 # 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} Globokim8 2020-10-28 Answered
Use the Laplace transform to solve the given initial-value problem.
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Step 1
The given differential equation is
${y}^{″}-6{y}^{\prime }+13y=0$
Step 2
Simplify the initial value problem using Laplace transform:
$L\left(y-6{y}^{\prime }+13y\right)=L\left(0\right)$
$L\left(y\right)-6L\left({y}^{\prime }\right)+13L\left(y\right)=0$
${s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)-6\left(sY\left(s\right)-y\left(0\right)\right)+13Y\left(s\right)=0$
${s}^{2}Y\left(s\right)+9-6sY\left(s\right)+13Y\left(s\right)=0$
$Y\left(s\right)=-\frac{9}{{s}^{2}-6s+9+4}$
Step 3
Simplify further,
$Y\left(s\right)=-\frac{9}{{\left(s-3\right)}^{2}+4}$
$Y\left(s\right)=-\frac{9}{{\left(s-3\right)}^{2}+{2}^{2}}$
Take inverse laplace on both sides,
${L}^{-1}\left(Y\left(s\right)\right)={L}^{-1}\left(-\frac{9}{{\left(s-3\right)}^{2}+{2}^{2}}\right)$
$y\left(t\right)=-\frac{9}{2}{L}^{-1}\left(\frac{2}{{\left(s-3\right)}^{2}+{2}^{2}}\right)$
$y\left(t\right)=-\frac{9}{2}{e}^{3t}\mathrm{sin}2t$
Hence , the solution of the initial value problem is $y\left(t\right)=-\frac{9}{2}{e}^{3t}\mathrm{sin}2t$