Answer to Question #295355 in Differential Equations for Divya

Question

Solve the following initial value problem using Laplace transform. y''+16y = 2 sin 4t, y(0)=-1/2; y'(0)=0 

Answer 1

$y''+16y=2\sin 4t, y(0)=-\frac{1}{2}, y'(0)=0\\$

Use Laplace transform

$L_t [y''+16y](p)=L_t[2\sin 4t](p)\\ L_t [y''](p)+L_t[16y](p)=2L_t[\sin 4t](p)\\ p^2L_t[y]-py(0)-y'(0)+16L_t[y]=2\cdot\frac{4}{p^2+16}\\ p^2L_t[y]-p(-\frac{1}{2})-0+16L_t[y]=2\cdot\frac{4}{p^2+16}\\ (p^2+16)L_t[y]=\frac{8}{p^2+16}-\frac{p}{2}\\ L_t[y]=\frac{8}{(p^2+16)^2}-\frac{p}{2(p^2+16)}\\ y=L_t^{-1}[\frac{8}{(p^2+16)^2}-\frac{p}{2(p^2+16)}](t)\\ y=L_t^{-1}[\frac{8}{(p^2+16)^2}](t)-L_t^{-1}[\frac{p}{2(p^2+16)}](t)\\ y=\frac{1}{16}(-4t\cos 4t+\sin 4t) -\frac{1}{2} \cos 4t$