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

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

2021-02-22
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
$$y=L^{-1}\left\{\frac{15s^3+52s^2+32s+128}{s^2(4+s)(s^2-16)}\right\}$$
$$y=-2t-2e^{-4t}+e^{-4t}t+2e^{4t}$$

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