Baixe o app para aproveitar ainda mais
Prévia do material em texto
Part III Systems of Differential Equations 1010 Systems of LinearDifferential Equations EXERCISES 10.1 Preliminary Theory 1. Let X = ( x y ) . Then X′ = ( 3 −5 4 8 ) X. 2. Let X = ( x y ) . Then X′ = ( 4 −7 5 0 ) X. 3. Let X = xy z . Then X′ = −3 4 −96 −1 0 10 4 3 X. 4. Let X = xy z . Then X′ = 1 −1 01 0 2 −1 0 1 X. 5. Let X = xy z . Then X′ = 1 −1 12 1 −1 1 1 1 X + 0−3t2 t2 + t0 −t + −10 2 . 6. Let X = xy z . Then X′ = −3 4 05 9 0 0 1 6 X + e −t sin 2t 4e−t cos 2t −e−t . 7. dx dt = 4x + 2y + et; dy dt = −x + 3y − et 8. dx dt = 7x + 5y − 9z − 8e−2t; dy dt = 4x + y + z + 2e5t; dz dt = −2y + 3z + e5t − 3e−2t 9. dx dt = x− y + 2z + e−t − 3t; dy dt = 3x− 4y + z + 2e−t + t; dz dt = −2x + 5y + 6z + 2e−t − t 10. dx dt = 3x− 7y + 4 sin t + (t− 4)e4t; dy dt = x + y + 8 sin t + (2t + 1)e4t 11. Since X′ = ( −5 −10 ) e−5t and ( 3 −4 4 −7 ) X = ( −5 −10 ) e−5t 551 10.1 Preliminary Theory we see that X′ = ( 3 −4 4 −7 ) X. 12. Since X′ = ( 5 cos t− 5 sin t 2 cos t− 4 sin t ) et and (−2 5 −2 4 ) X = ( 5 cos t− 5 sin t 2 cos t− 4 sin t ) et we see that X′ = (−2 5 −2 4 ) X. 13. Since X′ = ( 3 2 −3 ) e−3t/2 and (−1 14 1 −1 ) X = ( 3 2 −3 ) e−3t/2 we see that X′ = (−1 1/4 1 −1 ) X. 14. Since X′ = ( 5 −1 ) et + ( 4 −4 ) tet and ( 2 1 −1 0 ) X = ( 5 −1 ) et + ( 4 −4 ) tet we see that X′ = ( 2 1 −1 0 ) X. 15. Since X′ = 00 0 and 1 2 16 −1 0 −1 −2 −1 X = 00 0 we see that X′ = 1 2 16 −1 0 −1 −2 −1 X. 16. Since X′ = cos t12 sin t− 12 cos t − cos t− sin t and 1 0 11 1 0 −2 0 −1 X = cos t12 sin t− 12 cos t − cos t− sin t we see that X′ = 1 0 11 1 0 −2 0 −1 X. 17. Yes, since W (X1,X2) = −2e−8t �= 0 the set X1, X2 is linearly independent on −∞ < t < ∞. 18. Yes, since W (X1,X2) = 8e2t �= 0 the set X1, X2 is linearly independent on −∞ < t < ∞. 19. No, since W (X1,X2,X3) = 0 the set X1, X2, X3 is linearly dependent on −∞ < t < ∞. 20. Yes, since W (X1,X2,X3) = −84e−t �= 0 the set X1, X2, X3 is linearly independent on −∞ < t < ∞. 21. Since X′p = ( 2 −1 ) and ( 1 4 3 2 ) Xp + ( 2 −4 ) t + ( −7 −18 ) = ( 2 −1 ) 552 10.1 Preliminary Theory we see that X′p = ( 1 4 3 2 ) Xp + ( 2 −4 ) t + ( −7 −18 ) . 22. Since X′p = ( 0 0 ) and ( 2 1 1 −1 ) Xp + (−5 2 ) = ( 0 0 ) we see that X′p = ( 2 1 1 −1 ) Xp + (−5 2 ) . 23. Since X′p = ( 2 0 ) et + ( 1 −1 ) tet and ( 2 1 3 4 ) Xp − ( 1 7 ) et = ( 2 0 ) et + ( 1 −1 ) tet we see that X′p = ( 2 1 3 4 ) Xp − ( 1 7 ) et. 24. Since X′p = 3 cos 3t0 −3 sin 3t and 1 2 3−4 2 0 −6 1 0 Xp + −14 3 sin 3t = 3 cos 3t0 −3 sin 3t we see that X′p = 1 2 3−4 2 0 −6 1 0 Xp + −14 3 sin 3t. 25. Let X1 = 6−1 −5 e−t, X2 = −31 1 e−2t, X3 = 21 1 e3t, and A = 0 6 01 0 1 1 1 0 . Then X′1 = −61 5 e−t = AX1, X′2 = 6−2 −2 e−2t = AX2, X′3 = 63 3 e3t = AX3, and W (X1,X2,X3) = 20 �= 0 so that X1, X2, and X3 form a fundamental set for X′ = AX on −∞ < t < ∞. 26. Let X1 = ( 1 −1−√2 ) e √ 2 t, X2 = ( 1 −1 +√2 ) e− √ 2 t, 553 10.1 Preliminary Theory Xp = ( 1 0 ) t2 + (−2 4 ) t + ( 1 0 ) , and A = (−1 −1 −1 1 ) . Then X′1 = ( √ 2 −2−√2 ) e √ 2 t = AX1, X′2 = ( −√2 −2 +√2 ) e− √ 2 t = AX2, X′p = ( 2 0 ) t + (−2 4 ) = AXp + ( 1 1 ) t2 + ( 4 −6 ) t + (−1 5 ) , and W (X1,X2) = 2 √ 2 �= 0 so that Xp is a particular solution and X1 and X2 form a fundamental set on −∞ < t < ∞. EXERCISES 10.2 Homogeneous Linear Systems 1. The system is X′ = ( 1 2 4 3 ) X and det(A− λI) = (λ− 5)(λ + 1) = 0. For λ1 = 5 we obtain(−4 2 0 4 −2 0 ) =⇒ ( 1 − 12 0 0 0 0 ) so that K1 = ( 1 2 ) . For λ2 = −1 we obtain ( 2 2 0 4 4 0 ) =⇒ ( 1 1 0 0 0 0 ) so that K2 = (−1 1 ) . Then X = c1 ( 1 2 ) e5t + c2 (−1 1 ) e−t. 2. The system is X′ = ( 2 2 1 3 ) X and det(A− λI) = (λ− 1)(λ− 4) = 0. For λ1 = 1 we obtain( 1 2 0 1 2 0 ) =⇒ ( 1 2 0 0 0 0 ) so that K1 = (−2 1 ) . For λ2 = 4 we obtain (−2 2 0 1 −1 0 ) =⇒ (−1 1 0 0 0 0 ) so that K2 = ( 1 1 ) . 554 10.2 Homogeneous Linear Systems Then X = c1 (−2 1 ) et + c2 ( 1 1 ) e4t. 3. The system is X′ = ( −4 2 − 52 2 ) X and det(A− λI) = (λ− 1)(λ + 3) = 0. For λ1 = 1 we obtain( −5 2 0 − 52 1 0 ) =⇒ (−5 2 0 0 0 0 ) so that K1 = ( 2 5 ) . For λ2 = −3 we obtain ( −1 2 0 − 52 5 0 ) =⇒ (−1 2 0 0 0 0 ) so that K2 = ( 2 1 ) . Then X = c1 ( 2 5 ) et + c2 ( 2 1 ) e−3t. 4. The system is X′ = (− 52 2 3 4 −2 ) X and det(A− λI) = 12 (λ + 1)(2λ + 7) = 0. For λ1 = −7/2 we obtain( 1 2 0 3 4 3 2 0 ) =⇒ ( 1 2 0 0 0 0 ) so that K1 = (−2 1 ) . For λ2 = −1 we obtain (− 32 2 0 3 4 −1 0 ) =⇒ (−3 4 0 0 0 0 ) so that K2 = ( 4 3 ) . Then X = c1 (−2 1 ) e−7t/2 + c2 ( 4 3 ) e−t. 5. The system is X′ = ( 10 −5 8 −12 ) X and det(A− λI) = (λ− 8)(λ + 10) = 0. For λ1 = 8 we obtain( 2 −5 0 8 −20 0 ) =⇒ ( 1 − 52 0 0 0 0 ) so that K1 = ( 5 2 ) . For λ2 = −10 we obtain ( 20 −5 0 8 −2 0 ) =⇒ ( 1 − 14 0 0 0 0 ) so that K2 = ( 1 4 ) . Then X = c1 ( 5 2 ) e8t + c2 ( 1 4 ) e−10t. 6. The system is X′ = (−6 2 −3 1 ) X 555 10.2 Homogeneous Linear Systems and det(A− λI) = λ(λ + 5) = 0. For λ1 = 0 we obtain(−6 2 0 −3 1 0 ) =⇒ ( 1 − 13 0 0 0 0 ) so that K1 = ( 1 3 ) . For λ2 = −5 we obtain (−1 2 0 −3 6 0 ) =⇒ ( 1 −2 0 0 0 0 ) so that K2 = ( 2 1 ) . Then X = c1 ( 1 3 ) + c2 ( 2 1 ) e−5t. 7. The system is X′ = 1 1 −10 2 0 0 1 −1 X and det(A− λI) = (λ− 1)(2− λ)(λ + 1) = 0. For λ1 = 1, λ2 = 2, and λ3 = −1 we obtain K1 = 10 0 , K2 = 23 1 , and K3 = 10 2 , so that X = c1 10 0 et + c2 23 1 e2t + c3 10 2 e−t. 8. The system is X′ = 2 −7 05 10 4 0 5 2 X and det(A− λI) = (2− λ)(λ− 5)(λ− 7) = 0. For λ1 = 2, λ2 = 5, and λ3 = 7 we obtain K1 = 40 −5 , K2 = −73 5 , and K3 = −75 5 , so that X = c1 40 −5 e2t + c2 −73 5 e5t + c3 −75 5 e7t. 9. We have det(A− λI) = −(λ + 1)(λ− 3)(λ + 2) = 0. For λ1 = −1, λ2 = 3, and λ3 = −2 we obtain K1 = −10 1 , K2 = 14 3 , and K3 = 1−1 3 , so that X = c1 −10 1 e−t + c2 14 3 e3t + c3 1−1 3 e−2t. 556 10.2 Homogeneous Linear Systems 10. We have det(A− λI) = −λ(λ− 1)(λ− 2) = 0. For λ1 = 0, λ2 = 1, and λ3 = 2 we obtain K1 = 10 −1 , K2 = 01 0 , and K3 = 10 1 , so that X = c1 10 −1 + c2 01 0 et + c3 10 1 e2t. 11. We have det(A− λI) = −(λ + 1)(λ + 1/2)(λ + 3/2) = 0. For λ1 = −1, λ2 = −1/2, and λ3 = −3/2 we obtain K1 = 40 −1 , K2 = −126 5 , and K3 = 42 −1 , so that X = c1 40 −1 e−t + c2 −126 5 e−t/2 + c3 42 −1 e−3t/2. 12. We have det(A− λI) = (λ− 3)(λ + 5)(6− λ) = 0. For λ1 = 3, λ2 = −5, and λ3 = 6 we obtain K1 = 11 0 , K2 = 1−1 0 , and K3 = 2−2 11 , so that X = c1 11 0 e3t + c2 1−1 0 e−5t + c3 2−2 11 e6t. 13. We have det(A− λI) = (λ + 1/2)(λ− 1/2) = 0. For λ1 = −1/2 and λ2 = 1/2 we obtain K1 = ( 0 1 ) and K2 = ( 1 1 ) , so that X = c1 ( 0 1 ) e−t/2 + c2 ( 1 1 ) et/2. If X(0) = ( 3 5 ) then c1 = 2 and c2 = 3. 14. We have det(A− λI) = (2− λ)(λ− 3)(λ + 1) = 0. For λ1 = 2, λ2 = 3, and λ3 = −1 we obtain K1 = 5−3 2 , K2 = 20 1 , and K3 = −20 1 , so that X = c1 5−3 2 e2t + c2 20 1 e3t + c3 −20 1 e−t. 557 10.2 Homogeneous Linear Systems If X(0) = 13 0 then c1 = −1, c2 = 5/2, and c3 = −1/2. 