MATH 285 E1 F1 GRADED HOMEWORK SET 4 DUE FRIDAY OCTOBER 24 IN LECTURE This time the homework has just one part Please staple your homework together and put your name and section on it Thank you 1 10 points Find the general solution of the differential equation D 3 y xe2x e3x Hint Use the annihilator method An annihilator of e3x is D 3 An annihilator of xe2x is D 2 2 Thus an annihilator of their sum is A D D 2 2 D 3 Applying this to both sides of the equation we get D 2 2 D 3 D 3 y D 2 2 D 3 xe2x e3x D 2 2 D 3 2 y 0 This means that the function y we want to find must have the form y Ae2x Bxe2x Ce3x Exe3x We now plug this in to the original equation and see what coefficients are determined by that condition To to this systematically let s apply the operator D 3 to each of the terms D 3 e2x 2e2x 3e2x e2x D 3 xe2x 2xe2x e2x 3xe2x xe2x e2x D 3 e3x 3e3x 3e3x 0 D 3 xe3x 3xe3x e3x 3xe3x e3x Putting it all together we get D 3 y Ae2x Bxe2x Be2x Ee3x B A e2x Bxe2x Ee3x Since this should be equal to xe2x e3x we must have B A 0 B 1 and E 1 while C can be anything Thus the general solution is y e2x xe2x Ce3x xe3x 2 5 points Consider the forced mass spring oscillator with mass m 4 no damping c 0 and a spring constant k that is adjustable Consider a driving forces that is the sum of two cosine functions F t cos 2t cos 5t The differential equation for the displacement x t is then 4x00 kx cos 2t cos 5t 1 2 GRADED HOMEWORK 4 For what values of k is it impossible to find a particular solution of the form A cos 2t B cos 5t These are the values of k for which resonance occurs We try to find a solution of the form x t A cos 2t B cos 5t The second derivative is x00 t 4A cos 2t 25B cos 5t Thus 4x00 kx k 16 A cos 2t k 100 B cos 5t Since this is to be equal to cos 2t cos 5t We must have 1 1 B A k 16 k 100 At this step we have to assume that k 16 is not zero and that k 100 is not zero Conversely this shows us the values of k for which resonance occurs k 16 and k 100 For these values of k the constants A and B cannot be found that make A cos 2t B cos 5t a solution of the equation 3 5 points Recall from the lecture the damped c 0 forced oscillator obeying the differential equation mx00 cx0 kx F0 cos t In the lecture we derived a particular solution of this equation xp t C cos t where F0 C p k m 2 2 c 2 tan c k m 2 Here is something we did not talk about in the lecture the solution xp t is called the steady state solution because any solution will asymptotically converge to it as t More precisely if x1 t is any solution of the same nonhomogeneous equation then lim x1 t xp t 0 t Explain why this is true You should use the description of the solutions of the homogeneous equation mx00 cx0 kx 0 on page 191 of the textbook Hint It is very important that c 0 The first thing to say is that the difference x1 t xp t is a solution of the homogeneous equation mx00 cx0 kx 0 This means that it falls into one of the three cases underdamped critically damped or overdamped shown on page 191 We want to show that in all three cases the solutions of the homogeneous equation tend to zero as t goes to infinity GRADED HOMEWORK 4 3 Underdamped case The general solution is x t e pt A cos 1 t B sin 1 t The thing to recognize is that the factor e pt is going to zero This is because p is an abbreviation for c 2m which is positive and therefore limt e pt 0 Thus limt x t 0 as well Critically damed case The general solution is x t e pt c1 c2 t Again p c 2m is positive so the factor e pt is exponentially decaying However this time there is a factor of t in the second term that goes to infinty We just need to use L Hopital s rule to see that the product actually goes to zero t 1 lim te pt lim pt lim pt 0 t t e t pe Overdamped case The general solution is x t c1 er1 t c2 er2 t where r1 and r2 are the solutions of the characteristic equation mr2 cr k 0 That is c2 4km 2m We just need to see that both r1 and r2 are actually negative The quantity c c2 4km 2m is clearly negative since both terms in the numerator are negative The other root c c2 4km 2m will be negative as long as c c2 4km But this is clear c2 4km is less than c2 so c2 4km is less than c This completes the proof 4 5 points Use the orthogonality properties of sine and cosine to compute the following integral Z 5 sin 2t cos 3t cos 2t 5 sin 6t sin 2t dt r1 r2 c When we expand out the integrand we get products of sin nt and cos mt namely 5 sin 2t cos 2t 25 sin 2t sin 6t 5 sin 2t sin 2t cos 3t cos 2t 5 cos 3t sin 6t cos 3t sin 2t When this is integrated from to the orthogonality properties say that a term such as cos mt cos nt will integrate to zero unless m n that a term such as sin mt sin nt will integrate to zero unless 4 GRADED HOMEWORK 4 m n and that a term such as cos mt sin nt will always integrate to zero no matter what m and n are If we look at what we have the only term that gives a nonzero integral is 5 sin 2t sin 2t Thus the integral we were asked to compute simplifies to Z 5 sin 2t sin 2t dt 5
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