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

Solve the initial value problem below using the method of Laplace transforms.
$y"-16y=32t-8{e}^{-4t}$
$y\left(0\right)=0$
${y}^{\prime }\left(0\right)=15$
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Step 1
Given, $y"-16y=32t-8{e}^{-4t}$
$y\left(0\right)=0$
${y}^{\prime }\left(0\right)=15$
Step 2
$y"-16y=32t-8{e}^{-4t}$
Take Laplace transform of both sides of the equation
$L\left\{{y}^{″}-16y\right\}=L\left\{32t-8{e}^{-4t}\right\}$
$L\left\{{y}^{″}-16y\right\}:{s}^{2}L\left\{y\right\}-sy\left(0\right)-{y}^{\prime }\left(0\right)-16L\left\{y\right\}$
$L\left\{32t-8{e}^{-4t}\right\}:\frac{32}{{s}^{2}}-\frac{8}{\left(s+4\right)}$
${s}^{2}L\left\{y\right\}-sy\left(0\right)-{y}^{\prime }\left(0\right)-16L\left\{y\right\}=\frac{32}{{s}^{2}}-\frac{8}{\left(s+4\right)}$
Plug in the initial conditions: $y\left(0\right)=0, {y}^{\prime }\left(0\right)=15$
${s}^{2}L\left\{y\right\}-s\cdot 0-15-16L\left\{y\right\}=\frac{32}{{s}^{2}}-\frac{8}{\left(s+4\right)}$
Simplify
${s}^{2}L\left\{y\right\}-16L\left\{y\right\}-15=\frac{32}{{s}^{2}}-\frac{8}{\left(s+4\right)}$
Step 3
Isolate $L\left\{y\right\}:$
$L\left\{y\right\}=\left\{\frac{15{s}^{3}+52{s}^{2}+32s+128}{{s}^{2}\left(4+s\right)\left({s}^{2}-16\right)}\right\}$
Take the inverse Laplace transform
$y={L}^{-1}\left\{\frac{15{s}^{3}+52{s}^{2}+32s+128}{{s}^{2}\left(4+s\right)\left({s}^{2}-16\right)}\right\}$
$y=-2t-2{e}^{-4t}+{e}^{-4t}t+2{e}^{4t}$