15. X = c1 0.3821750.851161 0.359815 e8.58979t + c2 0.405188−0.676043 0.615458 e2.25684t + c3 −0.923562−0.132174 0.35995 e−0.0466321t 16. X = c1 0.0312209 0.949058 0.239535 0.195825 0.0508861 e5.05452t + c2 −0.280232 −0.836611 −0.275304 0.176045 0.338775 e4.09561t + c3 0.262219 −0.162664 −0.826218 −0.346439 0.31957 e−2.92362t +c4 0.313235 0.64181 0.31754 0.173787 −0.599108 e2.02882t + c5 −0.301294 0.466599 0.222136 0.0534311 −0.799567 e−0.155338t 17. (a) (b) Letting c1 = 1 and c2 = 0 we get x = 5e8t, y = 2e8t. Eliminating the parameter we find y = 25x, x > 0. When c1 = −1 and c2 = 0 we find y = 25x, x < 0. Letting c1 = 0 and c2 = 1 we get x = e−10t, y = 4e−10t. Eliminating the parameter we find y = 4x, x > 0. Letting c1 = 0 and c2 = −1 we find y = 4x, x < 0. (c) The eigenvectors K1 = (5, 2) and K2 = (1, 4) are shown in the figure in part (a). 18. In Problem 2, letting c1 = 1 and c2 = 0 we get x = −2et, y = et. Eliminating the parameter we find y = − 12x, x < 0. When c1 = −1 and c2 = 0 we find y = − 12x, x > 0. Letting c1 = 0 and c2 = 1 we get x = e4t, y = e4t. Eliminating the parameter we find y = x, x > 0. When c1 = 0 and c2 = −1 we find y = x, x < 0. 558 10.2 Homogeneous Linear Systems In Problem 4, letting c1 = 1 and c2 = 0 we get x = −2e−7t/2, y = e−7t/2. Eliminating the parameter we find y = − 12x, x < 0. When c1 = −1 and c2 = 0 we find y = − 12x, x > 0. Letting c1 = 0 and c2 = 1 we get x = 4e−t, y = 3e−t. Eliminating the parameter we find y = 34x, x > 0. When c1 = 0 and c2 = −1 we find y = 34x, x < 0. 19. We have det(A− λI) = λ2 = 0. For λ1 = 0 we obtain K = ( 1 3 ) . A solution of (A− λ1I)P = K is P = ( 1 2 ) so that X = c1 ( 1 3 ) + c2 [( 1 3 ) t + ( 1 2 )] . 20. We have det(A− λI) = (λ + 1)2 = 0. For λ1 = −1 we obtain K = ( 1 1 ) . A solution of (A− λ1I)P = K is P = ( 0 1 5 ) so that X = c1 ( 1 1 ) e−t + c2 [( 1 1 ) te−t + ( 0 1 5 ) e−t ] . 21. We have det(A− λI) = (λ− 2)2 = 0. For λ1 = 2 we obtain K = ( 1 1 ) . A solution of (A− λ1I)P = K is P = (− 13 0 ) so that X = c1 ( 1 1 ) e2t + c2 [( 1 1 ) te2t + (− 13 0 ) e2t ] . 22. We have det(A− λI) = (λ− 6)2 = 0. For λ1 = 6 we obtain K = ( 3 2 ) . A solution of (A− λ1I)P = K is P = ( 1 2 0 ) 559 10.2 Homogeneous Linear Systems so that X = c1 ( 3 2 ) e6t + c2 [( 3 2 ) te6t + ( 1 2 0 ) e6t ] . 23. We have det(A− λI) = (1− λ)(λ− 2)2 = 0. For λ1 = 1 we obtain K1 = 11 1 . For λ2 = 2 we obtain K2 = 10 1 and K3 = 11 0 . Then X = c1 11 1 et + c2 10 1 e2t + c3 11 0 e2t. 24. We have det(A− λI) = (λ− 8)(λ + 1)2 = 0. For λ1 = 8 we obtain K1 = 21 2 . For λ2 = −1 we obtain K2 = 0−2 1 and K3 = 1−2 0 . Then X = c1 21 2 e8t + c2 0−2 1 e−t + c3 1−2 0 e−t. 25. We have det(A− λI) = −λ(5− λ)2 = 0. For λ1 = 0 we obtain K1 = −4−5 2 . For λ2 = 5 we obtain K = −20 1 . A solution of (A− λ1I)P = K is P = 5 2 1 2 0 560 10.2 Homogeneous Linear Systems so that X = c1 −4−5 2 + c2 −20 1 e5t + c3 −20 1 te5t + 5 2 1 2 0 e5t . 26. We have det(A− λI) = (1− λ)(λ− 2)2 = 0. For λ1 = 1 we obtain K1 = 10 0 . For λ2 = 2 we obtain K = 0−1 1 . A solution of (A− λ2I)P = K is P = 0−1 0 so that X = c1 10 0 et + c2 0−1 1 e2t + c3 0−1 1 te2t + 0−1 0 e2t . 27. We have det(A− λI) = −(λ− 1)3 = 0. For λ1 = 1 we obtain K = 01 1 . Solutions of (A− λ1I)P = K and (A− λ1I)Q = P are P = 01 0 and Q = 1 2 0 0 so that X = c1 01 1 et + c2 01 1 tet + 01 0 et + c3 01 1 t22 et + 01 0 tet + 1 2 0 0 et . 28. We have det(A− λI) = (λ− 4)3 = 0. For λ1 = 4 we obtain K = 10 0 . Solutions of (A− λ1I)P = K and (A− λ1I)Q = P are P = 01 0 and Q = 00 1 561 10.2 Homogeneous Linear Systems so that X = c1 10 0 e4t + c2 10 0 te4t + 01 0 e4t + c3 10 0 t22 e4t + 01 0 te4t + 00 1 e4t . 29. We have det(A− λI) = (λ− 4)2 = 0. For λ1 = 4 we obtain K = ( 2 1 ) . A solution of (A− λ1I)P = K is P = ( 1 1 ) so that X = c1 ( 2 1 ) e4t + c2 [( 2 1 ) te4t + ( 1 1 ) e4t ] . If X(0) = (−1 6 ) then c1 = −7 and c2 = 13. 30. We have det(A− λI) = −(λ + 1)(λ− 1)2 = 0. For λ1 = −1 we obtain K1 = −10 1 . For λ2 = 1 we obtain K2 = 10 1 and K3 = 01 0 so that X = c1 −10 1 e−t + c2 10 1 et + c3 01 0 et. If X(0) = 12 5 then c1 = 2, c2 = 3, and c3 = 2. 31. In this case det(A − λI) = (2 − λ)5, and λ1 = 2 is an eigenvalue of multiplicity 5. Linearly independent eigenvectors are K1 = 1 0 0 0 0 , K2 = 0 0 1 0 0 , and K3 = 0 0 0 1 0 . 32. In Problem 20 letting c1 = 1 and c2 = 0 we get x = et, y = et. Eliminating the parameter we find y = x, x > 0. When c1 = −1 and c2 = 0 we find y = x, x < 0. 562 10.2 Homogeneous Linear Systems In Problem 21 letting c1 = 1 and c2 = 0 we get x = e2t, y = e2t. Eliminating the parameter we find y = x, x > 0. When c1 = −1 and c2 = 0 we find y = x, x < 0. In Problems 33-46 the form of the answer will vary according to the choice of eigenvector. For example, in Problem 33, if K1 is chosen to be ( 1 2− i ) the solution has the form X = c1 ( cos t 2 cos t + sin t ) e4t + c2 ( sin t 2 sin t− cos t ) e4t. 33. We have det(A− λI) = λ2 − 8λ + 17 = 0. For λ1 = 4 + i we obtain K1 = ( 2 + i 5 ) so that X1 = ( 2 + i 5 ) e(4+i)t = ( 2 cos t− sin t 5 cos t ) e4t + i ( cost + 2 sin t 5 sin t ) e4t. Then X = c1 ( 2 cos t− sin t 5 cos t ) e4t + c2 ( cos t + 2 sin t 5 sin t ) e4t. 34. We have det(A− λI) = λ2 + 1 = 0. For λ1 = i we obtain K1 = (−1− i 2 ) so that X1 = (−1− i 2 ) eit = ( sin t− cos t 2 cos t ) + i (− cos t− sin t 2 sin t ) . Then X = c1 ( sin t− cos t 2 cos t ) + c2 (− cos t− sin t 2 sin t ) . 35. We have det(A− λI) = λ2 − 8λ + 17 = 0. For λ1 = 4 + i we obtain K1 = (−1− i 2 ) so that X1 = (−1− i 2 ) e(4+i)t = ( sin t− cos t 2 cos t ) e4t + i (− sin t− cos t 2 sin t ) e4t. 563 10.2 Homogeneous Linear Systems Then X = c1 ( sin t− cos t 2 cos t ) e4t + c2 (− sin t− cos t 2 sin t ) e4t. 36. We have det(A− λI) = λ2 − 10λ + 34 = 0. For λ1 = 5 + 3i we obtain K1 = ( 1− 3i 2 ) so that X1 = ( 1− 3i 2 ) e(5+3i)t = ( cos 3t + 3 sin 3t 2 cos 3t ) e5t + i ( sin 3t− 3 cos 3t 2 sin 3t ) e5t. Then X = c1 ( cos 3t + 3 sin 3t 2 cos 3t ) e5t + c2 ( sin 3t− 3 cos 3t 2 sin 3t ) e5t. 37. We have det(A− λI) = λ2 + 9 = 0. For λ1 = 3i we obtain K1 = ( 4 + 3i 5 ) so that X1 = ( 4 + 3i 5 ) e3it = ( 4 cos 3t− 3 sin 3t 5 cos 3t ) + i ( 4 sin 3t + 3 cos 3t 5 sin 3t ) . Then X = c1 ( 4 cos 3t− 3 sin 3t 5 cos 3t ) + c2 ( 4 sin 3t + 3 cos 3t 5 sin 3t ) . 38. We have det(A− λI) = λ2 + 2λ + 5 = 0. For λ1 = −1 + 2i we obtain K1 = ( 2 + 2i 1 ) so that X1 = ( 2 + 2i 1 ) e(−1+2i)t = ( 2 cos 2t− 2 sin 2t cos 2t ) e−t + i ( 2 cos 2t + 2 sin 2t sin 2t ) e−t. Then X = c1 ( 2 cos 2t− 2 sin 2t cos 2t ) e−t + c2 ( 2 cos 2t + 2 sin 2t sin 2t ) e−t. 39. We have det(A− λI) = −λ (λ2 + 1) = 0. For λ1 = 0 we obtain K1 = 10 0 . For λ2 = i we obtain K2 = −ii 1 so that X2 = −ii 1 eit = sin t− sin t cos t + i − cos tcos t sin t . 564 10.2 Homogeneous Linear Systems Then X = c1 10 0 + c2 sin t− sin t cos t + c3 − cos tcos t sin t . 40. We have det(A− λI) = −(λ + 3)(λ2 − 2λ + 5) = 0. For λ1 = −3 we obtain K1 = 0−2 1 . For λ2 = 1 + 2i we obtain K2 = −2− i−3i 2 so that X2 = −2 cos 2t + sin 2t3 sin 2t 2 cos 2t et + i − cos 2t− 2 sin 2t−3 cos 2t 2 sin 2t et. Then X = c1 0−2 1 e−3t + c2 −2 cos 2t + sin 2t3 sin 2t 2 cos 2t et + c3 − cos 2t− 2 sin 2t−3 cos 2t 2 sin 2t et. 41. We have det(A− λI) = (1− λ)(λ2 − 2λ + 2) = 0. For λ1 = 1 we obtain K1 = 02 1 . For λ2 = 1 + i we obtain K2 = 1i i so that X2 = 1i i e(1+i)t = cos t− sin t − sin t et + i sin tcos t cos t et. Then X = c1 02 1 et + c2 cos t− sin t − sin t et + c3 sin tcos t cos t et. 42. We have det(A− λI) = −(λ− 6)(λ2 − 8λ + 20) = 0. For λ1 = 6 we obtain K1 = 01 0 . 