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# Theory of Ordinary Differential Equations I MATH 5330

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This 8 page Class Notes was uploaded by Ms. Ally Koelpin on Thursday October 22, 2015. The Class Notes belongs to MATH 5330 at Texas Tech University taught by Staff in Fall. Since its upload, it has received 27 views. For similar materials see /class/226468/math-5330-texas-tech-university in Mathematics (M) at Texas Tech University.

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Date Created: 10/22/15

The Method of Undetermined Coe icients and the Shifting Rule Math 5330 Spring 1996 In these notes we will show how to use operator polynomials and the shifting rule to nd a particular solution for a linear constant coefficient differential equation This method works for the same class of forcing functions right hand sides as the method discussed in the book This method is computationally more ef cient than the method in the book It does not require you to solve the homogeneous equation rst To use the method you have to write the equation in operator form We use D to stand for the differen tiation operator so Dy y DQy DDy y and so forth Thus we can write the equation y 33 53 sinx in the form DQy 3Dy 53 sinxi We can rewrite this in the form D2 3D 5 sinx or just PDy sinx where PD is the differential operator PD D2 3D 5 This operator can be thought of as a polynomial in the differential operator D In fact it can be thought of as the result of substituting D for in the polynomial P A2 3 5 which is of course the characteristic polynomial of the equation With this notation we can state the Shifting Rule as follows Let PD be a polynomial in D and let oz be a real or complex number Then for any sufficiently differentiable function y we ewPDy PD 7 aewy Note that if we replace oz by 7a we get the alternate form e wPDy PD ae wy of the shifting rule which is often helpful We ll give some examples of how to apply the shifting rule to nd a particular solution in these notes Polynomial Right Hand Sides We rst consider the case of an equation of the form P D 1196 where 1195 is a polynomial of degree 71 If PD has a nonzero constant term we look for a solution which is a polynomial of the same degree as 1 Example D2 2D ly 952 1 We look for a solution which is a polynomial of degree 2 Thus we try yAx2BxCi 1 For this to be a solution we must have z21D22D1y D22 2Dy y 2A 22Ax B A952 Bx C 2A4AxZBAx2BxC Ax24ABx2A2BC Equating coefficients of the powers of x on both sides we get A 1 4A B 0 2A 2B C 1 The solution of this system of equations is clearly A1 374 07 and so our particular solution is y x2 7 4x 7 The other case to consider is where PD has no constant termi In this case we take out as many factors of D as we can which reduces the problem to solving a lower order equation and then integratingi The procedure is shown in the next example 2 Example D2 3Dy 952 We rewrite the equation as D 3Dy 952 Temporarily let 2 Dyi Then 2 satis es D 32 952 We can t pull out any more factors of D so we look for a solution which is a polynomial of the same degree as the right hand side Thus we try 2 A952 Bar Cl Plugging this into the equation for 2 yields x2D232 2AxB3Ax2BxC 2AxB3Ax2SBx3C 314952 2ASBx B30 This yields the system of equations 3A1 2A3B0 B3C0 which have the solution 1 2 2 A 37 627 Thus we get 1 2 2 2 Z x 7 W 27 Now recall that 2 Dy Thus 1 2 2 Dy x2 x27 we can take the constant of integration to be zero because we re looking for a particular solution iiei just one function that satis es the equation Right Hand Sides of the form 11950quot Suppose we have an equation of the form PDy we where 1195 is a polynomial To solve this equation we rewrite it as emPltDgty am By the shifting rule this is the same as PD a e wyl 1196 If we let PD a and 2 e o xy then 2 satis es the equation QltDgtz gm Since the right hand side is a polynomial we know how to solve this equation Once we ve found 2 we can solve 2 e o xy for y name y y eo xzi 3 Example D2 2D 1y 95521 We write the equation in the form 1 PDy 95521 where 2 PD D2 2D 1 We rewrite equation 1 in the form 5 21PDy 95 By the shifting rule this is equivalent to PD 2e 2y x 3 Temporarily let 2 672131 50 PD 22 95 Now we calculate PD 2 by substituting D 2 for D in equation This gives PD2 D222D21 D24D42D41 D2 6D 9 Thus 2 satis es the equation 3 D2 6D 92 as For 2 we try a polynomial of the same degree as the right hand side so we try 2AxBi Plugging this into 3 yields xD226D292 06A9AxB 6A9Az93 This yields the equations 9A1 6AQB0 and so we have the solution 2 This gives us 1 2 2 7x 7 7 9 27 Recalling that 2 e hy we get y 252 Thus nally we get the particular solution 1 2 21 317995 27e 4 Example D2 7 4D 3 25 For the solution write the equation as PDy 