Question

Use the Laplace transform to solve the given initial-value problem.

y' − y = 2 cos(4t),  y(0) = 0

y(t) =

Answer 1

Let us use the Laplace transform to solve the given initial-value problem:

$y' − y = 2 \cos(4t),\ y(0) = 0$.

Taking into account that $y(t)\to Y(p),\ y'(t)\to pY(p)-y(0), \cos 4t\to\frac{p}{p^2+16}$, we have the equation:

$pY(p)-Y(p)=\frac{2p}{p^2+16}$ or $Y(p)(p-1)=\frac{2p}{p^2+16}$ or $Y(p)=\frac{2p}{(p-1)(p^2+16)}$.

It follows that $\frac{2p}{(p-1)(p^2+16)}=\frac{A}{p-1}+\frac{Bp+C}{p^2+16}$ and hence

$2p=Ap^2+16A+Bp^2+Cp-Bp-C$

Therefore, we have the following system:

$\begin{cases} A+B=0\\ C-B=2\\ 16A-C=0 \end{cases}$

$\begin{cases} A+C=2\\ B=C-2\\ 16A-C=0 \end{cases}$

$\begin{cases} 17A=2\\ B=C-2\\ C=16A \end{cases}$

$\begin{cases} A=\frac{2}{17}\\ B=C-2\\ C=\frac{32}{17} \end{cases}$

$\begin{cases} A=\frac{2}{17}\\ B=-\frac{2}{17}\\ C=\frac{32}{17} \end{cases}$

We conclude that $Y(p)=\frac{2}{17}\frac{1}{p-1}-\frac{2}{17}\frac{p}{p^2+16}+\frac{32}{17}\frac{1}{p^2+16}$, and using $\frac{1}{p-1}\to e^t,\ \frac{4}{p^2+16}\to\sin 4t$ we have the solution:

$y(t)=\frac{2}{17}e^t-\frac{2}{17}\cos 4t+\frac{8}{17}\sin 4t.$