565 10.2 Homogeneous Linear Systems For λ2 = 4 + 2i we obtain K2 = −i0 2 so that X2 = −i0 2 e(4+2i)t = sin 2t0 2 cos 2t e4t + i − cos 2t0 2 sin 2t e4t. Then X = c1 01 0 e6t + c2 sin 2t0 2 cos 2t e4t + c3 − cos 2t0 2 sin 2t e4t. 43. We have det(A− λI) = (2− λ)(λ2 + 4λ + 13) = 0. For λ1 = 2 we obtain K1 = 28−5 25 . For λ2 = −2 + 3i we obtain K2 = 4 + 3i−5 0 so that X2 = 4 + 3i−5 0 e(−2+3i)t = 4 cos 3t− 3 sin 3t−5 cos 3t 0 e−2t + i 4 sin 3t + 3 cos 3t−5 sin 3t 0 e−2t. Then X = c1 28−5 25 e2t + c2 4 cos 3t− 3 sin 3t−5 cos 3t 0 e−2t + c3 4 sin 3t + 3 cos 3t−5 sin 3t 0 e−2t. 44. We have det(A− λI) = −(λ + 2)(λ2 + 4) = 0. For λ1 = −2 we obtain K1 = 0−1 1 . For λ2 = 2i we obtain K2 = −2− 2i1 1 so that X2 = −2− 2i1 1 e2it = −2 cos 2t + 2 sin 2tcos 2t cos 2t + i −2 cos 2t− 2 sin 2tsin 2t sin 2t . 566 10.2 Homogeneous Linear Systems Then X = c1 0−1 1 e−2t + c2 −2 cos 2t + 2 sin 2tcos 2t cos 2t + c3 −2 cos 2t− 2 sin 2tsin 2t sin 2t . 45. We have det(A− λI) = (1− λ)(λ2 + 25) = 0. For λ1 = 1 we obtain K1 = 25−7 6 . For λ2 = 5i we obtain K2 = 1 + 5i1 1 so that X2 = 1 + 5i1 1 e5it = cos 5t− 5 sin 5tcos 5t cos 5t + i sin 5t + 5 cos 5tsin 5t sin 5t . Then X = c1 25−7 6 et + c2 cos 5t− 5 sin 5tcos 5t cos 5t + c3 sin 5t + 5 cos 5tsin 5t sin 5t . If X(0) = 46 −7 then c1 = c2 = −1 and c3 = 6. 46. We have det(A− λI) = λ2 − 10λ + 29 = 0. For λ1 = 5 + 2i we obtain K1 = ( 1 1− 2i ) so that X1 = ( 1 1− 2i ) e(5+2i)t = ( cos 2t cos 2t + 2 sin 2t ) e5t + i ( sin 2t sin 2t− 2 cos 2t ) e5t. and X = c1 ( cos 2t cos 2t + 2 sin 2t ) e5t + c3 ( sin 2t sin 2t− 2 cos 2t ) e5t. If X(0) = (−2 8 ) , then c1 = −2 and c2 = 5. 567 10.2 Homogeneous Linear Systems 47. 48. (a) From det(A− λI) = λ(λ− 2) = 0 we get λ1 = 0 and λ2 = 2. For λ1 = 0 we obtain( 1 1 0 1 1 0 ) =⇒ ( 1 1 0 0 0 0 ) so that K1 = (−1 1 ) . For λ2 = 2 we obtain(−1 1 0 1 −1 0 ) =⇒ (−1 1 0 0 0 0 ) so that K2 = ( 1 1 ) . Then X = c1 (−1 1 ) + c2 ( 1 1 ) e2t. The line y = −x is not a trajectory of the system. Trajectories are x = −c1 + c2e2t, y = c1 +c2e2t or y = x+2c1. This is a family of lines perpendicular to the line y = −x. All of the constant solutions of the system do, however, lie on the line y = −x. (b) From det(A− λI) = λ2 = 0 we get λ1 = 0 and K = (−1 1 ) . A solution of (A− λ1I)P = K is P = (−1 0 ) so that X = c1 (−1 1 ) + c2 [(−1 1 ) t + (−1 0 )] . All trajectories are parallel to y = −x, but y = −x is not a trajectory. There are con- stant solutions of the system, however, that do lie on the line y = −x. 568 10.2 Homogeneous Linear Systems 49. The system of differential equations is x′1 = 2x1 + x2 x′2 = 2x2 x′3 = 2x3 x′4 = 2x4 + x5 x′5 = 2x5. We see immediately that x2 = c2e2t, x3 = c3e2t, and x5 = c5e2t. Then x′1 = 2x1 + c2e 2t so x1 = c2te2t + c1e2t, and x′4 = 2x4 + c5e 2t so x4 = c5te2t + c4e2t. The general solution of the system is X = c2te 2t + c1e2t c2e 2t c3e 2t c5te 2t + c4e2t c5e 2t = c1 1 0 0 0 0 e2t + c2 1 0 0 0 0 te2t + 0 1 0 0 0 e2t + c3 0 0 1 0 0 e2t + c4 0 0 0 1 0 e2t + c5 0 0 0 1 0 te2t + 0 0 0 0 1 e2t = c1K1e2t + c2 K1te2t + 0 1 0 0 0 e2t + c3K2e2t + c4K3e2t + c5 K3te2t + 0 0 0 0 1 e2t . There are three solutions of the form X = Ke2t, where K is an eigenvector, and two solutions of the form X = Kte2t + Pe2t. See (12) in the text. From (13) and (14) in the text (A− 2I)K1 = 0 and 569 10.2 Homogeneous Linear Systems (A− 2I)K2 = K1. This implies 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 p1 p2 p3 p4 p5 = 1 0 0 0 0 , so p2 = 1 and p5 = 0, whilep1, p3, and p4 are arbitrary. Choosing p1 = p3 = p4 = 0 we have P = 1 0 0 0 0 . Therefore a solution is X = 1 0 0 0 0 te2t + 1 0 0 0 0 e2t. Repeating for K3 we find P = 0 0 0 0 1 , so another solution is X = 0 0 0 1 0 te2t + 0 0 0 0 1 e2t. 50. From x = 2 cos 2t− 2 sin 2t, y = − cos 2t we find x + 2y = −2 sin 2t. Then (x + 2y)2 = 4 sin2 2t = 4(1− cos2 2t) = 4− 4 cos2 2t = 4− 4y2 and x2 + 4xy + 4y2 = 4− 4y2 or x2 + 4xy + 8y2 = 4. This is a rotated conic section and, from the discriminant b2 − 4ac = 16 − 32 < 0, we see that the curve is an ellipse. 51. Suppose the eigenvalues are α ± iβ, β > 0. In Problem 36 the eigenvalues are 5 ± 3i, in Problem 37 they are ±3i, and in Problem 38 they are −1± 2i. From Problem 47 we deduce that the phase portrait will consist of a family of closed curves when α = 0 and spirals when α �= 0. The origin will be a repellor when α > 0, and an attractor when α < 0. 570 10.3 Solution by Diagonalization 52. (a) The given system can be written as x′′1 = − k1 + k2 m1 x1 + k2 m1 x2, x ′′ 2 = k2 m2 x1 − k2 m2 x2. In terms of matrices this is X′′ = AX where X = ( x1 x2 ) and A = − k1 + k2 m1 k2 m1 k2 m2 − k2 m2 . (b) If X = Keωt then X′′ = ω2Keωt and AX = AKeωt so that X′′ = AX becomes ω2Keωt = AKeωt or (A− ω2I)K = 0. Now let ω2 = λ. (c) When m1 = 1, m2 = 1, k1 = 3, and k2 = 2 we obtain A = (−5 2 2 −2 ) . The eigenvalues and corresponding eigenvectors of A are λ1 = −1, λ2 = −6, K1 = ( 1 2 ) , K2 = (−2 1 ) . Since ω1 = i, ω2 = −i, ω3 = √ 6 i, and ω4 = − √ 6 i a solution is X = c1 ( 1 2 ) eit + c2 ( 1 2 ) e−it + c3 (−2 1 ) e √ 6 it + c4 (−2 1 ) e− √ 6 it. (d) Using eit = cos t + i sin t and e √ 6 it = cos √ 6 t + i sin √ 6 t the preceding solution can be rewritten as X = c1 ( 1 2 ) (cos t + i sin t) + c2 ( 1 2 ) (cos t− i sin t) + c3 (−2 1 ) (cos √ 6 t + i sin √ 6 t) + c4 (−2 1 ) (cos √ 6 t + i sin √ 6 t) = (c1 + c2) ( 1 2 ) cos t + i(c1 − c2) ( 1 2 ) sin t + (c3 + c4) (−2 1 ) cos √ 6 t + i(c3 − c4) (−2 1 ) sin √ 6 t = b1 ( 1 2 ) cos t + b2 ( 1 2 ) sin t + b3 (−2 1 ) cos √ 6 t + b4 (−2 1 ) sin √ 6 t where b1 = c1 + c2, b2 = i(c1 − c2), b3 = c3 + c4, and b4 = i(c3 − c4). EXERCISES 10.3 Solution by Diagonalization 1. λ1 = 7, λ2 = −4, K1 = ( 3 1 ) , K2 = (−2 3 ) , P = ( 3 −2 1 3 ) ; X = PY = ( 3 −2 1 3 ) ( c1e 7t c2e −4t ) = ( 3c1e7t − 2c2e−4t c1e 7t + 3c2e−4t ) = c1 ( 3 1 ) e7t + c2 (−2 3 ) e−4t 571 10.3 Solution by Diagonalization 2. λ1 = 0, λ2 = 1, K1 = ( 1 −1 ) , K2 = ( 1 1 ) , P = ( 1 1 −1 1 ) ; X = PY = ( 1 1 −1 1 ) ( c1 c2e t ) = ( c1 + c2et −c1 + c2et ) = c1 ( 1 −1 ) + c2 ( 1 1 ) et 3. λ1 = 12 , λ2 = 3 2 , K1 = ( 1 −2 ) , K2 = ( 1 2 ) , P = ( 1 1 −2 2 ) ; X = PY = ( 1 1 −2 2 ) ( c1e t/2 c2e 3t/2 ) = ( c1e t/2 + c2e3t/2 −2c1et/2 + 2c2e3t/2 ) = c1 ( 1 −2 ) et/2 + c2 ( 1 2 ) e3t/2 4. λ1 = − √ 2 , λ2 = √ 2 , K1 = ( −1 1 + √ 2 ) , K2 = ( −1 1−√2 ) , P = ( −1 −1 1 + √ 2 1−√2 ) ; X = PY = ( −1 −1 1 + √ 2 1−√2 ) ( c1e −√2 t c2e √ 2 t ) = ( −c1e−√2 t − c2e√2 t (1 + √ 2 )c1e− √ 2 t + (1−√2 )c2e √ 2 t ) = c1 ( −1 1 + √ 2 ) e− √ 2 t + c2 ( −1 1−√2 ) e √ 2 t 5. λ1 = −4, λ2 = 2, λ3 = 6, K1 = −11 0 , K2 = 11 1 , K3 = 00 1 , P = −1 1 01 1 0 0 1 1 ; X = PY = −1 1 01 1 0 0 1 1 c1e −4t c2e 2t c3e 6t = −c1e −4t + c2e2t c1e −4t + c2e2t c2e 2t + c3e6t = c1 −11 0 e−4t + c2 11 0 e2t + c3 00 1 e6t 6. λ1 = −1, λ2 = 1, λ3 = 