# Use the Laplace transform to solve the given initial-value problem. displaystyle{y}{''}+{8}{y}'+{41}{y}=delta{left({t}-piright)}+delta{left({t}-{3}piright)}, {y}{left({0}right)}={1}, {y}'{left({0}right)}={0} {y}{left({t}right)}=?

Use the Laplace transform to solve the given initial-value problem.
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Step 1
Given initial value problem,
$y{}^{″}+8{y}^{\prime }+41y=\delta \left(t-\pi \right)+\delta \left(t-3\pi \right)$,
$y\left(0\right)=1$,
${y}^{\prime }\left(0\right)=0$
Solve this problem using Laplace transform method.
Step 2
Now
$y{}^{″}+8{y}^{\prime }+41y=\delta \left(t-\pi \right)+\delta \left(t-3\pi \right)$
$L\left[y{}^{″}+8{y}^{\prime }+41y\right]=L\left[\delta \left(t-\pi \right)+\delta \left(t-3\pi \right)\right]$
$L\left[y{}^{″}\right]+8L\left[{y}^{\prime }\right]+41L\left[y\right]=L\left[\delta \left(t-\pi \right)\right]+L\left[\delta \left(t-3\pi \right)\right]$
Use the formula such that
$L\left[y{}^{″}\right]={s}^{2}L\left[y\right]-sy\left(0\right)-{y}^{\prime }\left(0\right)$
$L\left[{y}^{\prime }\right]=sL\left[y\right]-y\left(0\right)$
$L\left[\delta \left(t-a\right)\right]={e}^{-as}$
Step 3
Then,
${s}^{2}L\left[y\right]-sy\left(0\right)-{y}^{\prime }\left(0\right)+8\left[sL\left[y\right]-y\left(0\right)\right]+41L\left[y\right]={e}^{-\pi s}+{e}^{-3\pi s}$
${s}^{2}L\left[y\right]-s\cdot 0-0+8\left[sL\left[y\right]-0\right]+41L\left[y\right]={e}^{-\pi s}+{e}^{-3\pi s}$
${s}^{2}L\left[y\right]+8sL\left[y\right]+41L\left[y\right]={e}^{-\pi s}+{e}^{-3\pi s}$
$\left[{s}^{2}+8s+41\right]L\left[y\right]={e}^{-\pi s}+{e}^{-3\pi s}$
$L\left[y\right]=\frac{{e}^{-\pi s}+{e}^{-3\pi s}}{{s}^{2}+8s+41}$
Taking inverse Laplace transform of both sides,
$y={L}^{-1}\left[\frac{{e}^{-\pi s}+{e}^{-3\pi s}}{{s}^{2}+8s+41}\right]$