s)$ and doing partial fraction expansions/simplifications yields:
$$y(s) = \dfrac{1}{20(s-3)} + \dfrac{3}{4(s+1)} + \dfrac{12}{5(2s-1)}$$
Now we want to find $\mathcal{L^{-1}}(y(s))$, yielding:
$$y(t) = \dfrac{1}{20}e^{3t} + \dfrac{3}{4}e^{-t} + \dfrac{6}{5}e^{t/2}$$
Notes:


*

*You can always easily validate $y(t)$ by plugging it back in to the DEQ and making sure it satisfies it.

*You can solve this problems using other methods like undetermined coefficients or variation of parameters.

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