4, K1 = −10 1 , K2 = 1−2 1 , K3 = 11 1 , P = −1 1 10 −2 1 1 1 1 ; X = PY = −1 1 10 −2 1 1 1 1 c1e −t c2e t c3e 4t = −c1 + c2e t + c3e4t −2c2et + c3e4t c1e −t + c2et + c3e4t = c1 −10 1 e−t+c2 1−2 1 et+c3 11 1 e4t 7. λ1 = −1, λ2 = 2, λ3 = 2, K1 = 11 1 , K2 = −11 0 , K3 = −10 1 , P = 1 −1 −11 1 0 1 0 1 ; X = PY = 1 −1 −11 1 0 1 0 1 c1e −t c2e 2t c3e 2t = c1e −t − c2e2t − c3e2t c1e −t + c2e2t c1e −t + c3e2t = c1 11 1 e−t + c2 −11 0 e2t + c3 −10 1 e2t 572 10.3 Solution by Diagonalization 8. λ1 = 0, λ2 = 0, λ3 = 0, λ4 = 4, K1 = −1 1 0 0 , K2 = −1 0 1 0 , K3 = −1 0 0 1 , K4 = 1 1 1 1 , P = −1 −1 −1 1 1 0 0 1 0 1 0 1 0 0 1 1 ; X = PY = −1 −1 −1 1 1 0 0 1 0 1 0 1 0 0 1 1 c1 c2 c3 c4e 4t = −c1 − c2 − c3 + c4e4t c1 + c4e4t c2 + c4e4t c3 + c4e4t = c1 −1 1 0 0 + c2 −1 0 1 0 + c3 −1 0 0 1 + c4 1 1 1 1 e4t 9. λ1 = 1, λ2 = 2, λ3 = 3, K1 = 11 1 , K2 = 22 3 , K3 = 34 5 , P = 1 2 31 2 4 1 3 5 ; X = PY = 1 2 31 2 4 1 3 5 c1e t c2e 2t c3e 3t = c1e t + 2c2e2t + 3c3e3t c1e t + 2c2e2t + 4c3e3t c1e t + 3c2e2t + 5c3e3t = c1 11 1 et + c2 22 3 e2t + c3 34 5 e3t 10. λ1 = 0, λ2 = −2 √ 2 , λ3 = 2 √ 2 , K1 = 10 −1 , K2 = 1−√2 1 , K3 = 1√2 1 , P = 1 1 10 −√2 √2 −1 1 1 ; X = PY = 1 1 10 −√2 √2 −1 1 1 c1c2e−2√2 t c3e 2 √ 2 t = c1 + c2e −2√2 t + c3e2 √ 2 t −√2 c2e−2 √ 2 t + √ 2 c3e2 √ 2 t −c1 + c2e−2 √ 2 t + c3e2 √ 2 t = c1 10 −1 + c2 1−√2 1 e−2√2 t + c3 1√2 1 e2√2 t 11. (a) Since M = ( m1 0 0 m2 ) is a diagonal matrix with nonzero diagonal entries, it has an inverse. Writing the system in the form m1x ′′ 1 + (k1 + k2)x1 − k2x2 = 0 m2x ′′ 2 − k2x1 + k2x2 = 0 we see that K = ( k1 + k2 −k2 −k2 k2 ) . (b) Since M has an inverse, MX′′ + KX = 0 can be written as X′′ + M−1KX = 0 or X′′ + BX = 0 where 573 10.3 Solution by Diagonalization B = M−1K = 1m1 0 0 1 m2 ( k1 + k2 −k2 −k2 k2 ) = k1 + k2 m1 − k2 m1 − k2 m2 k2 m2 . (c) With m1 = 1,m2 = 1, k1 = 3, and k2 = 2 we have B = ( 5 −2 −2 2 ) . The eigenvalues of B are λ1 = 1 and λ2 = 6 with corresponding eigenvectors ( 1 2 ) and (−2 1 ) . Letting X = PY the system can be written PY′′ + BPY = 0 or Y′′ + P−1BPY = 0 where ( 1 −2 2 1 ) and P−1BP = ( 1 0 0 6 ) . The system is then Y′′ + ( 1 0 0 6 ) Y = 0, which is uncoupled and equivalent to y′′1 + y1 = 0 and y ′′ 2 + 6y2 = 0. The solutions are y1 = c1 cos t + c2 sin t and y2 = c3 cos √ 6 t + c4 sin √ 6 t. (d) From X = PY = ( 1 −2 2 1 ) ( y1 y2 ) = ( y1 − 2y2 2y1 + y2 ) we have x1 = c1 cos t + c2 sin t− 2c3 cos √ 6 t− 2c4 sin √ 6 t x1 = 2c1 cos t + 2c2 sin t + c3cos √ 6 t + c4 sin √ 6 t which is the same as X = c1 ( 1 2 ) cos t + c2 ( 1 2 ) sin t + c3 (−2 1 ) cos √ 6 t + c4 (−2 1 ) sin √ 6 t. EXERCISES 10.4 Nonhomogeneous Linear Systems 1. Solving det(A− λI) = ∣∣∣∣ 2− λ 3−1 −2− λ ∣∣∣∣ = λ2 − 1 = (λ− 1)(λ + 1) = 0 we obtain eigenvalues λ1 = −1 and λ2 = 1. Corresponding eigenvectors are K1 = (−1 1 ) and K2 = (−3 1 ) . Thus Xc = c1 (−1 1 ) e−t + c2 (−3 1 ) et. Substituting Xp = ( a1 b1 ) 574 10.4 Nonhomogeneous Linear Systems into the system yields 2a1 + 3b1 = 7 −a1 − 2b1 = −5, from which we obtain a1 = −1 and b1 = 3. Then X(t) = c1 (−1 1 ) e−t + c2 (−3 1 ) et + (−1 3 ) . 2. Solving det(A− λI) = ∣∣∣∣ 5− λ 9−1 11− λ ∣∣∣∣ = λ2 − 16λ + 64 = (λ− 8)2 = 0 we obtain the eigenvalue λ = 8. A corresponding eigenvector is K = ( 3 1 ) . Solving (A− 8I)P = K we obtain P = ( 2 1 ) . Thus Xc = c1 ( 3 1 ) e8t + c2 [( 3 1 ) te8t + ( 2 1 ) e8t ] . Substituting Xp = ( a1 b1 ) into the system yields 5a1 + 9b1 = −2 −a1 + 11b1 = −6, from which we obtain a1 = 1/2 and b1 = −1/2. Then X(t) = c1 ( 3 1 ) e8t + c2 [( 3 1 ) te8t + ( 2 1 ) e8t ] + ( 1 2 − 12 ) . 3. Solving det(A− λI) = ∣∣∣∣ 1− λ 33 1− λ ∣∣∣∣ = λ2 − 2λ− 8 = (λ− 4)(λ + 2) = 0 we obtain eigenvalues λ1 = −2 and λ2 = 4. Corresponding eigenvectors are K1 = ( 1 −1 ) and K2 = ( 1 1 ) . Thus Xc = c1 ( 1 −1 ) e−2t + c2 ( 1 1 ) e4t. Substituting Xp = ( a3 b3 ) t2 + ( a2 b2 ) t + ( a1 b1 ) into the system yields a3 + 3b3 = 2 3a3 + b3 = 0 a2 + 3b2 = 2a3 3a2 + b2 + 1 = 2b3 a1 + 3b1 = a2 3a1 + b1 + 5 = b2 575 10.4 Nonhomogeneous Linear Systems from which we obtain a3 = −1/4, b3 = 3/4, a2 = 1/4, b2 = −1/4, a1 = −2, and b1 = 3/4. Then X(t) = c1 ( 1 −1 ) e−2t + c2 ( 1 1 ) e4t + (− 14 3 4 ) t2 + ( 1 4 − 14 ) t + (−2 3 4 ) . 4. Solving det(A− λI) = ∣∣∣∣ 1− λ −44 1− λ ∣∣∣∣ = λ2 − 2λ + 17 = 0 we obtain eigenvalues λ1 = 1 + 4i and λ2 = 1− 4i. Corresponding eigenvectors are K1 = ( i 1 ) and K2 = (−i 1 ) . Thus Xc = c1 [( 0 1 ) cos 4t + (−1 0 ) sin 4t ] et + c2 [(−1 0 ) cos 4t− ( 0 1 ) sin 4t ] et = c1 (− sin 4t cos 4t ) et + c2 (− cos 4t − sin 4t ) et. Substituting Xp = ( a3 b3 ) t + ( a2 b2 ) + ( a1 b1 ) e6t into the system yields a3 − 4b3 = −4 4a3 + b3 = 1 a2 − 4b2 = a3 4a2 + b2 = b3 −5a1 − 4b1 = −9 4a1 − 5b1 = −1 from which we obtain a3 = 0, b3 = 1, a2 = 4/17, b2 = 1/17, a1 = 1, and b1 = 1. Then X(t) = c1 (− sin 4t cos 4t ) et + c2 (− cos 4t − sin 4t ) et + ( 0 1 ) t + ( 4 17 1 17 ) + ( 1 1 ) e6t. 5. Solving det(A− λI) = ∣∣∣∣ 4− λ 139 6− λ ∣∣∣∣ = λ2 − 10λ + 21 = (λ− 3)(λ− 7) = 0 we obtain the eigenvalues λ1 = 3 and λ2 = 7. Corresponding eigenvectors are K1 = ( 1 −3 ) and K2 = ( 1 9 ) . Thus Xc = c1 ( 1 −3 ) e3t + c2 ( 1 9 ) e7t. Substituting Xp = ( a1 b1 ) et into the system yields 3a1 + 1 3 b1 = 3 9a1 + 5b1 = −10 from which we obtain a1 = 55/36 and b1 = −19/4. Then X(t) = c1 ( 1 −3 ) e3t + c2 ( 1 9 ) e7t + ( 55 36 − 194 ) et. 576 10.4 Nonhomogeneous Linear Systems 6. Solving det(A− λI) = ∣∣∣∣−1− λ 5−1 1− λ ∣∣∣∣ = λ2 + 4 = 0 we obtain the eigenvalues λ1 = 2i and λ2 = −2i. Corresponding eigenvectors are K1 = ( 5 1 + 2i ) and K2 = ( 5 1− 2i ) . Thus Xc = c1 ( 5 cos 2t cos 2t− 2 sin 2t ) + c2 ( 5 sin 2t 2 cos 2t + sin 2t ) . Substituting Xp = ( a2 b2 ) cos t + ( a1 b1 ) sin t into the system yields −a2 + 5b2 − a1 = 0 −a2 + b2 − b1 − 2 = 0 −a1 + 5b1 + a2 + 1 = 0 −a1 + b1 + b2 = 0 from which we obtain a2 = −3, b2 = −2/3, a1 = −1/3, and b1 = 1/3. Then X(t) = c1 ( 5 cos 2t cos 2t− 2 sin 2t ) + c2 ( 5 sin 2t 2 cos 2t + sin 2t ) + ( −3 − 23 ) cos t + (− 13 1 3 ) sin t. 7. Solving det(A− λI) = ∣∣∣∣∣∣∣ 1− λ 1 1 0 2− λ 3 0 0 5− λ ∣∣∣∣∣∣∣ = (1− λ)(2− λ)(5− λ) = 0 we obtain the eigenvalues λ1 = 1, λ2 = 2, and λ3 = 5. Corresponding eigenvectors are K1 = 10 0 , K2 = 11 0 and K3 = 12 2 . Thus Xc = C1 10 0 et + C2 11 0 e2t + C3 12 2 e5t. Substituting Xp = a1b1 c1 e4t into the system yields −3a1 + b1 + c1 = −1 −2b1 + 3c1 = 1 c1 = −2 577 10.4 Nonhomogeneous Linear Systems from which we obtain c1 = −2, b1 = −7/2, and a1 = −3/2. Then X(t) = C1 10 0 et + C2 11 0 e2t + C3 12 2 e5t + − 3 2 − 72 −2 e4t. 8. Solving det(A− λI) = ∣∣∣∣∣∣∣ −λ 0 5 0 5− λ 0 5 0 −λ ∣∣∣∣∣∣∣ = −(λ− 5)2(λ + 5) = 0 we obtain the eigenvalues λ1 = 5, λ2 = 5, and λ3 = −5. Corresponding eigenvectors are K1 = 10 0 , K2 = 11 1 and K3 = 10 −1 . Thus Xc = C1 10 1 e5t + C2 11 1 e5t + C3 10 −1 e−5t. Substituting Xp = a1b1 c1 into the system yields 5c1 = −5 5b1 = 10 5a1 = −40 from which we obtain c1 = −1, b1 = 2, and a1 = −8. Then X(t) = C1 10 1 e5t + C2 11 1 e5t + C3 10 −1 e−5t + −82 −1 . 