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# Engineering Mathematics MATH 3321

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This 15 page Class Notes was uploaded by Alvena McDermott on Saturday September 19, 2015. The Class Notes belongs to MATH 3321 at University of Houston taught by Staff in Fall. Since its upload, it has received 24 views. For similar materials see /class/208410/math-3321-university-of-houston in Mathmatics at University of Houston.

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Date Created: 09/19/15

64 Homogeneous Systems with Constant Coef cients A homogeneous system with constant coefficients is a linear differential system having the form m l a111 a122 l a1nn mZ a211 a222 l a2nn m an11 an22 l annn Where all 112 7 am are constants The system in vector matrix form is x l an 112 am 951 902 121 122 am 902 or X AX 1 m anl anZ arm m Example 1 Consider the 3Td order linear homogeneous differential equation y 2y 7 517 6y The characteristic equation is rs27275ri6 772r 1T3 0 and 5 5 2 e gt is a solution basis for the equation The corresponding linear homogeneous system is 0 1 X 0 0 1 X 6 5 2 and 521 V1t 25 ezt 2 45 is a solution vector see Problem 147 Exercises 63 Similarly7 5 1 573 s 1 V2t is 5 71 and V3t 735 673 f 73 at 1 95 9 are solution vectors I 279 Solutions Eigenvalues and Eigenvectors Example 1 suggests that homogeneous systems with constant coefficients might have solution vectors of the form vt a C7 for some number A and some constant vector c Set vt e c Then v t AeAtc Substituting into 17 we get AeAtc AeAt c which implies Ac A c The latter equation is an eigenvalueeigenvector equation for A Thus7 we look for solutions of the form vt eAtc where A is an eigenvalue of A and c is a corresponding eigenvector Example 2 Returning to Example 17 note that 0 1 0 1 1 0 1 0 1 1 0 0 1 2 2 2 7 0 0 1 71 1 71 7 6 5 72 4 4 6 5 72 1 1 and 0 1 0 1 1 0 1 73 3 73 6 5 72 9 9 0 1 1 2 is an eigenvalue of A 0 0 1 with corresponding eigenvector 2 7 71 6 5 72 4 1 is an eigenvalue of A with corresponding eigenvector 71 7 and 73 is an eigenvalue 1 1 of A with corresponding eigenvector 73 I 9 Example 3 Find a fundamental set of solution vectors of 7 1 5 x x 3 3 and give the general solution of the system SOLUTION First we nd the eigenvalues 17A 5 detA7AI 3 3 A A7 6A 2 280 The eigenvalues are 1 6 and A2 72 Next7 we nd corresponding eigenvectors For 1 6 we have A 7 6X T5 5 x1 0 which implies 1 2 2 arbitrary 3 73 m2 0 1 Setting 2 17 we get the eigenvector lt 1 i i 5 Repeating the process for A2 727 we get the e1genvector 3 1 5 Thus V1t e lt 1 gt and V2t 5 lt 3 gt are solution vectors of the system The Wronskian of V1 and V2 is 66t 5e72t Wt 785 7g 0 66t 73672t Thus V1 and V2 are linearly independent they form a fundamental set of solutions The general solution of the system is Xt 0156tlt 1 gt 025 lt 7 gt I Example 4 Find a fundamental set of solution vectors of 3 71 71 x 712 0 5 X 4 72 71 1 and nd the solution that satis es the initial condition X0 0 1 SOLUTION 3 7 71 71 detA7I 712 7 5 732272 4 72 717 Now detA 7 AI 0 implies A3 7 2A2 7 2 A 7 27 1 1 0 The eigenvalues are 1 27 A2 17 A3 71 281 As you can check7 corresponding eigenvectors are 1 3 C1 71 7 C2 71 7 C3 2 2 7 2 A fundamental set of solution vectors is 1 3 1 V1 25 5 71 7 V2t it 71 7 V3 75 aft 2 2 7 2 since distinct exponential vector functions are linearly independent