# Use Laplace transforms to solve the following initial value problem. x"+x=3 cos 3t , x(0)=1 , x'(0)=0

Use Laplace transforms to solve the following initial value problem.
${x}^{″}+x=3\mathrm{cos}3t,x\left(0\right)=1,{x}^{\prime }\left(0\right)=0$
The solution is $x\left(t\right)=?$

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Step 1
The given differential equation is
$x{}^{″}+x=3\mathrm{cos}\left(3t\right)$
Initial condition: $x\left(0\right)=1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{x}^{\prime }\left(0\right)=0$
Solve the differential equation by using laplace transform.
step 2
Consider
$x{}^{″}+x=3\mathrm{cos}\left(3t\right)$
Operate laplace operator on the both sides, then
$L\left\{x"\left(t\right)\right\}+L\left\{x\left(t\right)\right\}=L\left\{3\mathrm{cos}\left(3t\right)\right\}$
${s}^{2}x\left(s\right)-s{x}^{\prime }\left(0\right)-{x}^{\prime }\left(0\right)+x\left(s\right)=\frac{3s}{{s}^{2}+9}$
${s}^{2}x\left(s\right)+x\left(s\right)-1=\frac{3s}{{s}^{2}+9}$
$\left({s}^{2}+1\right)x\left(s\right)=\frac{3s}{{s}^{2}+9}+1$
$x\left(s\right)=\frac{{s}^{2}+3s+9}{\left({s}^{2}+9\right)\left({s}^{2}+1\right)}$
Step 3
By using partial fraction method,
$x\left(s\right)=\frac{{s}^{2}+3s+9}{\left({s}^{2}+9\right)\left({s}^{2}+1\right)}$
$=\frac{3}{8}\left(\frac{s}{\left({s}^{2}+9\right)}\right)+\frac{3}{8}\left(\frac{s}{\left({s}^{2}+1\right)}\right)+\frac{1}{\left({s}^{2}+1\right)}$
Operate laplace inverse on the both sides,then
${L}^{-1}\left\{x\left(s\right)\right\}=\frac{3}{8}{L}^{-1}\left(\frac{s}{\left({s}^{2}+9\right)}\right)+\frac{3}{8}{L}^{-1}\left(\frac{s}{\left({s}^{2}+1\right)}\right)+{L}^{-1}\left(\frac{1}{\left({s}^{2}+1\right)}\right)$
$=\frac{3}{8}\mathrm{cos}\left(3t\right)+\frac{3}{8}\mathrm{cos}\left(t\right)+\mathrm{sin}\left(t\right)$
$⇒x\left(t\right)=\frac{3}{8}\mathrm{cos}\left(3t\right)+\frac{3}{8}\mathrm{cos}\left(t\right)+\mathrm{sin}\left(t\right)$
Hence,the required solution is