Question

Solve the initial value problem below using the method of Laplace transforms.
\(y"-16y=32t-8e^{-4t}\)
\(y(0)=0\)
\(y'(0)=15\)

Answer 1

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