calculate the Wronskian to verify and 1 3 1 xt 0152 5 71 of 71 0354 2 2 7 2 is the general solution To nd the solution vector satisfying the initial condition7 solve 01V10 02V20 03V30 0 which is 1 3 1 Cl 71 02 71 03 2 0 2 7 or 1 1 C1 1 71 71 2 C2 2 7 2 C3 1 Note The matrix of coefficients here is the fundamental matrix evaluated at t 0 Using the solution method of your choice row reduction7 inverse7 Cramer s rule7 the solution is C1 37 C2 717 C3 1 The solution of the initial value problem is 1 3 1 Xt 352 71 7 5t 71 aft 2 I 2 7 2 TWO Di iculties There are two difficulties that can arise 282 1 A has complex eigenvalues lf a bi is a complex eigenvalue of A with corresponding complex eigenvector u iv then X a 7 bi the complex conjugate of A is also an eigenvalue of A and u 7 iv is a corresponding eigenvector The corresponding linearly independent complex solutions of x Ax are and w1t eltabitu iv e tcos bt isin btu iv 5 cos bt u 7 sin bt v icos btv sin bt u wzt 5a bitu 7 iv e tcos bt 7 isin btu 7 iv 5 cos bt u 7 sin bt v 7 icos btv sin bt u x1t w1t w2t e tcos bt u 7 sin bt v x2t w1t 7 w2t e tcos btv sin bt u are linearly independent solutions of the system7 and they are real valued vector functions Note that x1 and x2 are simply the real and imaginary parts of W1 or of w2 Review Section 33 where you were shown how to convert complex exponential solutions into real valued solutions involving sine and cosine Example 5 Determine the general solution of 2 75 X X 1 0 SOLUTION 27A 75 17 detA7I 2725 The eigenvalues are 1 1 2i7 A2 1 7 2i The corresponding eigenvectors are c7lt1igt7lt1gtiltsgtv c27ltkigt7lt1gtiltsgt Now elt1wilt1gtiltsgtr etcos2tisin2t etcos2tlt1gt7sin2tlt3gt cos2tlt gtsin2tlt1gt iet 283 A fundamental set of solution vectors for the system is t 1 i 2 V1te cos2tlt1gt7sin2tlt0gt7 t 2 i 1 V2te cos2tlt0gtsin2tlt1gt The general solution of the system is 2 1 cos2tlt gtsin2tlt I 0 1 cos2t 1 7sin2t 2 1 0 Example 6 Determine a fundamental set of solution vectors of Xt 01 it 02 at 1 74 71 X 3 2 3 X 1 1 3 SOLUTION 17 74 71 detA7I 3 27A 3 73627212677227413 1 137 The eigenvalues are 1 27 A2 2 1 32 7 A3 2 7 32 The corresponding eigenvectors are 1 75 3i 75 3 c1 0 C2 3 32 3 3 3 71 2 2 0 75 7 3i 75 3 C3 3 7 3i 3 7 i 3 2 2 0 Now 75 3 e23it 3 i 3 2 0 7 3 52tcos 3t 1 i sin 3t 3 H 3 3 3 75 5 loos 3t 3 7 sin 3t 3 l New loos 3t 3 1 sin 3t 3 l 0 l 0 2 284 A fundamental set of solution vectors for the system is 1 75 3 v1t 52 5 0 vzt 52 5 cos 3t 3 7 sin 3t 3 7 71 0 l 3 5 l V3t 52 5 cos 3t 3 1 sin 3t 3 J I 0 2 2 A has an eigenvalue of multiplicity greater than 1 We ll treat the case Where A has an eigenvalue of multiplicity 2 Example 7 Determine a fundamental set of solution vectors of 1 73 3 x 3 75 3 x 6 76 4 SOLUTION 17 73 3 detA7I 3 757A 3 7A312A71677422 6 76 4 7 The eigenvalues are 1 47 A2 3 72 1 As you can check7 an eigenvector corresponding to 1 4 is C1 1 2 We ll carry out the details involved in nding an eigenvector corresponding to the dou ble77 eigenvalue 72 3 73 3 01 0 A772Ic 3 73 3 02 0 6 76 6 03 0 The augmented matrix for this system of equations is 3 73 3 0 1 71 1 0 3 73 3 0 which row reduces to 0 0 0 0 6 76 6 0 0 0 0 0 The solutions of this system are 01 02 7 03 02 03 arbitrary We can assign values to 02 and 03 independently and obtain two linearly independent eigenvectors For example7 285 setting 62 17 Cg 07 we get the eigenvector C2 1 Reversing