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

Question

Solve the initial value problem below using the method of Laplace transforms.

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

y (0) = 0

y^{'} (0) = 15

Answer 1

Step 1
Given,

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

y (0) = 0

y^{'} (0) = 15

Step 2

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

Take Laplace transform of both sides of the equation

L {y^{″} - 16 y} = L {32 t - 8 e^{- 4 t}}

L {y^{″} - 16 y} : s^{2} L {y} - s y (0) - y^{'} (0) - 16 L {y}

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

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

Plug in the initial conditions:

y (0) = 0, y^{'} (0) = 15

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

Simplify

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

Step 3
Isolate

L {y} :

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

Take the inverse Laplace transform

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

y = - 2 t - 2 e^{- 4 t} + e^{- 4 t} t + 2 e^{4 t}

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