Brittney Lord

2020-12-05

Solve the equation.$\delta \left(t-{t}_{0}\right)$ is the Dirac-Delta function.

Dora

Step 1
Consider the given IVP as follows.

$\text{Apply Laplace transform on both sides as follows.}$
$L\left\{y"+4{y}^{\prime }+5y\right\}=L\left\{\delta \left(t-2\pi \right)+\delta \left(t-4\pi \right)\right\}$
$L\left\{y"\right\}+4L\left\{{y}^{\prime }\right\}+5L\left\{y\right\}=L\left\{\delta \left(t-2\pi \right)\right\}+L\left\{\delta \left(t-4\pi \right)\right\}$
${s}^{2}L\left\{y\right\}-sy\left(0\right)-{y}^{\prime }\left(0\right)+4sL\left\{y\right\}-4y\left(0\right)+5L\left\{y\right\}={e}^{-2\pi s}+{e}^{-4\pi s}$
${s}^{2}L\left\{y\right\}+4sL\left\{y\right\}+5L\left\{y\right\}={e}^{-2\pi s}+{e}^{-4\pi s}$
$L\left\{y\right\}\left[{s}^{2}+4s+5\right]={e}^{-2\pi s}+{e}^{-4\pi s}$
$L\left\{y\right\}=\frac{{e}^{-2\pi s}+{e}^{-4\pi s}}{{s}^{2}+4s+5}$
$L\left\{y\right\}=\frac{{e}^{-2\pi s}}{\left(s+2{\right)}^{2}+1}+\frac{{e}^{-4\pi s}}{\left(s+2{\right)}^{2}+1}$
$\text{Step 2}$
$\text{Apply inverse Laplace transform to solve the given IVP as follows.}$
$L\left\{y\right\}=\frac{{e}^{-2\pi s}}{\left(s+2{\right)}^{2}+1}+\frac{{e}^{-4\pi s}}{\left(s+2{\right)}^{2}+1}$
$y\left(t\right)={L}^{-1}\left\{\frac{{e}^{-2\pi s}}{\left(s+2{\right)}^{2}+1}+\frac{{e}^{-4\pi s}}{\left(s+2{\right)}^{2}+1}\right\}$
$={L}^{-1}\left\{\frac{{e}^{-2\pi s}}{\left(s+2{\right)}^{2}+1}\right\}+{L}^{-1}\left\{\frac{{e}^{-4\pi s}}{\left(s+2{\right)}^{2}+1}\right\}$
$=u\left(t-2\pi \right){e}^{-2\left(t-2\pi \right)}\mathrm{sin}\left(t-2\pi \right)+u\left(t-4\pi \right){e}^{-2\left(t-4\pi \right)}\mathrm{sin}\left(t-4\pi \right)$

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