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hw4Sol1 - HOMEWORK 4 MATH 265 Due in class on Oct 20...

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HOMEWORK 4: MATH 265 Due in class on Oct 20 PARTIAL SOLUTIONS Problem 1: (a) Solve the nonhomogeneous ODE y ′′ +16 y = cos( ωt ) with y (0) = 0, y (0) = 0 and express your solution in terms of the frequency of the forcing term ω > 0. Sketch the solution when ω = 3 and when ω = 4. (b) NEW (Can be added at end of your HW pages, if you have already done other problems) In Part (a), you may have gotten a solution in the form of a sum or difference of two trig functions such as sines or cosines. Use a trigonometric identity to re-express that solution as a product of two trig functions (such as K sin( w 1 t ) sin( w 2 t ) - but you must find the values of K, w 1 , w 2 in terms of other quantities in the problem). (c) Based on part (b) explain the behaviour of the solution when ω is close to but not equal to 4. That is, explain the phenomenon of beats . (You may find it useful to read about this in your book.) (d) Use your favorite graphics software (or calculator) to plot the solution in two cases, ω = 3 . 5 and ω = 3 . 8. What is the period of the envelope of the oscillations in each case? What is the frequency of the oscillations within that envelope in each case? Solution to Problem 1: First solve the homogeneous problem. Let y = e rt . Then, the characteristic polynomial is r 2 + 16 = 0, so that r = ± 4 i . Thus, y = sin(4 t ) and y = cos(4 t ) are solutions to the homogeneous problem. Find the particular solution when ω 0 negationslash = 4. The particular solution is easily found to be y p = cos( ω 0 t ) 16 ω 2 0 . The general solution, when ω 0 negationslash = 4, is y = c 1 cos(4 t ) + c 2 sin(4 t ) + cos( ω 0 t ) 16 ω 2 0 . Satisfying the initial conditions y (0) = y (0) = 0, we get c 1 = 0, c 2 = 1 / (16 ω 2 0 ). Thus, the solution to the IVP is y ( t ) = 1 (16 ω 2 0 ) [cos( ω 0 t ) cos(4 t )] , when ω 0 negationslash = 4 . (2) When ω 0 = 4 we have resonance and y p = (
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