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# Differential Equations MATH 33B

UCLA

GPA 3.55

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This 8 page Class Notes was uploaded by Kaylin Wehner on Friday September 4, 2015. The Class Notes belongs to MATH 33B at University of California - Los Angeles taught by Staff in Fall. Since its upload, it has received 153 views. For similar materials see /class/177836/math-33b-university-of-california-los-angeles in Mathematics (M) at University of California - Los Angeles.

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Date Created: 09/04/15

Math 33B Lecture 1 Midterm II Name UCLA lD Discussion session circle one 1A Faizal Sainal7 Tuesday 1B Faizal Sainal7 Thursday 10 Chris Vogl7 Tuesday 1D Chris Vogl7 Thursday Directions Fill in your name and circle your section above No outside materials are allowed Show all the necessary steps involved in nding your solutions PROBLEM SCORE TOTAL 1 25 pt Find the general solution of y 4y cos 2t t We rst solve y 4y cos 225 Since cos 2t solves the homogeneous equation 3 4y 07 one should put yp At cos 225 Bt sin 2t as the form of particular solution Alternatively one can try the complex method After carrying out the computation7 we obtain y it sin 225 Next we solve y 4y t to obtain y it Thus the particular solution of the original equation is 1 i 1 ypyzl7y Zts1n2t 1t Finally7 general solution of the given equation is y 01 cos 2t Cg sin2t itsin 2t it 2 12 pt a For what real values of k do some solutions of z 6m km 0 grow without bound as t a 00 This will be the case when the characteristic equation7 A2 6 k 0 has at least one positive real root By quadratic formula7 17397kgt273797k Hence the bigger root7 A1 is positive if and only if k lt O 13 pt b It is observed that a certain solution to z 6x km 0 satis es Mt 0 for two values oft differing by 7r2 and nowhere in between It may also vanish for other values of 25 What is k This is the case when the characteristic equation has complex roots7 which is when k gt 9 In this case 73i xk 7 9i and the general solution is of the form 0153 cos bt 0253 sin bt 053 cosbt 7 b k 7 9 To satisfy the description in b7 b 2 Hence xk 7 9 27 yielding k 13 3 A 1 kg mass is attached to a spring having the spring constant 9kg52 The system is displaced x3m upward from its equilibrium position and released with downward velocity of 3m s 12 pt a Suppose there is no damping Find the frequency and phase of the resulting motion The equation is my Iuy kg y 9y 0 with initial condition y0 7 3 and y 0 3 remember that y is the downward displacement from the equilibrium position If we solve the equation7 we get y 7V cos 3tsin 3t 13 pt b Suppose u 6 How many times does the resulting motion cross the equilibrium position If we solve the corresponding equation 3 63 9y 0 with above initial conditions7 we obtain y i 317 2553 above function is strictly negative for all 257 hence the resulting motion never crosses the equilibrium position Math 33B Lecture 2 Midterm II Solutions 1 Set 3 C1 sin 2x C2 cos 2x and plug into the DE to obtain C1 713 and C2 i a Thus yp 713sin2w 7 c0s2w 723sin2w 7r3 23sin2w 47r3 0 To obtain 6 for example with A 723 we look for 6 E 027r such that sinO g and cosO This gives you 6 7r3 Note that solving for tan give you two values of 6 7r347r3 and thus this is not enough to decide on 6 2 Two homogeneous solutions are 31 x and 32 15402 set 3 yiui y2u2 to obtain amt1 w ZuZ 07 714 Qw BuZ em I 36x and 71 76 It follows that 22 3 Hence one obtains y 0146 025372 7 mew 1 72 3 zd Math 33B Lecture 2 Midterm I Solutions 1 Integration factor v t2 7 3 Equation becomes d 3t 7 t2 3 dtW 3 f 73 Thus the general solution is 31 t2 3 4 Plug in y 1t 4 and one gets C 712V 13 2 25 pts a Since ODE above must be exact the following equation holds 8M 7 3534 7 31517153 1 a 7 8y So7 752m after integrating we have 00590 c where c is a constant Using initial condition 1 M0 0050 c gt c 0 Hence M 00553 2 b Let us denote by fwy the solution of the ODE7 then 7 00553 3 34 7 ysdnw 7 iysz nw gw 1me equation 13 we have that fwy fcosdy gw ycosw And equation 14 implies that g w 34 iysinwysmw 34 gt gw 3 Hence7 A 4 5 fvy ycosw ac gt A 01 V 5 0 ycosw E 6 a k 1 since outgoing amount of mixed water is k 7 2t and 120 k72gtlt1200 b dAt 727 2At dt 7 120439

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