# How to solve for the equation y'''+4y''+5y'+2y = 4x+16 using laplace transform method given that y(0) = 0, y'(0) = 0, text{and } y''(0) = 0

How to solve for the equation
${y}^{‴}+4{y}^{″}+5{y}^{\prime }+2y=4x+16$
using laplace transform method given that
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Step 1
We use identities of Laplace transform then we take Inverse Laplace Transform.
Step 2
Given ODE is,
${y}^{‴}+4{y}^{″}+5{y}^{\prime }+2y=4x+16$
We know that $L\left[{y}^{‴}\right]={s}^{3}\cdot y\left(s\right)-{s}^{2}y\left(0\right)-s{y}^{\prime }\left(0\right)-y"\left(0\right)$
$L\left[y"\right]={s}^{2}y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)$
$L\left[{y}^{\prime }\right]=sy\left(s\right)-y\left(0\right),L\left[y\right]=y\left(s\right)$
$L\left[x\right]=\frac{1}{{s}^{2}},L\left[1\right]=\frac{1}{s}$
Given conditions all $y\left(0\right)=0,{y}^{\prime }\left(0\right)=0,y"\left(0\right)=0$
Taking Laplace tranform of ${y}^{‴}+4{y}^{″}+5{y}^{\prime }+2y=4x+16$ from
$L\left[{y}^{‴}\right]+4L\left[y"\right]+5L\left[{y}^{\prime }\right]+2L\left[y\right]=4L\left[x\right]+16L\left[1\right]$
${s}^{3}y\left(s\right)-{s}^{2}y\left(0\right)-s{y}^{\prime }\left(0\right)-y"\left(0\right)+4\left[{s}^{2}y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)\right]+5\left[s\left(y\left(s\right)\right)-y\left(0\right)\right]+2y\left(s\right)=\frac{4}{{s}^{2}}+\frac{16}{s}$
$\left({s}^{3}+4{s}^{2}+5s+2\right)y\left(s\right)=4\left(\frac{1+4s}{{s}^{2}}\right)$
$\left(s+1\right)\left(s+1\right)\left(s+2\right)y\left(s\right)=4\left[\frac{4s+1}{{s}^{2}}\right]$
$⇒y\left(s\right)=\frac{4\left[4s+1\right]}{{s}^{2}\left(s+1{\right)}^{2}\left(s+2\right)}$
Taking inverse Laplace transform$y\left(t\right)=4{L}^{-1}\left\{\frac{4s+1}{{s}^{2}\left(s+1{\right)}^{2}\left(s+2\right)}\right\}$
$=3H\left(t\right)+2t+4{e}^{-t}-{e}^{-t}\cdot 12t-7{e}^{-2t}$
$y\left(t\right)=3H\left(t\right)+2t+\left(4-12t\right){e}^{-t}-7{e}^{-2t}$
This is required solution