Given, \(y"-16y=32t-8e^{-4t}\)

\(y(0)=0\)

\(y'(0)=15\)

Step 2

\(y"-16y=32t-8e^{-4t}\)

Take Laplace transform of both sides of the equation

\(L\left\{y''-16y\right\}=L\left\{32t-8e^{-4t}\right\}\)

\(L\left\{y''-16y\right\} : s^2L\left\{y\right\}-sy(0)-y'(0)-16L\left\{y\right\}\)

\(L\left\{32t-8e^{-4t}\right\} : \frac{32}{s^2}-\frac{8}{(s+4)}\)

\(s^2L\left\{y\right\}-sy(0)-y'(0)-16L\left\{y\right\}=\frac{32}{s^2}-\frac{8}{(s+4)}\)

Plug in the initial conditions: \(y(0)=0, y'(0)=15\)

\(s^2L\left\{y\right\}-s \cdot 0-15-16L\left\{y\right\}=\frac{32}{s^2}-\frac{8}{(s+4)}\)

Simplify

\(s^2L\left\{y\right\}-16L\left\{y\right\}-15=\frac{32}{s^2}-\frac{8}{(s+4)}\)

Step 3

Isolate \(L\left\{y\right\}:\)

\(L\left\{y\right\}=\left\{\frac{15s^3+52s^2+32s+128}{s^2(4+s)(s^2-16)}\right\}\)

Take the inverse Laplace transform

\(y=L^{-1}\left\{\frac{15s^3+52s^2+32s+128}{s^2(4+s)(s^2-16)}\right\}\)

\(y=-2t-2e^{-4t}+e^{-4t}t+2e^{4t}\)