"Solve equation using Laplace transform y¨+9y=δ(t−π6)sin⁡(t),y(0)=0,y˙(0)=0" was answered by our experts

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usagirl007A

Answered question

2021-09-03

Solve equation using Laplace transform
$\ddot{y} + 9 y = δ (t - \frac{π}{6}) \sin (t), y (0) = 0, \dot{y} (0) = 0$

Answer & Explanation

unessodopunsep

Skilled2021-09-04Added 105 answers

Step 1
Given
the given equation is
$\ddot{y} + 9 y = δ (t - \frac{π}{6}) \sin (t), y (0) = 0, \dot{y} (0) = 0$
Use the property of delta function.
$\ddot{y} + 9 y = δ (t - \frac{π}{6}) \sin (t), y (0) = 0, \dot{y} (0) = 0$
$\int_{- \infty}^{\infty} δ (t - t^{'}) ϕ (t) d t = ϕ (t^{'})$
Step 2
Calculation
$y " + 9 y = δ (t - \frac{π}{6}) \sin (t), y (0) = 0, \dot{y} (0) = 0$
Take the Laplace in both sides.
$L [y] + 9 L [y] = L [δ (t - \frac{π}{6}) \sin (t)]$
$[s^{2} Y (s) - s y (0) - y^{'} (0)] + 9 Y (s) = L [δ (t - \frac{π}{6}) \sin (t)]$
$L [δ (t - \frac{π}{6}) \sin (t)] = \int_{0}^{\infty} [δ (t - \frac{π}{6}) e^{- s t} \sin (t)] d t = \sin (\frac{π}{6}) e^{- s \frac{π}{6}} = \frac{1}{2} e^{- \frac{π s}{6}}$
$\Rightarrow [s^{2} Y (s) - 0 - 0] + 9 Y (s) = \frac{1}{2} e^{- \frac{π s}{6}}$
$\Rightarrow s^{2} Y (s) + 9 Y (s) = \frac{1}{2} e^{- \frac{π s}{6}}$
$\Rightarrow Y (s) (s^{2} + 9) = \frac{1}{2} e^{- \frac{π s}{6}}$

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