usagirl007A

2021-09-03

Solve equation using Laplace transform
$\stackrel{¨}{y}+9y=\delta \left(t-\frac{\pi }{6}\right)\mathrm{sin}\left(t\right),y\left(0\right)=0,\stackrel{˙}{y}\left(0\right)=0$

unessodopunsep

Step 1
Given
the given equation is
$\stackrel{¨}{y}+9y=\delta \left(t-\frac{\pi }{6}\right)\mathrm{sin}\left(t\right),y\left(0\right)=0,\stackrel{˙}{y}\left(0\right)=0$
Use the property of delta function.
$\stackrel{¨}{y}+9y=\delta \left(t-\frac{\pi }{6}\right)\mathrm{sin}\left(t\right),y\left(0\right)=0,\stackrel{˙}{y}\left(0\right)=0$
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\delta \left(t-{t}^{\prime }\right)\varphi \left(t\right)dt=\varphi \left({t}^{\prime }\right)$
Step 2
Calculation
$y"+9y=\delta \left(t-\frac{\pi }{6}\right)\mathrm{sin}\left(t\right),y\left(0\right)=0,\stackrel{˙}{y}\left(0\right)=0$
Take the Laplace in both sides.
$L\left[y\right]+9L\left[y\right]=L\left[\delta \left(t-\frac{\pi }{6}\right)\mathrm{sin}\left(t\right)\right]$
$\left[{s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)\right]+9Y\left(s\right)=L\left[\delta \left(t-\frac{\pi }{6}\right)\mathrm{sin}\left(t\right)\right]$
$L\left[\delta \left(t-\frac{\pi }{6}\right)\mathrm{sin}\left(t\right)\right]={\int }_{0}^{\mathrm{\infty }}\left[\delta \left(t-\frac{\pi }{6}\right){e}^{-st}\mathrm{sin}\left(t\right)\right]dt=\mathrm{sin}\left(\frac{\pi }{6}\right){e}^{-s\frac{\pi }{6}}=\frac{1}{2}{e}^{-\frac{\pi s}{6}}$
$⇒\left[{s}^{2}Y\left(s\right)-0-0\right]+9Y\left(s\right)=\frac{1}{2}{e}^{-\frac{\pi s}{6}}$
$⇒{s}^{2}Y\left(s\right)+9Y\left(s\right)=\frac{1}{2}{e}^{-\frac{\pi s}{6}}$
$⇒Y\left(s\right)\left({s}^{2}+9\right)=\frac{1}{2}{e}^{-\frac{\pi s}{6}}$

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