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Calculus II

by: Braeden Lind

Calculus II MA 241

Braeden Lind
GPA 3.93


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Class Notes
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This 5 page Class Notes was uploaded by Braeden Lind on Thursday October 15, 2015. The Class Notes belongs to MA 241 at North Carolina State University taught by Staff in Fall. Since its upload, it has received 27 views. For similar materials see /class/223725/ma-241-north-carolina-state-university in Mathematics (M) at North Carolina State University.

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Date Created: 10/15/15
SecondOrder Linear Equations Example 1 section 78 pg 1172 problem 10 Solve the initial value problem using the method of undetermined coef cients 2 y 7 2y 96 sin 2967 110 17 40 0 Solution 1 First we solve the corresponding homogeneous equation y y 72y 0 The characteristic equation is r2 r 7 2 7quot 7 1r 2 0 with roots 7 1 72 So the general solution of the homogeneous equation is ycz Clem Cze h 2 Second we look for a particular solution of the given di erential equation Since G z sin 2x we seek a particular solution of the form ypx AzBCcos2zDsin2z Then A 7 2C sin 2x 2D cos 2x y 740 cos 2x 7 4D sin 2x So substituting into the given di erential equation we have 740 cos 2x74D sin 2xA72C sin 2x2D cos 2x72 Ax B 0 cos 2x D sin 2x sin 2x or 72Az A 7 2B 740 2D 7 20 cos 2x 74D 7 2C 7 2D sin 2x z sin 2x or 72Az A 7 2B 760 2D cos 2x 76D 7 2C sin 2x z sin 2x This is true if 72A1 A72B0 76C2D0 76D7201 The solution of the system is 1 1 1 3 A77 B77 077 D77 2 4 20 20 A particular solution is therefore i 1 1 1 2 3 2 yp m 7 2x 4 20cos m 20sin m 3 By the superposition principle the general solution is l l l 3 ycz ypz Clem 025 7 am 7 Z 7 cos 2x 7 sin 2x The nal step is to nd 01 and Cg such that the general solution satis es the initial conditions y0 1y 0 0 lmposing the initial condition y0 17 we get 1 1 y001027171 or 13 0102 E To impose the other initial condition we rst di erentiate the solution 1 2 6 y z 016m 7 2026 2m 7 E 7 sin 2x 7 E cos 2x 20 So 1 6 ymyqi27 7 6o or 4 01 7 202 g The solution of the system 13 4 0102i 017202g 10 is 3 7 07 07 15 2 m Therefore 3m 7 1 1 1 3 yz7e 76 77x7777cos2z77s1n2x 5 10 2 4 20 20 Example 2 section 78 pg 1172 problem 26 Solve the initial value problem using the method of variation of parameters 721 MM 3 1 First7 we solve the corresponding homogeneous equation y 4y 4y 0 The charac teristic equation is r2 47quot 4 7quot 22 0 with roots r12 72 So the general solution of the homogeneous equation is yew Cle h C e zm 2 Second7 we look for a particular solution of the given di erential equation Using variation of parameters7 we seek a solution of the form ypx 111xe 2m u x e zm Then uQLe Zm u zxe zw 0 uQLe Zm u zze h 0 6 O r 72m u e zm u 2e 2 113726 25 u 2e 2m 7 2x64 6 3 Solving the above systern7 we get 7 1 7 1 ul uz 7 E or 1 1 11195 E 295 Therefore 1 1 1 7 72m 72m 7 72m 7 7 72m 7 72m ypx i u1e u2ze x6 2 e 2ze 3 By the superposition principle7 the general solution is l MW pr Cleizm ngeizm Ee zm 93 Examp 85 Lenyjk 010 wow 2mm 515 243 5 d l Q X L H a A p x Q X K 6 p af al 39PWX 05 71 5 A 3 I Q39Z fl 1leH 393 S 4169 anqlaL1u3t1qm 3 491A 1 n3 0K 3ZQ 3 Ltadff39k O X z 43 Curved a ram 39KCaZ 39 P R gut 7Cy OgiS 9 42291 521 it At 13 b L mmgu J 1 f3 J3 O 3 V act t9 0H n 74 Hi 0 aha 939115 4 40 HM Lt Woks L1 1 43 iJE Mm u 3 155 Lf K 4k 51 u u3kll1 2 39 3 A 3 quots w ampE


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