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# Review for Exam Sample 3363

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This 5 page Study Guide was uploaded by Shahmeer Baweja on Tuesday September 2, 2014. The Study Guide belongs to 3363 at University of Houston taught by a professor in Fall. Since its upload, it has received 341 views. For similar materials see Intro to PDE in Math at University of Houston.

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Date Created: 09/02/14

MATH 3321 Sample Questions for Exam 2 Second Order Nonhomogeneous Differential Equations Section 34 35 1 z1cc 2cc2 tan cc z2cc cc2 2cc tan cc z3cc cc2 3cc tan cc are solutions of a second order linear nonhomogeneous equation Ly f a Give a fundamental set of solutions of the corresponding reduced equation Ly 0 b Give the general solution of the nonhomogeneous equation Ly f Answer a y1cc cc2 y2cc cc b y C1cc2 C2cc tan cc 3 are solutions of a second order linear nonhomo 2 is a solution of the corresponding reduced 2 z1cc 2cc3 ccln cc z2cc ccln cc cc geneous equation Ly y1cc cc equation Ly 0 a Give a fundamental set of solutions of the reduced equation Ly 0 b Give the general solution of the nonhomogeneous equation Ly f Answer a y1cc cc 2 y2cc cc3 b y C1cc 2 C2cc3 cc ln cc 4 6 3 Find the general solution of yquot E y E y Answer y C1 cc2 C2 cc3 4 Find the general solution of y 4y 2 tan 2cc Answer y C1 cos 2cc C2 sin 2cc cos 2cc ln sec 2cc tan 2cc 39c 5 Find the general solution of y 6 y 9 y 4 e3 6 Answer y C1 63 C2 cc 63 2cc2e3 quot cc e3 ln cc 6 Find the general solution of y 93 4 cos 2cc Answer y C1 cos 3cc C2 sin 3cc g cos 2cc 7 Find the general solution of y 43 2 sin 2cc Answer y C1 cos 2cc C2 sin 2cc cc cos 2cc 8 Find the general solution of y 6 y 8 y 2 6439 6 Answer y C1 64 C2 62 cc 6495 g 9 A particular solution of the nonhomogeneous differential equation yquot 2y 153 2 cos 3c5e5 2 will have the form Answer 2 A cos 356 B sin 356 Cve5 D 10 A particular solution of the nonhomogeneous differential equation 3 8y39 163 62 sin 453 2e4 I556 will have the form Answer A e2 cos 46 B e295 sin 43 C 32 64 D53 E Higher Order Linear Equations Section 37 1 The general solution of y 4 y y 6 y 0 is Hint 7quot 2 is a root of the characteristic equation Answer y C162 C263 C3e 2 The general solution of y y 8 y 12 y 0 is Hint 7quot 3 is a root of the characteristic equation Answer y C163 C264 C3ce 2 3 The general solution of y4 2 y 4y 2 y 5 y 0 is Hint 7quot 1 273 is a root of the characteristic equation Answer y C1e quot cos 2513 C2e sin 256 C36 C4e quot 4 The homogeneous equation with constant coefficients that has y C1e2 C2ce2 C3 cos 256 C4 sin 256 C5 as its general solution is Answer y5 4y4 83 163 163 0 5 The homogeneous equation with constant coefficients of least order that has y 2e3 3sin 2c2c as a solution is Answer y5 334 4y 123 0 6 A particular solution of y 2y 33 26 5663 2 will have the form Answer 2 Ace quot B532 C39ce3 D56 7 A particular solution of y4 16 y 2e 2 364 cos 2513 5 will have the form Answer 2 Ave 2 B649 056 cos 2513 Dzc sin 256 E 8 The general solution of y4 53 363 2 cos 3513 35662 will have the form Answer y C1 cos 3cC392 sin 3vC393e2 C394e 2 quotAc cos 3cBc sin 3cC39c2Dce2 9 The general solution of y y y y 5 sin cc 2e e quot 456 will have the form Answer y C1 cos cc C2 cos cc C393e Ax cos cc Bzc sin cc C39e Dce E53 F Laplace Transformations Chapter 4 1 Find the Laplace transform of fc 2 e 3 cos 2513 5513 2 3 5 Answer Fs 83 82 4 p 2 Find the Laplace transform of f 356 62 6 sin 3513 3 3 9 22 s2 2s10 Answer Fs 2 3 3 8 2 then C1Fs is Answer fx 256 cos 256 sin 256 1 32 s 32 s2 4s 13 4 If Fs then C 1Fs is Answer fc 5c 63 e2 cos 3513 e2 sin 3513 10 11 Find the Laplace transform of the solution of the initial Value problem y 23 3 cos 256 y0 3 9 38 6 A Y nswer 3 48 2 482 4 Find the Laplace transform of the solution of the initial Value problem yquot 5y39 6y4sin 356 y00 30 2 12 2 s2 9s2 5s 6 32 5s6 Answer Fs Find the Laplace transform of the solution of the initial Value problem yquot 25y 263 2 23 3 3s2 25 32 25 Answer Fs Use the Laplace transform method to nd the solution of the initial Value problem 621 y 3y2 y01 Answer 3 3e3 262 Use the Laplace transform method to nd the solution of the initial Value problem y394y 3 cos 2513 y0 3 12 Answer 3 e 4 quot cos 256 110 sin 253 Use the Laplace transform method to nd the solution of the initial Value problem 2 32 2y 26 1 20 2 20 1 Answer y 26 2627 5c 2 Find the Values of 7 such that the solution of the initial Value problem y 4y sin 6 y0 Y 20 0 is bounded on 0 oo 1 Answer 7 E 12 13 14 15 16 17 18 Find the Value of 6 such that the solution of the initia1 Va1ue problem y 32 2639 210 5 has limit 0 as cc gt 00 Answer 6 4 38 2 If Fs 82 829 the C 1Fs 1s Answer fcc 4cc 3 cos 3cc sin 3cc 23 3 If then P 1S2 Answer fcc 193 e3 216 sin 2cc 193 cos 2cc If cc2 0 3 cc lt 3 fx 2 cc 2 3 then Cfcc 2 3 1 3 1 3 1 Answer Fs 83 2e 583 6 582 7 5 If 2 0 3 cc lt 2 fcc cc 2 3 cc lt 5 3 cc 2 5 then Cfcc 2 1 1 1 1 Answer Fs e25 4e25 e55 2 e55 3 32 3 32 3 If Fs 3 1 gig 2e358L2 46358 391 2 then C 1Fs is 2 2512 0 3 cc lt 3 Answer x 8 2cc 2cc2 4e 2 3 cc 2 3 33 3 2e 7T5 If Fs 829 then C 1Fs fcc is 3 cos 3cc 0 3 cc lt 7r Answen x 2 cos 3cc sin 3cc cc 2 7r

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