HW7_Solution - EEL3105 Fall2011 Homework7Solution 1 d2y dy dy k 16 y 0 y(0 1(0 0 2 dt dt dt

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EEL 3105 Fall 2011 Homework 7 Solution 1. Consider the 2 nd order differential equation 2 2 16 0, (0) 1, (0) 0 d y dy dy ky y dt dt dt  Find solution y(t) for the following values of k: k=12, k=6, k=5.6, k=2, k=0, k= 1. Choose a suitable time interval and plot these solutions. You can use MATLAB or some other computational math tool for plotting y(t). Comment on your answers. Answer: First, let us calculate the roots of the polynomial corresponding to the differential equation: 2 16 k  The roots are 2 64 2 kk  There are four distinct cases: k >8 in which case we have two distinct real roots, k = 8 in which case we have two roots at 4 and 8 < k < 8 in which case we have a pair of complex conjugate roots and k < 8 when we again have two real roots. We will show solution for three values of k: k=12, k = 6, and k = 1 as these exhibit different behaviors. a. k=12: the roots are 1.52 and 10.48. The general form of solution is 1.52 10.48 () tt y t Ae Be   We need to find coefficients A and B. Applying initial conditions gives us: (0) 1 (0) 1.52 10.48 0 yA B dy AB dt  We can solve for A and B to get A = 1.171 B = 0.171 >> t=0:.1:2*pi; >> a=1.171*exp( 1.52*t) 0.171*exp( 10.48*t); >> plot(t,a)
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b. k=6: the roots are 3+j2.65 and 3 j2.65. There are two ways to proceed here. We can use either
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This note was uploaded on 12/27/2011 for the course EEL 3105 taught by Professor Boykins during the Fall '10 term at University of Florida.

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HW7_Solution - EEL3105 Fall2011 Homework7Solution 1 d2y dy dy k 16 y 0 y(0 1(0 0 2 dt dt dt

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