the roles7 we set 02 O7 03 71 to get the eigenvector C3 0 Clearly c2 and C3 are linearly 71 independent You should understand that there is nothing magic about our two choices for 02 03 any choice which produces two independent vectors will do The important thing to note here is that this eigenvalue of multiplicity 2 produced two independent eigenvectors Based on our work above7 a fundamental set of solutions for the differential system 1 73 3 x 3 75 3 X 6 76 4 is 1 V1t EM 1 7 V2lttgt 67 1 7 V3lttgt 67 0 I 71 0 1 0 Example 8 Let A 0 0 1 12 8 71 7A 1 0 detA7I 0 7A 1 737287127A7322 12 8 717A The eigenvalues are 1 37 A2 A3 72 As you can check7 an eigenvector corresponding to 1 3 is c1 3 We ll carry out the details involved in nding an eigenvector corresponding to the dou ble77 eigenvalue 72 2 1 0 01 0 A772Ic 0 2 1 02 0 12 8 1 03 0 The augmented matrix for this system of equations is 2 1 0 0 1 0 0 0 2 1 0 which row reduces to 0 2 1 0 12 8 1 0 0 0 0 286 The solutions of this system are 01 03 62 7 cg Cg arbitrary Here there is only one parameter and so we ll get only one eigenvector Setting 03 4 we get the eigenvector In contrast to the preceding example7 the double eigenvalue here has only one inde pendent eigenvector Suppose that we were asked to nd a fundamental set of solutions ofthe linear differential system 0 1 0 X 0 0 1 X 12 8 71 By our work above7 we have two independent solutions 1 1 V1 53 5 3 and V2 572 f 72 9 4 We need a third solution which is independent of these two Our system has a special form it is equivalent to the third order equation y y 7 837123 0 The characteristic equation is 7371772787712 r73r220 compare with detA 7 AD The roots are 71 37 r2 r3 72 and a fundamental set of solutions is yl eSt yg 5 22 yg te Zt The correspondence between these solutions and the solution vectors we found above should be clear 1 1 eat eat 3 7 5721 5721 72 4 As we saw in Section 627 the solution 243t 15572 5 of the equation produces the solution vector y3t 25572 0 1 v3t ygt 5 27 7 2255 5 quot 1 255 72 745 2 7 4te72t 74 4 of the correspondng system 287 The appearance of the te Ztcz term should not be unexpected since we know that a characteristic root r of multiplicity 2 produces a solution of the form t5quot You can check that V3 is independent of V1 and V2 Therefore7 the solution vectors vl7 V27 V3 are a fundamental set of solutions of the system The question is What is the significance of the vector w 1 7 How is it related 74 to the eigenvalue 72 which generated it7 and to the corresponding eigenvector Let s look at A 7 72Iw A 2w 2 1 0 1 w i C2 A 2 0 2 1 1 2 12 8 1 74 4 A 7 72 maps w onto the eigenvector Q The corresponding solution of the system has the form V3t e 2tw teTZtcz where C2 is the eigenvector corresponding to 72 and w satis es A772Iwcz I General Result Given the linear differential system xl Ax Suppose that A has an eigenvalue A of multiplicity 2 Then exactly one of the following holds 1 A has two linearly independent eigenvectors7 c1 and c2 Corresponding linearly independent solution vectors of the differential system are V1t eAtcl and V2t EMU to A has only one independent eigenvector G Then a linearly independent pair of solution vectors corresponding to A are At V1t e c and V2t eAtw teAtc where w is a vector that satis es A 7 AIw c The vector w is called a y I J LULLU puiidiiig t0 the i l A 1 71 Example 9 Find a fundamental