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

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