9. Solving det(A− λI) = ∣∣∣∣−1− λ −23 4− λ ∣∣∣∣ = λ2 − 3λ + 2 = (λ− 1)(λ− 2) = 0 we obtain the eigenvalues λ1 = 1 and λ2 = 2. Corresponding eigenvectors are K1 = ( 1 −1 ) and K2 = (−4 6 ) . Thus Xc = c1 ( 1 −1 ) et + c2 (−4 6 ) e2t. Substituting Xp = ( a1 b1 ) 578 10.4 Nonhomogeneous Linear Systems into the system yields −a1 − 2b1 = −3 3a1 + 4b1 = −3 from which we obtain a1 = −9 and b1 = 6. Then X(t) = c1 ( 1 −1 ) et + c2 (−4 6 ) e2t + (−9 6 ) . Setting X(0) = (−4 5 ) we obtain c1 − 4c2 − 9 = −4 −c1 + 6c2 + 6 = 5. Then c1 = 13 and c2 = 2 so X(t) = 13 ( 1 −1 ) et + 2 (−4 6 ) e2t + (−9 6 ) . 10. (a) Let I = ( i2 i3 ) so that I′ = (−2 −2 −2 −5 ) I + ( 60 60 ) and Ic = c1 ( 2 −1 ) e−t + c2 ( 1 2 ) e−6t. If Ip = ( a1 b1 ) then Ip = ( 30 0 ) so that I = c1 ( 2 −1 ) e−t + c2 ( 1 2 ) e−6t + ( 30 0 ) . For I(0) = ( 0 0 ) we find c1 = −12 and c2 = −6. (b) i1(t) = i2(t) + i3(t) = −12e−t − 18e−6t + 30. 11. From X′ = ( 3 −3 2 −2 ) X + ( 4 −1 ) we obtain Xc = c1 ( 1 1 ) + c2 ( 3 2 ) et. Then Φ = ( 1 3et 1 2et ) and Φ−1 = ( −2 3 e−t −e−t ) so that U = ∫ Φ−1F dt = ∫ ( −11 5e−t ) dt = ( −11t −5e−t ) and Xp = ΦU = (−11 −11 ) t + (−15 −10 ) . 579 10.4 Nonhomogeneous Linear Systems 12. From X′ = ( 2 −1 3 −2 ) X + ( 0 4 ) t we obtain Xc = c1 ( 1 1 ) et + c2 ( 1 3 ) e−t. Then Φ = ( et e−t et 3e−t ) and Φ−1 = ( 3 2e −t − 12e−t − 12et 12et ) so that U = ∫ Φ−1F dt = ∫ (−2te−t 2tet ) dt = ( 2te−t + 2e−t 2tet − 2et ) and Xp = ΦU = ( 4 8 ) t + ( 0 −4 ) . 13. From X′ = ( 3 −5 3 4 −1 ) X + ( 1 −1 ) et/2 we obtain Xc = c1 ( 10 3 ) e3t/2 + c2 ( 2 1 ) et/2. Then Φ = ( 10e3t/2 2et/2 3e3t/2 et/2 )and Φ−1 = ( 1 4e −3t/2 − 12e−3t/2 − 34e−t/2 52e−t/2 ) so that U = ∫ Φ−1F dt = ∫ ( 3 4e −t − 134 ) dt = (− 34e−t − 134 t ) and Xp = ΦU = (− 132 − 134 ) tet/2 + (− 152 − 94 ) et/2. 14. From X′ = ( 2 −1 4 2 ) X + ( sin 2t 2 cos 2t ) we obtain Xc = c1 (− sin 2t 2 cos 2t ) e2t + c2 ( cos 2t 2 sin 2t ) e2t. Then Φ = (−e2t sin 2t e2t cos 2t 2e2t cos 2t 2e2t sin 2t ) and Φ−1 = (− 12e−2t sin 2t 14e−2t cos 2t 1 2e −2t cos 2t 14e −2t sin 2t ) so that U = ∫ Φ−1F dt = ∫ ( 1 2 cos 4t 1 2 sin 4t ) dt = ( 1 8 sin 4t − 18 cos 4t ) and Xp = ΦU = (− 18 sin 2t cos 4t− 18 cos 2t cos 4t 1 4 cos 2t sin 4t− 14 sin 2t cos 4t ) e2t. 580 10.4 Nonhomogeneous Linear Systems 15. From X′ = ( 0 2 −1 3 ) X + ( 1 −1 ) et we obtain Xc = c1 ( 2 1 ) et + c2 ( 1 1 ) e2t. Then Φ = ( 2et e2t et e2t ) and Φ−1 = ( e−t −e−t −e−2t 2e−2t ) so that U = ∫ Φ−1F dt = ∫ ( 2 −3e−t ) dt = ( 2t 3e−t ) and Xp = ΦU = ( 4 2 ) tet + ( 3 3 ) et. 16. From X′ = ( 0 2 −1 3 ) X + ( 2 e−3t ) we obtain Xc = c1 ( 2 1 ) et + c2 ( 1 1 ) e2t. Then Φ = ( 2et e2t et e2t ) and Φ−1 = ( e−t −e−t −e−2t 2e−2t ) so that U = ∫ Φ−1F dt = ∫ ( 2e−t − e−4t −2e−2t + 2e−5t ) dt = (−2e−t + 14e−4t e−2t − 25e−5t ) and Xp = ΦU = ( 1 10e −3t − 3 − 320e−3t − 1 ) . 17. From X′ = ( 1 8 1 −1 ) X + ( 12 12 ) t we obtain Xc = c1 ( 4 1 ) e3t + c2 (−2 1 ) e−3t. Then Φ = ( 4e3t −2e−3t e3t e−3t ) and Φ−1 = ( 1 6e −3t 1 3e −3t − 16e3t 23e3t ) so that U = ∫ Φ−1F dt = ∫ ( 6te−3t 6te3t ) dt = (−2te−3t − 23e−3t 2te3t − 23e3t ) and Xp = ΦU = (−12 0 ) t + (− 43 − 43 ) . 581 10.4 Nonhomogeneous Linear Systems 18. From X′ = ( 1 8 1 −1 ) X + ( e−t tet ) we obtain Xc = c1 ( 4 1 ) e3t + c2 (−2 1 ) e−3t. Then Φ = ( 4e3t −2e3t e3t e−3t ) and Φ−1 = ( 1 6e −3t 1 3e −3t − 16e3t 23e3t ) so that U = ∫ Φ−1F dt = ∫ ( 1 6e −4t + 13 te −2t − 16e2t + 23 te4t ) dt = (− 124e−4t − 16 te−2t − 112e−2t − 112e2t + 16 te4t − 124e4t ) and Xp = ΦU = ( −tet − 14et − 18e−t − 18et ) . 19. From X′ = ( 3 2 −2 −1 ) X + ( 2 1 ) e−t we obtain Xc = c1 ( 1 −1 ) et + c2 [( 1 −1 ) tet + ( 0 1 2 ) et ] . Then Φ = ( et tet −et 12et − tet ) and Φ−1 = ( e−t − 2te−t −2te−t 2e−t 2e−t ) so that U = ∫ Φ−1F dt = ∫ ( 2e−2t − 6te−2t 6e−2t ) dt = ( 1 2e −2t + 3te−2t −3e−2t ) and Xp = ΦU = ( 1 2 −2 ) e−t. 20. From X′ = ( 3 2 −2 −1 ) X + ( 1 1 ) we obtain Xc = c1 ( 1 −1 ) et + c2 [( 1 −1 ) tet + ( 0 1 2 ) et ] . Then Φ = ( et tet −et 12et − tet ) and Φ−1 = ( e−t − 2te−t −2te−t 2e−t 2e−t ) so that U = ∫ Φ−1F dt = ∫ ( e−t − 4te−t 2e−t ) dt = ( 3e−t + 4te−t −2e−t ) and Xp = ΦU = ( 3 −5 ) . 582 10.4 Nonhomogeneous Linear Systems 21. From X′ = ( 0 −1 1 0 ) X + ( sec t 0 ) we obtain Xc = c1 ( cos t sin t ) + c2 ( sin t − cos t ) . Then Φ = ( cos t sin t sin t − cos t ) and Φ−1 = ( cos t sin t sin t − cos t ) so that U = ∫ Φ−1F dt = ∫ ( 1 tan t ) dt = ( t − ln | cos t| ) and Xp = ΦU = ( t cos t− sin t ln | cos t| t sin t + cos t ln | cos t| ) . 22. From X′ = ( 1 −1 1 1 ) X + ( 3 3 ) et we obtain Xc = c1 (− sin t cos t ) et + c2 ( cos t sin t ) et. Then Φ = (− sin t cos t cos t sin t ) et and Φ−1 = (− sin t cos t cos t sin t ) e−t so that U = ∫ Φ−1F dt = ∫ (−3 sin t + 3 cos t 3 cos t + 3 sin t ) dt = ( 3 cos t + 3 sin t 3 sin t− 3 cos t ) and Xp = ΦU = (−3 3 ) et. 23. From X′ = ( 1 −1 1 1 ) X + ( cos t sin t ) et we obtain Xc = c1 (− sin t cos t ) et + c2 ( cos t sin t ) et. Then Φ = (− sin t cos t cos t sin t ) et and Φ−1 = (− sin t cos t cos t sin t ) e−t so that U = ∫ Φ−1F dt = ∫ ( 0 1 ) dt = ( 0 t ) and Xp = ΦU = ( cos t sin t ) tet. 583 10.4 Nonhomogeneous Linear Systems 24. From X′ = ( 2 −2 8 −6 ) X + ( 1 3 ) 1 t e−2t we obtain Xc = c1 ( 1 2 ) e−2t + c2 [( 1 2 ) te−2t + ( 1 2 1 2 ) e−2t ] . Then Φ = ( 1 t + 12 2 2t + 12 ) e−2t and Φ−1 = (−4t− 1 2t + 1 4 −2 ) e2t so that U = ∫ Φ−1F dt = ∫ ( 2 + 2/t −2/t ) dt = ( 2t + 2 ln t −2 ln t ) and Xp = ΦU = ( 2t + ln t− 2t ln t 4t + 3 ln t− 4t ln t ) e−2t. 25. From X′ = ( 0 1 −1 0 ) X + ( 0 sec t tan t ) we obtain Xc = c1 ( cos t − sin t ) + c2 ( sin t cos t ) . Then Φ = ( cos t sin t − sin t cos t ) t and Φ−1 = ( cos t − sin t sin t cos t ) so that U = ∫ Φ−1F dt = ∫ (− tan2 t tan t ) dt = ( t− tan t − ln | cos t| ) and Xp = ΦU = ( cos t − sin t ) t + ( − sin t sin t tan t ) − ( sin t cos t ) ln | cos t|. 26. From X′ = ( 0 1 −1 0 ) X + ( 1 cot t ) we obtain Xc = c1 ( cos t − sin t ) + c2 ( sin t cos t ) . Then Φ = ( cos t sin t − sin t cos t ) and Φ−1 = ( cos t − sin t sin t cos t ) so that U = ∫ Φ−1F dt = ∫ ( 0 csc t ) dt = ( 0 ln | csc t− cot t| ) and Xp = ΦU = ( sin t ln | csc t− cot t| cos t ln | csc t− cot t| ) . 584 10.4 Nonhomogeneous Linear Systems 27. From X′ = ( 1 2 − 12 1 ) X + ( csc t sec t ) et we obtain Xc = c1 ( 2 sin t cos t ) et + c2 ( 2 cos t − sin t ) et. Then Φ = ( 2 sin t 2 cos t cos t − sin t ) et and Φ−1 = ( 1 2 sin t cos t 1 2 cos t − sin t ) e−t so that U = ∫ Φ−1F dt = ∫ ( 3 2 1 2 cot t− tan t ) dt = ( 3 2 t 1 2 ln | sin t|+ ln | cos t| ) and Xp = ΦU = ( 3 sin t 3 2 cos t ) tet + ( cos t − 12 sin t ) et ln | sin t|+ ( 2 cos t − sin t ) et ln | cos t|. 28. From X′ = ( 1 −2 1 −1 ) X + ( tan t 1 ) we obtain Xc = c1 ( cos t− sin t cos t ) + c2 ( cos t + sin t sin t ) . Then Φ = ( cos t− sin t cos t + sin t cos t sin t ) and Φ−1 = (− sin t cos t + sin t cos t sin t− cos t ) so that U = ∫ Φ−1F dt = ∫ ( 2 cos t + sin t− sec t 2 sin t− cos t ) dt = ( 2 sin t− cos t− ln | sec t + tan t| −2 cos t− sin t ) and Xp = ΦU = ( 3 sin t cos t− cos2 t− 2 sin2 t + (sin t− cos t) ln | sec t + tan t| sin2 t− cos2 t− cos t(ln | sec t + tan t|) ) . 