Use the Laplace transform to solve the given initial-value problem. displaystyle{y}{''}-{6}{y}'+{13}{y}={0}, {y}{left({0}right)}={0}, {y}'{left({0}right)}=-{9}

Step 1
The given differential equation is
$y^{″}-6y^{\prime }+13y=0$
Step 2
Simplify the initial value problem using Laplace transform:
${\displaystyle L(y-6y^{\prime }+13y)=L\left(0\right)}$
${\displaystyle L\left(y\right)-6L\left(y^{\prime }\right)+13L\left(y\right)=0}$
${\displaystyle s^{2}Y\left(s\right)-sy\left(0\right)-y^{\prime }\left(0\right)-6(sY\left(s\right)-y\left(0\right))+13Y\left(s\right)=0}$
${\displaystyle s^{2}Y\left(s\right)+9-6sY\left(s\right)+13Y\left(s\right)=0}$
${\displaystyle Y\left(s\right)=-\frac{9}{s^2-6s+9+4}}$
Step 3
Simplify further,
${\displaystyle Y\left(s\right)=-\frac{9}{{(s-3)}^2+4}}$
${\displaystyle Y\left(s\right)=-\frac{9}{{(s-3)}^2+2^2}}$
Take inverse laplace on both sides,
${\displaystyle L^{-1}\left(Y\left(s\right)\right)=L^{-1}(-\frac{9}{{(s-3)}^2+2^2})}$
${\displaystyle y\left(t\right)=-\frac{9}{2}{L}^{-1}\left(\frac{2}{{(s-3)}^2+2^2}\right)}$
${\displaystyle y\left(t\right)=-\frac{9}{2}{e}^{3t}\sin 2t}$
Hence , the solution of the initial value problem is ${\displaystyle y\left(t\right)=-\frac{9}{2}{e}^{3t}\sin 2t}$