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Ordinary and Partial Differential Equations

by: Elaina Osinski

Ordinary and Partial Differential Equations MATH 251

Marketplace > Pennsylvania State University > Mathematics (M) > MATH 251 > Ordinary and Partial Differential Equations
Elaina Osinski
Penn State
GPA 3.88


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This 0 page Class Notes was uploaded by Elaina Osinski on Sunday November 1, 2015. The Class Notes belongs to MATH 251 at Pennsylvania State University taught by Staff in Fall. Since its upload, it has received 11 views. For similar materials see /class/233005/math-251-pennsylvania-state-university in Mathematics (M) at Pennsylvania State University.

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Date Created: 11/01/15
Differential Equations LECTURE 11 Complex and Repeated Roots 1 Complex Roots Last time we saw that when the characteristic equation has complex roots r12 04 i i3 the general solution of ay 1 by 1 cy 0 is yt Clem cos t 3250quot sin t Let s do some examples EXAMPLE 111 Solve the VP y 4y 9y0 The characteristic equation is r2 7 47 9 0 which has roots r12 2 i Thus the general solution and its derivative are Mt Clezt COS5t 62621 sin5t Zt 20152 COS5t 7 V5016 sin5t 20252 sin5t V5025 cosh5t If we apply the initial conditions7 we get 0 Cl 72 261 V362 which is solved by 01 0 and 62 7 So the particular solution is 2 21 t fie sin t y EXAMPLE 112 Solve the VP y 7 83 17y 0 The characteristic equation is r2 7 ST 17 0 which has roots r12 4 i 2 Hence the general solution and its derivative are yt 3154 t cost 3254 sint yt 4015 cost 7 3154 sint 4025 sint 3254 cost and plugging in initial conditions yields the system 2 Cl 5 461 627 Differential Equations Lecture 11 Complex and Repeated Roots so we conclude 01 2 and 62 73 and the particular solution is yt 254 cost 7 354 sint EXAMPLE 113 Solve the VP 4y 1 121 10y 0 y0 71 yO 3 The characteristic equation is 4 12r 10 0 which has roots r1 7 i So the general solution and its derivative are t t yt 015 cos 325 sin t 3 gt cos t 1 gt sin t 3 gt sin 3 1 gt cos t C 5 i 7 C 5 i C 5 i C 5 i y 2 1 2 2 1 2 2 2 2 2 2 2 Plugging in the initial conditions yields 1 Cl 3 1 3 561 562 which has solution 01 71 and 02 9 The particular solution is t t yt 75 cos 1 gegtsin EXAMPLE 114 Solve the VP y 4y0 y 7104 4 The characteristic equation is r2 4 0 which has roots r12 i2i The general solution and its derivative are yt 01 cos2t 02 sin2t yt 7201 sin2t 202 cos2t The initial conditions gives the system 710 32 4 7261 so we conclude 01 72 and 62 710 and the particular solution is yt 72cos2t 7 10 sin2t Differential Equations Lecture 11 Complex and Repeated Roots 2 Repeated Roots The last case we have to consider is when the characteristic equation has a repeated root r1 r2 r This is problematic though our usual method of nding solutions to constant coef cient equations would lead us to form the two solutions 27105 e39rlt e39rt 27205 ergt art But these are the same and de nitely not different in any sense let alone different enough to form a general solution So we re left needing a second solution which is different in a way I promise I will make precise later from y1t equot What do we do Let s start by recalling that if the quadratic equation arz br c 0 has a repeated root r it must be r 7271 Thus our solution is more precisely y1t e t We know that any constant multiple of y1 is a solution These though won t help us nd a different enough second solution The question that might come to mind is if it s possible that we can nd a solution of the form L y2t vty1t U036 2amp5 239e yg is the product of some other function oft and y1 In the past when we ve guessed at the form of a solution we ve checked our guess and derived speci cs by plugging the guess into the differential equation This is no different So we ll need to differentiate 34225 m v lttgte 7 Wm 2 gig7 may 7 game t 7 imagi t tme t 7 fit b fit 52 it 71 25 2a 7amp1 te 2a mvte 2a For the rest of this calculation I m going to stop explicitly denoting the parameter of u but don t forget that v is a function not a constant Now plugging in 2 a U 67t 7 EU67t Lveit I U 57 7 iveit c veitgt 0 a 402 2a 2 2 e t 01 71 bv 7 7 S 311 0 7i 1 2 5 mt lt04 7 4a b 7 406 1 0 Since we re in the repeated root case we know that the discriminant 2 7 4m 0 As exponentials are never zero we re left with the condition 0w 0 a U 0 We can drop the a because a can t be zero if a were zero our equation wouldn t be second order So what form can u take It must be that v is linear Le Ut ct k for some constants c and k Thus for any such vt y2t vte t will be a solution The most general possible vt that will work for us is ct k We can take 0 1 and k 0 to get a speci c 1 which is nice and simple and then our second solution is it y2t 755 2a and the general solution is yt 615 t 2502577233 Differential Equations Lecture 11 Complex and Repeated Roots REMARK Here s another way of looking at this choice of constants Suppose we don t make it Then we have7 for our general solution7 b b yt 61575 62ct k57t it fit fit 015 2m Cgcte 2m Cgke 2a b b cl 62k67t Cgctei t Since 01 02 c and k are just constants7 we ll just roll them together and write 2 b yt 615 it Cgtei t To summarize the previous discussion if the characteristic equation has repeated roots r1 r2 r the general solution is yt 615 62756 Let s work some examples


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