29. From X′ = 1 1 01 1 0 0 0 3 X + e t e2t te3t we obtain Xc = c1 1−1 0 + c2 11 0 e2t + c3 00 1 e3t. Then Φ = 1 e 2t 0 −1 e2t 0 0 0 e3t and Φ−1 = 1 2 − 12 0 1 2e −2t 1 2e −2t 0 0 0 e−3t so that U = ∫ Φ−1F dt = ∫ 1 2e t − 12e2t 1 2e −t + 12 t dt = 1 2e t − 14e2t − 12e−t + 12 t 1 2 t 2 585 10.4 Nonhomogeneous Linear Systems and Xp = ΦU = − 14e2t + 12 te2t −et + 14e2t + 12 te2t 1 2 t 2e3t . 30. From X′ = 3 −1 −11 1 −1 1 −1 1 X + 0t 2et we obtain Xc = c1 11 1 et + c2 11 0 e2t + c3 10 1 e2t. Then Φ = e t e2t e2t et e2t 0 et 0 e2t and Φ−1 = −e −t e−t e−t e−2t 0 −e−2t e−2t −e−2t 0 so that U = ∫ Φ−1F dt = ∫ te −t + 2 −2e−t −te−2t dt = −te −t − e−t + 2t 2e−t 1 2 te −2t + 14e −2t and Xp = ΦU = − 1 2 −1 − 12 t + − 3 4 −1 − 34 + 22 0 et + 22 2 tet. 31. From X′ = ( 3 −1 −1 3 ) X + ( 4e2t 4e4t ) we obtain Φ = (−e4t e2t e4t e2t ) , Φ−1 = (− 12e−4t 12e−4t 1 2e −2t 1 2e −2t ) , and X = ΦΦ−1(0)X(0) + Φ ∫ t 0 Φ−1F ds = Φ · ( 0 1 ) + Φ · ( e−2t + 2t− 1 e2t + 2t− 1 ) = ( 2 2 ) te2t + (−1 1 ) e2t + (−2 2 ) te4t + ( 2 0 ) e4t. 32. From X′ = ( 1 −1 1 −1 ) X + ( 1/t 1/t ) we obtain Φ = ( 1 1 + t 1 t ) , Φ−1 = (−t 1 + t 1 −1 ) , and X = ΦΦ−1(1)X(1) + Φ ∫ t 1 Φ−1F ds = Φ · (−4 3 ) + Φ · ( ln t 0 ) = ( 3 3 ) t− ( 1 4 ) + ( 1 1 ) ln t. 586 10.4 Nonhomogeneous Linear Systems 33. Let I = ( i1 i2 ) so that I′ = (−11 3 3 −3 ) I + ( 100 sin t 0 ) and Ic = c1 ( 1 3 ) e−2t + c2 ( 3 −1 ) e−12t. Then Φ = ( e−2t 3e−12t 3e−2t −e−12t ) , Φ−1 = ( 1 10e 2t 3 10e 2t 3 10e 12t − 110e12t ) , U = ∫ Φ−1F dt = ∫ ( 10e2t sin t 30e12t sin t ) dt = ( 2e2t(2 sin t− cos t) 6 29e 12t(12 sin t− cos t) ) , and Ip = ΦU = ( 332 29 sin t− 7629 cos t 276 29 sin t− 16829 cos t ) so that I = c1 ( 1 3 ) e−2t + c2 ( 3 −1 ) e−12t + Ip. If I(0) = ( 0 0 ) then c1 = 2 and c2 = 629 . 34. (a) The eigenvalues are 0, 1, 3, and 4, with corresponding eigenvectors −6 −4 1 2 , 2 1 0 0 , 3 1 2 1 , and −1 1 0 0 . (b) Φ = −6 2et 3e3t −e4t −4 et e3t e4t 1 0 2e3t 0 2 0 e3t 0 , Φ−1 = 0 0 − 13 23 1 3 e −t 1 3 e −t −2e−t 83 e−t 0 0 23 e −3t − 13 e−3t − 13 e−4t 23 e−4t 0 13 e−4t (c) Φ−1(t)F(t) = 2 3 − 13 e2t 1 3 e −2t + 83 e −t − 2et + 13 t − 13 e−3t + 23e−t 2 3 e −5t + 13 e −4t − 13 te−3t , ∫ Φ−1(t)F(t)dt = − 16 e2t + 23 t − 16 e−2t − 83 e−t − 2et + 16 t2 1 9 e −3t − 23 e−t − 215 e−5t − 112 e−4t + 127 e−3t + 19 te−3t , Xp(t) = Φ(t) ∫ Φ−1(t)F(t)dt = −5e2t − 15 e−t − 127 et − 19 tet + 13 t2et − 4t− 5912 −2e2t − 310 e−t + 127et + 19 tet + 16 t2et − 83 t− 9536 − 32 e2t + 23 t + 29 −e2t + 43 t− 19 , 587 10.4 Nonhomogeneous Linear Systems Xc(t) = Φ(t)C = −6c1 + 2c2et + 3c3e3t − c4e4t −4c1 + c2et + c3e3t + c4e4t c1 + 2c3e3t 2c1 + c3e3t , X(t) = Φ(t)C + Φ(t) ∫ Φ−1(t)F(t)dt = −6c1 + 2c2et + 3c3e3t − c4e4t −4c1 + c2et + c3e3t + c4e4t c1 + 2c3e3t 2c1 + c3e3t + −5e2t − 15 e−t − 127 et − 19 tet + 13 t2et − 4t− 5912 −2e2t − 310 e−t + 127et + 19 tet + 16 t2et − 83 t− 9536 − 32 e2t + 23 t + 29 −e2t + 43 t− 19 (d) X(t) = c1 −6 −4 1 2 + c2 2 1 0 0 et + c3 3 1 2 1 e3t + c4 −1 1 0 0 e4t + −5e2t − 15 e−t − 127 et − 19 tet + 13 t2et − 4t− 5912 −2e2t − 310 e−t + 127et + 19 tet + 16 t2et − 83 t− 9536 − 32 e2t + 23 t + 29 −e2t + 43 t− 19 35. λ1 = −1, λ2 = −2, K1 = ( 1 3 ) , K2 = ( 2 7 ) , P = ( 1 2 3 7 ) , P−1 = ( 7 −2 −3 1 ) , P−1F = ( 34 −14 ) ; Y′ = (−1 0 0 −2 ) Y + ( 34 −14 ) y1 = 34 + c1e−t, y2 = −7 + c2e−2t X = PY = ( 1 2 3 7 ) ( 34 + c1e−t −7 + c2e−2t ) = ( 20 + c1e−t + 2c2e−2t 53 + 3c1e−t + 7c2e−2t ) = c1 ( 1 3 ) e−t + c2 ( 2 7 ) e−2t + ( 20 53 ) 36. λ1 = −1, λ2 = 4, K1 = ( 3 −2 ) , K2 = ( 1 1 ) , P = ( 3 1 −2 1 ) , P−1 = 1 5 ( 1 −1 2 3 ) , P−1F = ( 0 et ) ; Y′ = (−1 0 0 4 ) Y + ( 0 et ) y1 = c1e−t, y2 = −13e t + c2e4t X = PY = ( 3 1 −2 1 ) ( c1e −t − 13et + c2e4t ) = (− 13et + 3c1e−t + c2e4t − 13et − 2c1e−t + c2e4t ) = c1 ( 3 −2 ) e−t + c2 ( 1 1 ) e4t − 1 3 ( 1 1 ) et 37. λ1 = 0, λ2 = 10, K1 = ( 1 −1 ) , K2 = ( 1 1 ) , P = ( 1 1 −1 1 ) , P−1 = 1 2 ( 1 −1 1 1 ) , P−1F = ( t− 4 t + 4 ) ; Y′ = ( 0 0 0 10 ) Y + ( t− 4 t + 4 ) y1 = 1 2 t2 − 4t + c1, y2 = − 110 t− 41 100 + c2e10t 588 10.5 Matrix Exponential X = PY = ( 1 1 −1 1 ) ( 1 2 t 2 − 4t + c1 − 110 t− 41100 + c2e10t ) = ( 1 2 t 2 − 4110 t− 41100 + c1 + c2e10t − 12 t2 + 3910 t− 41100 − c1 + c2e10t ) = c1 ( 1 −1 ) + c2 ( 1 1 ) e10t + 1 2 ( 1 −1 ) t2 + 1 10 (−41 39 ) t− 41 100 ( 1 1 ) 38. λ1 = −1, λ2 = 1, K1 = ( 1 −1 ) , K2 = ( 1 1 ) , P = ( 1 1 −1 1 ) , P−1 = 1 2 ( 1 −1 1 1 ) , P−1F = ( 2− 4e−2t 2 + 4e−2t ) ; Y′ = (−1 0 0 1 ) Y + ( 2− 4e−2t 2 + 4e−2t ) y1 = 2 + 4e−2t + c1e−t, y2 = −2− 43e −2t + c2et X = PY = ( 1 1 −1 1 ) ( 2 + 4e−2t + c1e−t −2− 43e−2t + c2et ) = ( 8 3e −2t + c1e−t + c2et −4− 163 e−2t − c1e−t + c2et ) = c1 ( 1 −1 ) e−t + c2 ( 1 1 ) et + 8 3 ( 1 −2 ) e−2t + ( 0 −4 ) EXERCISES 10.5 Matrix Exponential 1. For A = ( 1 0 0 2 ) we have A2 = ( 1 0 0 2 ) ( 1 0 0 2 ) = ( 1 0 0 4 ) , A3 = AA2 = ( 1 0 0 2 ) ( 1 0 0 4 ) = ( 1 0 0 8 ) , A4 = AA3 = ( 1 0 0 2 ) ( 1 0 0 8 ) = ( 1 0 0 16 ) , and so on. In general Ak = ( 1 0 0 2k ) for k = 1, 2, 3, . . . . Thus eAt = I + A 1! t + A2 2! t2 + A3 3! t3 + · · · = ( 1 0 0 1 ) + 1 1! ( 1 0 0 2 ) t + 1 2! ( 1 0 0 4 ) t2 + 1 3! ( 1 0 0 8 ) t3 + · · · = 1 + t + t2 2! + t3 3! + · · · 0 0 1 + t + (2t)2 2! + (2t)3 3! + · · · = ( et 0 0 e2t ) 589 10.5 Matrix Exponential and e−At = ( e−t 0 0 e−2t ) . 2. For A = ( 0 1 1 0 ) we have A2 = ( 0 1 1 0 ) ( 0 1 1 0 ) = ( 1 0 0 1 ) = I A3 = AA2 = ( 0 1 1 0 ) I = ( 0 1 1 0 ) = A A4 = (A2)2 = I A5 = AA4 = AI = A, and so on. In general, Ak = { A, k = 1, 3, 5, . . . I, k = 2, 4, 6, . . . . Thus eAt = I + A 1! t + A2 2! t2 + A3 3! t3 + · · · = I + At + 1 2! It2 + 1 3! At3 + · · · = I ( 1 + 1 2! t2 + 1 4! t4 + · · · ) + A ( t + 1 3! t3 + 1 5! t5 + · · · ) = I cosh t + A sinh t = ( cosh t sinh t sinh t cosh t ) and e−At = ( cosh(−t) sinh(−t) sinh(−t) cosh(−t) ) = ( cosh t − sinh t − sinh t cosh t ) . 3. For A = 1 1 11 1 1 −2 −2 −2 we have A2 = 1 1 11 1 1 −2 −2 −2 1 1 11 1 1 −2 −2 −2 = 0 0 00 0 0 0 0 0 . Thus, A3 = A4 = A5 = · · · = 0 and eAt = I + At = 1 0 00 1 0 0 0 1 + t t tt t t −2t −2t −2t = t + 1 t tt t + 1 t −2t −2t −2t + 1 . 590 10.5 Matrix Exponential 4. For A = 0 0 03 0 0 5 1 0 we have A2 = 0 0 03 0 0 5 1 0 0 0 03 0 0 5 1 0 = 0 0 00 0 0 3 0 0 A3 = AA2 = 0 0 03 0 0 5 1 0 0 0 00 0 0 3 0 0 = 0 0 00 00 0 0 0 . Thus, A4 = A5 = A6 = · · · = 0 and eAt = I + At + 1 2 A2t2 = 1 0 00 1 0 0 0 1 + 0 0 03t 0 0 5t t 0 + 0 0 00 0 0 3 2 t 2 0 0 = 1 0 03t 1 0 3 2 t 2 + 5t t 1 . 5. Using the result of Problem 1, X = ( et 0 0 e2t ) ( c1 c2 ) = c1 ( et 0 ) + c2 ( 0 et ) . 6. Using the result of Problem 2, X = ( cosh t sinh t sinh t cosh t ) ( c1 c2 ) = c1 ( cosh t sinh t ) + c2 ( sinh t cosh t ) . 7. Using the result of Problem 3, X = t + 1 t tt t + 1 t −2t −2t −2t + 1 c1c2 c3 = c1 t + 1t −2t + c2 tt + 1 −2t + c3 tt −2t + 1 . 8. Using the result of Problem 4, X = 1 0 03t 1 0 3 2 t 2 + 5t t 1 c1c2 c3 = c1 13t 3 2 t 2 + 5t + c2 01 t + c3 00 1 . 