Answered: Consider the initial value problem…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Consider the initial value problem

y′′+γy′+y=kδ(t−1),y(0)=0,y′(0)=0

where k is the magnitude of an impulse at t = 1, and γ is the damping coefficient (or resistance).

Determine how k₁ varies as γ decreases. What is the value of k₁ when γ = 0?

Expert Solution

Step 1

Given : $y'' + γ y' + y = k δ (t - 1), y (0) = 0, y' (0) = 0$

To Find : $k_{1} (γ = 0)$

Step 2

$y'' + γ y' + y = k δ (t - 1) . . . (1)$

Take the Laplace transform of both side of (1)

$L \{y'' + γ y' + y\} = L \{k δ (t - 1)\} L \{y''} + γ L {y'} + L {y\} = L \{k δ (t - 1)\} [s^{2} Y (s) - s y (0) - y' (0)] + γ [s Y (s) - y (0)] + Y (s) = k \int_{0}^{\infty} e^{- s t} δ (t - 1) d t$

Since $y (0) = 0, y' (0) = 0$

$[s^{2} Y (s)] + γ [s Y (s)] + Y (s) = k e^{- s (1)} Y (s) = \frac{k e^{- s}}{[s^{2} + γ s + 1]} Y (s) = k \frac{1}{[(s^{2} + γ s + \frac{γ^{2}}{4}) + (1 - \frac{γ^{2}}{4})]} e^{- s} = k \frac{1}{[{(s^{2} + \frac{γ}{2})}^{2} + (\frac{4 - γ^{2}}{4})]} e^{- s} Y (s) = \frac{2 k}{\sqrt{4 - γ^{2}}} \frac{\frac{\sqrt{4 - γ^{2}}}{2}}{[{(s^{2} + \frac{γ}{2})}^{2} + (\frac{4 - γ^{2}}{4})]} e^{- s}$

Step 3

Take the inverse Laplace transform to get $y (t)$

$y (t) = \frac{2 k}{\sqrt{4 - γ^{2}}} e^{\frac{- γ (t - 1)}{2}} \sin [\frac{\sqrt{4 - γ^{2}}}{2} (t - 1)] H (t - 1)$

For Value of t>1 the Heaviside function is 1

$y (t) = \frac{2 k}{\sqrt{4 - γ^{2}}} e^{\frac{- γ (t - 1)}{2}} \sin [\frac{\sqrt{4 - γ^{2}}}{2} (t - 1)]$ , t>1

Take derivative and set it equal to zero to find the value of t for which y(t) is maximum

$y' (t) = \frac{γ k}{\sqrt{4 - γ^{2}}} e^{\frac{- γ (t - 1)}{2}} \sin [\frac{\sqrt{4 - γ^{2}}}{2} (t - 1)] + k e^{\frac{- γ (t - 1)}{2}} c o s [\frac{\sqrt{4 - γ^{2}}}{2} (t - 1)] = 0 \tan [\frac{\sqrt{4 - γ^{2}}}{2} (t - 1)] = \frac{\sqrt{4 - γ^{2}}}{γ}$

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