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).

a.Let γ=12γ=12. Find the value of k for which the response has a peak value of 2; call this value k₁.

b.Repeat part (a) for γ=14γ=14.

Expert Solution

Step 1

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

To Find : $k_{1} (12)$

: $k_{2} (14)$

Step 2

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

The Given ODE is Linear , So we can use Laplace transform can be used to solve

$Y (s) = L \{y (t)\} = \int_{0}^{\infty} e^{- s t} y (t) d t$

Similarly first and Second derivatives transform as follows .

$L \{\frac{d y}{d t}\} = s Y (s) - y (0) L \{\frac{d^{2} y}{d t^{2}}\} = s^{2} Y (s) - s y (0) - y' (0)$

Take the Laplace transform of both sides of ODE

$L \{y'' + γ γ' + y\} = L \{k δ (t - 1)\}$

$L \{y''} + γ L {γ'} + L {y\} = L \{k δ (t - 1)\}$

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

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

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

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

Step 3

Suppose that $0 < γ < 2$

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

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 Values 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

Step 4

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

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

$t a n [\frac{\sqrt{4 - γ^{2}}}{2} (t - 1)] = \frac{\sqrt{4 - γ^{2}}}{γ}$

$t_{m a x} = 1 + \frac{2}{\sqrt{4 - γ^{2}}} \tan^{- 1} (\frac{\sqrt{4 - γ^{2}}}{γ})$

Now putting this value of t into y(t) to find the maximum value of y.

$y (t_{m a x}) = \frac{2 k}{\sqrt{4 - γ^{2}}} e x p [- \frac{γ}{\sqrt{4 - γ^{2}}} \tan^{- 1} (\frac{\sqrt{4 - γ^{2}}}{γ})] \sin \tan^{- 1} [\frac{\sqrt{4 - γ^{2}}}{γ}]$

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