9. To solve X′ = ( 1 0 0 2 ) X + ( 3 −1 ) 591 10.5 Matrix Exponential we identify t0 = 0, F(t) = ( 3 −1 ) , and use the results of Problem 1 and equation (6) in the text. X(t) = eAtC + eAt ∫ t t0 e−AsF(s) ds = ( et 0 0 e2t ) ( c1 c2 ) + ( et 0 0 e2t ) ∫ t 0 ( e−s 0 0 e−2s ) ( 3 −1 ) ds = ( c1e t c2e 2t ) + ( et 0 0 e2t ) ∫ t 0 ( 3e−s −e−2s ) ds = ( c1e t c2e 2t ) + ( et 0 0 e2t ) (−3e−s 1 2e −2s ) ∣∣∣∣t 0 = ( c1e t c2e 2t ) + ( et 0 0 e2t ) (−3e−t + 3 1 2e −2t − 12 ) = ( c1e t c2e 2t ) + (−3 + 3et 1 2 − 12e2t ) = c3 ( 1 0 ) et + c4 ( 0 1 ) e2t + ( −3 1 2 ) . 10. To solve X′ = ( 1 0 0 2 ) X + ( t e4t ) we identify t0 = 0, F(t) = ( t e4t ) , and use the results of Problem 1 and equation (6) in the text. X(t) = eAtC + eAt ∫ t t0 e−AsF(s) ds = ( et 0 0 e2t ) ( c1 c2 ) + ( et 0 0 e2t ) ∫ t 0 ( e−s 0 0 e−2s ) ( s e4s ) ds = ( c1e t c2e 2t ) + ( et 0 0 e2t ) ∫ t 0 ( se−s e2s ) ds = ( c1e t c2e 2t ) + ( et 0 0 e2t ) (−se−s − e−s 1 2e 2s ) ∣∣∣∣t 0 = ( c1e t c2e 2t ) + ( et 0 0 e2t ) (−te−t − e−t + 1 1 2e 2t − 12 ) = ( c1e t c2e 2t ) + (−t− 1 + et 1 2e 4t − 12e2t ) = c3 ( 1 0 ) et + c4 ( 0 1 ) e2t + (−t− 1 1 2e 4t ) . 11. To solve X′ = ( 0 1 1 0 ) X + ( 1 1 ) 592 10.5 Matrix Exponential we identify t0 = 0, F(t) = ( 1 1 ) , and use the results of Problem 2 and equation (6) in the text. X(t) = eAtC + eAt ∫ t t0 e−AsF(s) ds = ( cosh t sinh t sinh t cosh t ) ( c1 c2 ) + ( cosh t sinh t sinh t cosh t ) ∫ t 0 ( cosh s − sinh s − sinh s cosh s ) ( 1 1 ) ds = ( c1 cosh t + c2 sinh t c1 sinh t + c2 cosh t ) + ( cosh t sinh t sinh t cosh t ) ∫ t 0 ( cosh s− sinh s − sinh s + cosh s ) ds = ( c1 cosh t + c2 sinh t c1 sinh t + c2 cosh t ) + ( cosh t sinh t sinh t cosh t ) ( sinh s− cosh s − cosh s + sinh s ) ∣∣∣∣t 0 = ( c1 cosh t + c2 sinh t c1 sinh t + c2 cosh t ) + ( cosh t sinh t sinh t cosh t ) ( sinh t− cosh t + 1 − cosh t + sinh t + 1 ) = ( c1 cosh t + c2 sinh t c1 sinh t + c2 cosh t ) + ( sinh2 t− cosh2 t + cosh t + sinh t sinh2 t− cosh2 t + sinh t + cosh t ) = c1 ( cosh t sinh t ) + c2 ( sinh t cosh t ) + ( cosh t sinh t ) + ( sinh t cosh t ) − ( 1 1 ) = c3 ( cosh t sinh t ) + c4 ( sinh t cosh t ) − ( 1 1 ) . 12. To solve X′ = ( 0 1 1 0 ) X + ( cosh t sinh t ) we identify t0 = 0, F(t) = ( cosh t sinh t ) , and use the results of Problem 2 and equation (6) in the text. X(t) = eAtC + eAt ∫ t t0 e−AsF(s) ds = ( cosh t sinh t sinh t cosh t ) ( c1 c2 ) + ( cosh t sinh t sinh t cosh t ) ∫ t 0 ( cosh s − sinh s − sinh s cosh s ) ( cosh s sinh s ) ds = ( c1 cosh t + c2 sinh t c1 sinh t + c2 cosh t ) + ( cosh t sinh t sinh t cosh t ) ∫ t 0 ( 1 0 ) ds = ( c1 cosh t + c2 sinh t c1 sinh t + c2 cosh t ) + ( cosh t sinh t sinh t cosh t ) ( s 0 ) ∣∣∣∣t 0 = ( c1 cosh t + c2 sinh t c1 sinh t + c2 cosh t ) + ( cosh t sinh t sinh t cosh t ) ( t 0 ) = ( c1 cosh t + c2 sinh t c1 sinh t + c2 cosh t ) + ( t cosh t t sinh t ) = c1 ( cosh t sinh t ) + c2 ( sinh t cosh t ) + t ( cosh t sinh t ) . 13. We have X(0) = c1 10 0 + c2 01 0 + c3 00 1 = c1c2 c3 = 1−4 6 . 593 10.5 Matrix Exponential Thus, the solution of the initial-value problem is X = t + 1t −2t − 4 tt + 1 −2t + 6 tt −2t + 1 . 14. We have X(0) = c3 ( 1 0 ) + c4 ( 0 1 ) + (−3 1 2 ) = ( c3 − 3 c4 + 12 ) = ( 4 3 ) . Thus, c3 = 7 and c4 = 52 , so X = 7 ( 1 0 ) et + 5 2 ( 0 1 ) e2t + (−3 1 2 ) . 15. From sI−A = ( s− 4 −3 4 s + 4 ) we find (sI−A)−1 = 3/2 s− 2 − 1/2 s + 2 3/4 s− 2 − 3/4 s + 2 −1 s− 2 + 1 s + 2 −1/2 s− 2 + 3/2 s + 2 and eAt = ( 3 2e 2t − 12e−2t 34e2t − 34e−2t −e2t + e−2t − 12e2t + 32e−2t ) . The general solution of the system is then X = eAtC = ( 3 2e 2t − 12e−2t 34e2t − 34e−2t −e2t + e−2t − 12e2t + 32e−2t )( c1 c2 ) = c1 ( 3 2 −1 ) e2t + c1 (− 12 1 ) e−2t + c2 ( 3 4 − 12 ) e2t + c2 (− 34 3 2 ) e−2t = (1 2 c1 + 1 4 c2 ) ( 3 −2 ) e2t + ( −1 2 c1 − 34c2 ) ( 1 −2 ) e−2t = c3 ( 3 −2 ) e2t + c4 ( 1 −2 ) e−2t. 16. From sI−A = ( s− 4 2 −1 s− 1 ) we find (sI−A)−1 = 2 s− 3 − 1 s− 2 − 2 s− 3 + 2 s− 2 1 s− 3 − 1 s− 2 −1 s− 3 + 2 s− 2 and eAt = ( 2e3t − e2t −2e3t + 2e2t e3t − e2t −e3t + 2e2t ) . The general solution of the system is then 594 10.5 Matrix Exponential X = eAtC = ( 2e3t − e2t −2e3t + 2e2t e3t − e2t −e3t + 2e2t ) ( c1 c2 ) = c1 ( 2 1 ) e3t + c1 (−1 −1 ) e2t + c2 (−2 −1 ) e3t + c2 ( 2 2 ) e2t = (c1 − c2) ( 2 1 ) e3t + (−c1 + 2c2) ( 1 1 ) e2t = c3 ( 2 1 ) e3t + c4 ( 1 1 ) e2t. 17. From sI−A = ( s− 5 9 −1 s + 1 ) we find (sI−A)−1 = 1 s− 2 + 3 (s− 2)2 − 9 (s− 2)2 1 (s− 2)2 1 s− 2 − 3 (s− 2)2 and eAt = ( e2t + 3te2t −9te2t te2t e2t − 3te2t ) . The general solution of the system is then X = eAtC = ( e2t + 3te2t −9te2t te2t e2t − 3te2t ) ( c1 c2 ) = c1 ( 1 0 ) e2t + c1 ( 3 1 ) te2t + c2 ( 0 1 ) e2t + c2 (−9 −3 ) te2t = c1 ( 1 + 3t t ) e2t + c2 ( −9t 1− 3t ) e2t. 18. From sI−A = ( s −1 2 s + 2 ) we find (sI−A)−1 = s + 1 + 1 (s + 1)2 + 1 1 (s + 1)2 + 1 −2 (s + 1)2 + 1 s + 1− 1 (s + 1)2 + 1 and eAt = ( e−t cos t + e−t sin t e−t sin t −2e−t sin t e−t cos t− e−t sin t ) . The general solution of the system is then X = eAtC = ( e−t cos t + e−t sin t e−t sin t −2e−t sin t e−t cos t− e−t sin t ) ( c1 c2 ) = c1 ( 1 0 ) e−t cos t + c1 ( 1 −2 ) e−t sin t + c2 ( 0 1 ) e−t cos t + c2 ( 1 −1 ) e−t sin t 595 10.5 Matrix Exponential = c1 ( cos t + sin t −2 sin t ) e−t + c2( sin t cos t− sin t ) e−t. 19. The eigenvalues are λ1 = 1 and λ2 = 6. This leads to the system et = b0 + b1 e6t = b0 + 6b1, which has the solution b0 = 65e t − 15e6t and b1 = − 15et + 15e6t. Then eAt = b0I + b1A = ( 4 5e t + 15e 6t 2 5e t − 25e6t 2 5e t − 25e6t 15et + 45e6t ) . The general solution of the system is then X = eAtC = ( 4 5e t + 15e 6t 2 5e t − 25e6t 2 5e t − 25e6t 15et + 45e6t )( c1 c2 ) = c1 ( 4 5 2 5 ) et + c1 ( 1 5 − 25 ) e6t + c2 ( 2 5 1 5 ) et + c2 (− 25 4 5 ) e6t = (2 5 c1 + 1 5 c2 ) ( 2 1 ) et + (1 5 c1 − 25c2 ) ( 1 −2 ) e6t = c3 ( 2 1 ) et + c4 ( 1 −2 ) e6t. 20. The eigenvalues are λ1 = 2 and λ2 = 3. This leads to the system e2t = b0 + 2b1 e3t = b0 + 3b1, which has the solution b0 = 3e2t − 2e3t and b1 = −e2t + e3t. Then eAt = b0I + b1A = ( 2e2t − e3t −2e2t + 2e3t e2t − e3t −e2t + 2e3t ) . The general solution of the system is then X = eAtC = ( 2e2t − e3t −2e2t + 2e3t e2t − e3t −e2t + 2e3t ) ( c1 c2 ) = c1 ( 2 1 ) e2t + c1 (−1 −1 ) e3t + c2 (−2 −1 ) e2t + c2 ( 2 2 ) e3t = (c1 − c2) ( 2 1 ) e2t + (−c1 + 2c2) ( 1 1 ) e3t = c3 ( 2 1 ) e2t + c4 ( 1 1 ) e3t. 596 10.5 Matrix Exponential 21. The eigenvalues are λ1 = −1 and λ2 = 3. This leads to the system e−t = b0 − b1 e3t = b0 + 3b1, which has the solution b0 = 34e −t + 14e 3t and b1 = − 14e−t + 14e3t. Then eAt = b0I + b1A = ( e3t −2e−t + 2e3t 0 e−t ) . The general solution of the system is then X = eAtC = ( e3t −2e−t + 2e3t 0 e−t )( c1 c2 ) = c1 ( 1 0 ) e3t + c2 (−2 1 ) e−t + c2 ( 2 0 ) e3t = c2 (−2 1 ) e−t + (c1 + 2c2) ( 1 0 ) e3t = c3 (−2 1 ) e−t + c4 ( 1 0 ) e3t. 22. The eigenvalues are λ1 = 14 and λ2 = 1 2 . This leads to the system et/4 = b0 + 1 4 b1 et/2 = b0 + 1 2 b1, which has the solution b0 = 2et/4 + et/2 and b1 = −4et/4 + 4et/2. Then eAt = b0I + b1A = (−2et/4 + 3et/2 6et/4 − 6et/2 −et/4 + et/2 3et/4 − 2et/2 ) . The general solution of the system is then X = eAtC = (−2et/4 + 3et/2 6et/4 − 6et/2 −et/4 + et/2 3et/4 − 2et/2 ) ( c1 c2 ) = c1 (−2 −1 ) et/4 + c1 ( 3 1 ) et/2 + c2 ( 6 3 ) et/4 + c2 (−6 −2 ) et/2 = (−c1 + 3c2) ( 2 1 ) et/4 + (c1 − 2c2) ( 3 1 ) et/2 = c3 ( 2 1 ) et/4 + c4 ( 3 1 ) et/2. 