Solve y′+y=4δ(t−2π)Subject toy(0)=0y'(0)=0

Step 1According to the given information, it is required to solve the given differential equation.y′+y=4δ(t−2π)Subject to y(0)=0 , y'(0)=0Step 2Use laplace transform to solve the above equation.y′+y=δ(t−2π)L(y′)+L(y)=4L(δ(t−2π))sY(s)−y(0)+Y(s)=4[e−2πs](L(δ(t−c))=e−cs)(s+1)Y(s)=4[e−2πs]Y(s)=4[e−2πs]s+1y(t)=4L−1([e−2πs]s+1)y(t)=4[u2π(t)e−(t−2π)](L(uc(t)f(t−c))=e−csF(s))

Step by step solution to "Solve y′+y=4δ(t−2π)Subject toy(0)=0y'(0)=0"

Caelan

Answered question

2021-09-03

Solve

y^{'} + y = 4 δ (t - 2 π)

Subject to
y(0)=0
y'(0)=0

Answer & Explanation

timbalemX

Skilled2021-09-04Added 108 answers

Step 1
According to the given information, it is required to solve the given differential equation.

y^{'} + y = 4 δ (t - 2 π)

Subject to y(0)=0 , y'(0)=0
Step 2
Use laplace transform to solve the above equation.

y^{'} + y = δ (t - 2 π)

L (y^{'}) + L (y) = 4 L (δ (t - 2 π))

s Y (s) - y (0) + Y (s) = 4 [e^{- 2 π s}] (L (δ (t - c)) = e^{- c s})

(s + 1) Y (s) = 4 [e^{- 2 π s}]

Y (s) = \frac{4 [e^{- 2 π s}]}{s + 1}

y (t) = 4 L^{- 1} (\frac{[e^{- 2 π s}]}{s + 1})

y (t) = 4 [u_{2 π} (t) e^{- (t - 2 π)}] (L (u_{c} (t) f (t - c)) = e^{- c s} F (s))

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