set of solution vectors for x lt 1 3 gtx SOLUTION 17A 71 detA7AI 1 3 A A274A4A722 288 Characteristic values 1 A2 2 Characteristic vectors WW1 112 3 71710 110 1 10 T 00039 The solutions are 01 702 02 arbitrary there is only one eigenvector Setting 02 71 1 we et c g 71 1 The vector V1 5 lt 1 gt is a solution of the system A second solution7 independent of V1 is V2 52tw t52tc where w is a solution of A 7 2z c A7 2z 1 1 21 1 1 1 22 71 71 71 1 1 1 71 A 1 1 1 0 0 0 The solutions of this system are 21 71 7 22 22 arbitrary If we choose 22 0 any choice for 22 will do7 we get 21 71 and w lt 71 Thus t 1 t 1 V2t62lt 0gtt52 71gt is a solution of the system independent of V1 The solutions v1t 52tlt j v2t 52 lt 71 gt 255 lt 71gt are a fundamental set of solutions of the system I 71 Example 10 Let A 2 2 71 Find a fundamental set of solutions of 0 x AX SOL U TI ON 289 37A 1 71 detA7I 2 27A 71 7352784771722 2 2 7A The eigenvalues are 1 17 A2 3 2 An eigenvector corresponding to 1 1 is c1 0 check this We ll show the details involved in nding an eigenvector or eigenvectors corresponding to the double eigenvalue 2 1 1 71 01 0 A72Ic 2 0 71 02 0 2 2 72 03 The augmented matrix for this system of equations is 1 1 71 0 1 1 71 0 2 0 71 0 which row reduces to 0 72 1 0 2 2 72 0 0 0 0 0 The solutions of this system are 01 702 03 cg 2127 62 arbitrary There is only 1 one eigenvector corresponding to the eigenvalue 2 Setting 02 17 we get c2 1 2 Thus7 two independent solutions of the given linear differential system are 1 V1 6t 0 7 V2 EZt 2 We need another solution corresponding to the eigenvalue 27 one which is independent of V2 We know that this solution has the form V3t 52tw tEZtCZ where w is a solution of A 7 2z c2 That is 1 1 71 21 2 0 71 22 2 2 72 23 The augmented matrix is 1 1 71 1 1 1 71 1 2 0 71 1 which row reduces to 0 72 1 71 2 2 72 2 0 0 0 0 290 The solutions of this system are 2 3 71 222 2 1 17 2 2 2 3 17 2 2 71 222 227 2 2 arbitrary If we choose 22 0 any choice for 22 will do7 we get 21 O7 22 O7 23 71 and 0 w 0 Thus 71 1 V3 52 s 0 tezt 1 71 2 1 1 1 V1 et 0 7 V2 52 5 1 7 V3 52 5 0 2552 1 2 2 71 2 are a fundamental set of solutions of the system I Exercises 64 Find the general solution of the system xl AX Where A is the given matrix If an initial condition is given7 also nd the solution that satis es the condition 1 24 11 732 17239 291 71 1 74 7539 8lt 3 0 i2 2 8 0 211 7 0 X 7 293 443 02 444 31 344 7 292 2711 200 371 0713 27171 1 21 2 171x0 72 0711 0 72171 22 373 3712 2276 23 27173 72711 8761 2410792 10770 Appendix Eigenvalues 0f Multiplicity 3 Given the differential system X AX Suppose that A is an eigenvalue of A of multiplicity 3 Then exactly one of the following holds 1 A has three linearly independent eigenvectors c17 Cg C3 Then three linearly inde pendent solution vectors of the system corresponding to A are V1t eAtcl V2t EMCZ7 V3lttgt EAth 2 A has two linearly independent eigenvectors c1 c2 Then two linearly independent solutions of the system corresponding to A are A A v1t e tcl v2t e tc2 A third solution7 independent of v1 and vz has the form V3t eAtw teAtv where v is an eigenvector corresponding to A and A 7 AIw v A has only one independent eigenvector c Then three linearly independent solu tions of the system have the form OJ v1 eAtc7 v2 eAtw tgeMc7 V3t 5M2 teAtw tzeAtc where AiAIwc and AiAIzw 293

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