25 PD D2 7 4D 3 Move the exponential over to the left hand side and write the equation as e xPDy 2 By the shifting rule this is the same as PD 1e xy 2 Let 2 e xy so we want to solve the equation P D 1z 2 4 Now calculate PD 1 as follows PD1D1274D13 D22D174D743 D2 7 2D Thus the equation for 2 becomes D2 7 2D2 2 Since we can factor out a D we do so and rewrite the equation as D 7 2D2 2 Let w D2 so we have the equation D 7 2w 2 for 11 Our trial solution should be a polynomial of the same degree as the right hand side In this case the right hand side has degree zero so we want a polynomial of degree zero iiei a constant Thus we try 11 A Plugging into the equation yields 2 Du 7 2w 7 0 7 2A 72A so clearly A 71 Thus we have 11 71 Since 11 D2 we have D2 71 Antidifferentiation yields 2 7x Recalling that 2 543 we nally have y 25 79555 Right Hand Sides of the Form qxew cos x or qxe sin x Consider the equations P D 11MB COSWI P D we sinwx where we assume that oz and are real numbers 1195 is a polynomial with real coefficients and that the coefficients of PD are real To deal with these problems recall that if a ai is the complex number with real part oz and imaginary part then 5 5w cos x 2150 sin xi Of course the independent variable x is always a real number so this formula is the expression of e is terms of its real and imaginary parts Thus 1195 1195 cos x iqxew sin xi 5 Since qx is always real we conclude that 1106 COSWI Relah ml mew sinwx Imqxe Now consider the equation 4 PM we coswx Reqxe l To solve this equation we nd a complex solution of 5 PDy we Once we ve found the complex solution we can get a solution of 4 by taking its real part Similarly the imaginary part of the solution of 5 is a solution of 6 PDy Iml11x5 l 1106 8111596 Thus by solving the complex equation 5 we solve the two real equations 4 and 6 at the same time We can solve 5 by exactly the same method we used above for exponential right hand sides the only difference is that we now have to deal with complex coef cients 5 Example D2 2D 1y xsin2x We want to solve 7 D2 2D 1y xsin2x The right hand side if of the form 1195quot sin x where oz 0 and 2 Thus xsin2x lrnxem The complex equation we want to solve is 8 D2 2D 1y 9552quot The solution of 7 will be the imaginary part of the solution of equation We write equation 8 as PDyx52m PD D22D1 Moving the exponential over gives 5 211PDy 95 By the shifting rule this can be rewritten as PD 2ie 2 y 95 Let 2 Jimmy so we want to solve PD 2239z 95 Next we calculate PD 2239 D 222 2D 2239 1 D24239D423922D42391 D24239D742D42391 D2 2 4239D 73 4239 6 Thus the equation for 2 is D2 2 4239D 73 42 2 7 as Our trial solution is 2 Ax B Plugging this into the equation yields 95 D22 2 4239Dz 73 4239z 7 0 2 42 A 73 4239Ax B 7 73 4239Ax 2 4239A 73 420B Equating coefficients gives the equations 9 734 A 71 10 242 A 734239B 0 From equation 9 we have 1 1 7374i 7374i A777 734 734 7374 732 42 7 7325 7 4225 Then from equation 10 we have 73 4293 7 72 4iA 7 72 7 4 7325 7 4225 7 625 71625 12i25 8225 7 725 4275 B 7 725 4 573 4239 7 725 4 57325 7 4225 7 22125 7 42125 Thus we get 2 7325 7 4i25x 22125 7 42125 7325x 22125 7 239425x 4125 721 Recalling that 2 e y we get y 2521 7325x 22125 7 239425x 4125cos2x 239 sin2x 7325x 22125 cos2x 425x 4125 sin2x i7325x 22125 sin2x 7 425x 4125 cos2x This is a particular solution of the complex equation Taking the imaginary part of this solution we nally get the particular solution y 7325x 22125 sin2x 7 425x 4125 cos2x for the original equation Notice that the real part 7325x 22125 cos2x 425x 4125 sin2x of the solution of 8 is a particular solution of the equation D2 2D 1y Rehem xcos2xi 7 6 Example D2 7 2D 2y ex cosxi 11 12 We want to solve D2 7 2D 2y ex cosx Ree1lx so we solve the complex equation D2 7 2D 2y 50 and then take the real part We can rewrite the complex equation as PDy BOW PD D2 7 21 2 Moving the exponential to the left side gives 571l1PDy 1 By the shifting rule this is the same as PD 1 ie 1lgt1y 1 Let 2 e 1 xy so we want to solve PD 1 2392 1 Now we calculate PD1239 D 12 72D 12 2 D221239D1i272D7272i2 D222iD22 72D7272i2 D2 22D Thus the equation for 2 is D2 22mg 1 We factor out a D and write this as D 2239D2 1 Thus 11 D2 satis es D 2239u 1 Since the right hand side is a polynomial of degree zero we try a polynomial of degree zero for 11 so let w A Plugging into the equation gives just 2239A 1 Thus A 1 w 7 72 2239 2 Since D2 w antidifferentiation gives 1 2 772x 2 Since 2 e 1 y we have 1 y ii 39xe 7326455 cosx 2151 sinx 2 1 E 1 E Ease s1nx 7 25955 cosxi This is a particular solution of the complex equation 12 The real part of this is a particular solution to the original equation so our nal answer for a particular solution of equation 11 is 1 y 7x51 sinxi 2

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