ankarskogC
2021-02-21
Answered

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

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

$y(0)=0$

${y}^{\prime}(0)=15$

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SoosteethicU

Answered 2021-02-22
Author has **102** answers

Step 1

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

$y(0)=0$

${y}^{\prime}(0)=15$

Step 2

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

Take Laplace transform of both sides of the equation

$L\{{y}^{\u2033}-16y\}=L\{32t-8{e}^{-4t}\}$

$L\{{y}^{\u2033}-16y\}:{s}^{2}L\left\{y\right\}-sy(0)-{y}^{\prime}(0)-16L\left\{y\right\}$

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

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

Plug in the initial conditions:$y(0)=0,\hspace{0.17em}{y}^{\prime}(0)=15$

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

Simplify

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

Step 3

Isolate$L\left\{y\right\}:$

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

Take the inverse Laplace transform

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

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

Given,

Step 2

Take Laplace transform of both sides of the equation

Plug in the initial conditions:

Simplify

Step 3

Isolate

Take the inverse Laplace transform

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