Question

To solve the initial value problem using the method of Laplace transforms:
y''+16y=8cos4t; y(0)=0; y'(0)=8

Answer 1

$y''+16y'=8\cos4t\\ y(0)=0, y'(0)=8\\$

Laplace transforms is

$y(t)\mapsto Y\\ y''(t)\mapsto p^2Y-py(0)-y'(0)=\\ =p^2Y-p•0-8=p^2Y-8 \\ \cos4t\mapsto\frac{p}{p^2+16}$

then

$p^2Y-8+16Y=\frac{8p}{p^2+16}\\ (p^2+16)Y=\frac{8p}{p^2+16}+8\\ Y=\frac{8p}{(p^2+16)^2}+\frac{8}{p^2+16}$

Then

$\frac{8p}{(p^2+16)^2}=\\ =\frac{2p•4}{(p^2+4^2)^2}\mapsto t\sin 4t\\$

$\frac{8}{p^2+16}=\frac{2•4}{p^2+4^2}\mapsto 2\sin 4t$

Solution is

$y=t\sin4t+2\sin4t\\ y=(t+2)\sin4t$