Solve the equation.δ(t−t0) is the Dirac-Delta function.y"+4y′+5y=δ(t−2π)+δ(t−4π) , y(0)=0 , y′(0)=0

Step 1Consider the given IVP as follows.y"+4y′+5y=δ(t−2π)+δ(t−4π) , y(0)=0 , y′(0)=0Apply Laplace transform on both sides as follows.L{y"+4y′+5y}=L{δ(t−2π)+δ(t−4π)}L{y"}+4L{y′}+5L{y}=L{δ(t−2π)}+L{δ(t−4π)}s2L{y}−sy(0)−y′(0)+4sL{y}−4y(0)+5L{y}=e−2πs+e−4πss2L{y}+4sL{y}+5L{y}=e−2πs+e−4πsL{y}[s2+4s+5]=e−2πs+e−4πsL{y}=e−2πs+e−4πss2+4s+5L{y}=e−2πs(s+2)2+1+e−4πs(s+2)2+1Step 2Apply inverse Laplace transform to solve the given IVP as follows.L{y}=e−2πs(s+2)2+1+e−4πs(s+2)2+1y(t)=L−1{e−2πs(s+2)2+1+e−4πs(s+2)2+1}=L−1{e−2πs(s+2)2+1}+L−1{e−4πs(s+2)2+1}=u(t−2π)e−2(t−2π)sin⁡(t−2π)+u(t−4π)e−2(t−4π)sin⁡(t−4π)

Key to "Solve the equation.δ(t−t0) is the Dirac-Delta function.y"+4y′+5y=δ(t−2π)+δ(t−4π) ,"

Brittney Lord

Answered question

2020-12-05

Solve the equation.

δ (t - t_{0})

is the Dirac-Delta function.

y " + 4 y^{'} + 5 y = δ (t - 2 π) + δ (t - 4 π), y (0) = 0, y^{'} (0) = 0

Answer & Explanation

Dora

Skilled2020-12-06Added 98 answers

Step 1
Consider the given IVP as follows.

y " + 4 y^{'} + 5 y = δ (t - 2 π) + δ (t - 4 π), y (0) = 0, y^{'} (0) = 0

Apply Laplace transform on both sides as follows.

L {y " + 4 y^{'} + 5 y} = L {δ (t - 2 π) + δ (t - 4 π)}

L {y "} + 4 L {y^{'}} + 5 L {y} = L {δ (t - 2 π)} + L {δ (t - 4 π)}

s^{2} L {y} - s y (0) - y^{'} (0) + 4 s L {y} - 4 y (0) + 5 L {y} = e^{- 2 π s} + e^{- 4 π s}

s^{2} L {y} + 4 s L {y} + 5 L {y} = e^{- 2 π s} + e^{- 4 π s}

L {y} [s^{2} + 4 s + 5] = e^{- 2 π s} + e^{- 4 π s}

L {y} = \frac{e^{- 2 π s} + e^{- 4 π s}}{s^{2} + 4 s + 5}

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

Step 2

Apply inverse Laplace transform to solve the given IVP as follows.

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

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

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

= u (t - 2 π) e^{- 2 (t - 2 π)} \sin (t - 2 π) + u (t - 4 π) e^{- 2 (t - 4 π)} \sin (t - 4 π)

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