23. From equation (3) in the text we have eDt = I + tD + t2 2! D2 + t3 3! D3 + · · · so that PeDtP−1 = PP−1 + t(PDP−1) + t2 2! (PD2P−1) + t3 3! (PD3P−1) + · · · . 597 10.5 Matrix Exponential But PP−1 + I, PDP−1 = A and PDnP−1 = An (see Problem 37, Exercises 8.12). Thus, PeDtP−1 = I + tA + t2 2! A2 + t3 3! A3 + · · · = eAt. 24. From equation (3) in the text eDt = 1 0 · · · 0 0 1 · · · 0 ... ... . . . ... 0 0 · · · 1 + λ1 0 · · · 0 0 λ2 · · · 0 ... ... . . . ... 0 0 · · · λn + 12! t2 λ21 0 · · · 0 0 λ22 · · · 0 ... ... . . . ... 0 0 · · · λ2n + 1 3! t3 λ31 0 · · · 0 0 λ32 · · · 0 ... ... . . . ... 0 0 · · · λ3n + · · · = 1 + λ1t + 12! (λ1t) 2 + · · · 0 · · · 0 0 1 + λ2t + 12! (λ2t) 2 + · · · · · · 0 ... ... . . . ... 0 0 · · · 1 + λnt + 12! (λnt)2 + · · · = eλ1t 0 · · · 0 0 eλ2t · · · 0 ... ... . . . ... 0 0 · · · eλnt 25. From Problems 23 and 24 and equation (1) in the text X = eAtC = PeDtP−1C = ( e3t e5t e3t 3e5t ) ( e3t 0 0 e5t ) ( 3 2e −3t − 12e−3t − 12e−5t 12e−5t )( c1 c2 ) = ( 3 2e 3t − 12e5t − 12e3t + 12e5t 3 2e 3t − 32e5t − 12e3t + 32e5t )( c1 c2 ) . 26. From Problems 23 and 24 and equation (1) in the text X = eAtC = PeDtP−1C = (−et e3t et e3t ) ( et 0 0 e3t ) (− 12e−t 12e−t 1 2e 3t 1 2e −3t )( c1 c2 ) = ( 1 2e t + 12e 9t − 12et + 12e3t − 12et + 12e9t 12et + 12e3t )( c1 c2 ) . 27. (a) The following commands can be used in Mathematica: A={{4, 2},{3, 3}}; c={c1, c2}; m=MatrixExp[A t]; sol=Expand[m.c] Collect[sol, {c1, c2}]//MatrixForm 598 10.5 Matrix Exponential The output gives x(t) = c1 ( 2 5 et + 3 5 e6t ) + c2 ( −2 5 et + 2 5 e6t ) y(t) = c1 ( −3 5 et + 3 5 e6t ) + c2 ( 3 5 et + 2 5 e6t ) . The eigenvalues are 1 and 6 with corresponding eigenvectors(−2 3 ) and ( 1 1 ) , so the solution of the system is X(t) = b1 (−2 3 ) et + b2 ( 1 1 ) e6t or x(t) = −2b1et + b2e6t y(t) = 3b1et + b2e6t. If we replace b1 with − 15c1 + 15c2 and b2 with 35c1 + 25c2, we obtain the solution found using the matrix exponential. (b) x(t) = c1e−2t cos t− (c1 + c2)e−2t sin t y(t) = c2e−2t cos t + (2c1 + c2)e−2t sin t 28. x(t) = c1(3e−2t − 2e−t) + c3(−6e−2t + 6e−t) y(t) = c2(4e−2t − 3e−t) + c4(4e−2t − 4e−t) z(t) = c1(e−2t − e−t) + c3(−2e−2t + 3e−t) w(t) = c2(−3e−2t + 3e−t) + c4(−3e−2t + 4e−t) 29. If det(sI−A) = 0, then s is an eigenvalue of A. Thus sI−A has an inverse if s is not an eigenvalue of A. For the purposes of the discussion in this section, we take s to be larger than the largest eigenvalue of A. Under this condition sI−A has an inverse. 30. Since A3 = 0, A is nilpotent. Since eAt = I + At + A2 t2 2! + · · ·+ Ak t k k! + · · · , if A is nilpotent and Am = 0, then Ak = 0 for k ≥ m and eAt = I + At + A2 t2 2! + · · ·+ Am−1 t m−1 (m− 1)! . In this problem A3 = 0, so eAt = I + At + A2 t2 2 = 1 0 00 1 0 0 0 1 + −1 1 1−1 0 1 −1 1 1 t + −1 0 10 0 0 −1 0 1 t22 = 1− t− t 2/2 t t + t2/2 −t 1 t −t− t2/2 t 1 + t + t2/2 and the solution of X′ = AX is X(t) = eAtC = eAt c1c2 c3 = c1(1− t− t 2/2) + c2t + c3(t + t2/2) −c1t + c2 + c3t c1(−t− t2/2) + c2t + c3(1 + t + t2/2) . 599 10.5 Matrix Exponential CHAPTER 10 REVIEW EXERCISES CHAPTER 10 REVIEW EXERCISES 1. If X = k ( 4 5 ) , then X′ = 0 and k ( 1 4 2 −1 ) ( 4 5 ) − ( 8 1 ) = k ( 24 3 ) − ( 8 1 ) = ( 0 0 ) . We see that k = 13 . 2. Solving for c1 and c2 we find c1 = − 34 and c2 = 14 . 3. Since 4 6 61 3 2 −1 −4 −3 31 −1 = 124 −4 = 4 31 −1 , we see that λ = 4 is an eigenvalue with eigenvector K3. The corresponding solution is X3 = K3e4t. 4. The other eigenvalue is λ2 = 1− 2i with corresponding eigenvector K2 = ( 1 −i ) . The general solution is X(t) = c1 ( cos 2t − sin 2t ) et + c2 ( sin 2t cos 2t ) et. 5. We have det(A− λI) = (λ− 1)2 = 0 and K = ( 1 −1 ) . A solution to (A− λI)P = K is P = ( 0 1 ) so that X = c1 ( 1 −1 ) et + c2 [( 1 −1 ) tet + ( 0 1 ) et ] . 6. We have det(A− λI) = (λ + 6)(λ + 2) = 0 so that X = c1 ( 1 −1 ) e−6t + c2 ( 1 1 ) e−2t. 7. We have det(A− λI) = λ2 − 2λ + 5 = 0. For λ = 1 + 2i we obtain K1 = ( 1 i ) and X1 = ( 1 i ) e(1+2i)t = ( cos 2t − sin 2t ) et + i ( sin 2t cos 2t ) et. Then X = c1 ( cos 2t − sin 2t ) et + c2 ( sin 2t cos 2t ) et. 8. We have det(A− λI) = λ2 − 2λ + 2 = 0. For λ = 1 + i we obtain K1 = ( 3− i 2 ) and X1 = ( 3− i 2 ) e(1+i)t = ( 3 cos t + sin t 2 cos t ) et + i (− cos t + 3 sin t 2 sin t ) et. 600 CHAPTER 10 REVIEW EXERCISES Then X = c1 ( 3 cos t + sint 2 cos t ) et + c2 (− cos t + 3 sin t 2 sin t ) et. 9. We have det(A− λI) = −(λ− 2)(λ− 4)(λ + 3) = 0 so that X = c1 −23 1 e2t + c2 01 1 e4t + c3 712 −16 e−3t. 10. We have det(A−λI) = −(λ+2)(λ2−2λ+3) = 0. The eigenvalues are λ1 = −2, λ2 = 1+ √ 2i, and λ2 = 1− √ 2i, with eigenvectors K1 = −75 4 , K2 = 11 2 √ 2 i 1 , and K3 = 1− 12√2 i 1 . Thus X = c1 −75 4 e−2t + c2 10 1 cos√2t− 012√2 0 sin√2t et + c3 012√2 0 cos√2t + 10 1 sin√2t et = c1 −75 4 e−2t + c2 cos √ 2t − 12 √ 2 sin √ 2t cos √ 2t et + c3 sin √ 2t 1 2 √ 2 cos √ 2t sin √ 2t et. 11. We have Xc = c1 ( 1 0 ) e2t + c2 ( 4 1 ) e4t. Then Φ = ( e2t 4e4t 0 e4t ) , Φ−1 = ( e−2t −4e−2t 0 e−4t ) , and U = ∫ Φ−1F dt = ∫ ( 2e−2t − 64te−2t 16te−4t ) dt = ( 15e−2t + 32te−2t −e−4t − 4te−4t ) , so that Xp = ΦU = ( 11 + 16t −1− 4t ) . 12. We have Xc = c1 ( 2 cos t − sin t ) et + c2 ( 2 sin t cos t ) et. Then Φ = ( 2 cos t 2 sin t − sin t cos t ) et, Φ−1 = ( 1 2 cos t − sin t 1 2 sin t cos t ) e−t, and U = ∫ Φ−1F dt = ∫ ( cos t− sec t sin t ) dt = ( sin t− ln | sec t + tan t| − cos t ) , 601 CHAPTER 10 REVIEW EXERCISES so that Xp = ΦU = ( −2 cos t ln | sec t + tan t| −1 + sin t ln | sec t + tan t| ) et. 13. We have Xc = c1 ( cos t + sin t 2 cos t ) + c2 ( sin t− cos t 2 sin t ) . Then Φ = ( cos t + sin t sin t− cos t 2 cos t 2 sin t ) , Φ−1 = ( sin t 12 cos t− 12 sin t − cos t 12 cos t + 12 sin t ) , and U = ∫ Φ−1F dt = ∫ ( 1 2 sin t− 12 cos t + 12 csc t − 12 sin t− 12 cos t + 12 csc t ) dt = (− 12 cos t− 12 sin t + 12 ln | csc t− cot t| 1 2 cos t− 12 sin t + 12 ln | csc t− cot t| ) , so that Xp = ΦU = (−1 −1 ) + ( sin t sin t + cos t ) ln | csc t− cot t|. 14. We have Xc = c1 ( 1 −1 ) e2t + c2 [( 1 −1 ) te2t + ( 1 0 ) e2t ] . Then Φ = ( e2t te2t + e2t −e2t −te2t ) , Φ−1 = (−te−2t −te−2t − e−2t e−2t e−2t ) , and U = ∫ Φ−1F dt = ∫ ( t− 1 −1 ) dt = ( 1 2 t 2 − t −t ) , so that Xp = ΦU = (− 12 1 2 ) t2e2t + (−2 1 ) te2t. 15. (a) Letting K = k1k2 k3 we note that (A− 2I)K = 0 implies that 3k1 + 3k2 + 3k3 = 0, so k1 = −(k2 + k3). Choosing k2 = 0, k3 = 1 and then k2 = 1, k3 = 0 we get K1 = −10 1 and K2 = −11 0 , respectively. Thus, X1 = −10 1 e2t and X2 = −11 0 e2t 602 CHAPTER 10 REVIEW EXERCISES are two solutions. (b) From det(A− λI) = λ2(3− λ) = 0 we see that λ1 = 3, and 0 is an eigenvalue of multiplicity two. Letting K = k1k2 k3 , as in part (a), we note that (A − 0I)K = AK = 0 implies that k1 + k2 + k3 = 0, so k1 = −(k2 + k3). Choosing k2 = 0, k3 = 1, and then k2 = 1, k3 = 0 we get K2 = −10 1 and K3 = −11 0 , respectively. Since the eigenvector corresponding to λ1 = 3 is K1 = 11 1 , the general solution of the system is X = c1 11 1 e3t + c2 −10 1 + c3 −11 0 . 16. For X = ( c1 c2 ) et we have X′ = X = IX. 603
Compartilhar