We know from the theor that when the matrix has a complex eigenvalue r ff ± ifi with eigenvector u a ± ib then the two linearl independent solutions t

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1 Solutions, 36-XXIII April 9, 23 23( 4/5) Complex eigenvalues 9.6[,2*,5*,6,3(a*,b,c,d)] CAUTION there ma be errors Problem 9.6. Find a general solution of the sstem x Ax for the matrix 2 4 A 2 2 Solution Form the characteristic polnomial P (r) det (A 2 ri) det r r (r + 2)(r 2) + 8 r 2 +4(r +2i)(r 2i) r +2i; the results for r 2 2i are found b complex conjugation of the answer (i.e. change the sign of i everwhere). We have 2 2i 4 x 2 2 2i x 2 i 2 + i

2 We know from the theor that when the matrix has a complex eigenvalue r ff ± ifi with eigenvector u a ± ib then the two linearl independent solutions to x Ax are x e fft cos fita e fft sin fitb, x 2 e fft sin fita+ e fft cos fitb. Here this gives, with r +2i 2 x cos 2t sin 2t x 2 sin 2t 2 + cos 2t so that the general solution is given b 2 cos 2t x gen (t) C + C2 cos 2t + sin 2t 2 sin 2t sin 2t cos 2t 2

3 2 Problem Find a general solution of the sstem x Ax for the matrix 2 2 A 4 2 Solution Form the characteristic polnomial P (r) det (A 2 ri) det r r (r + 2)(r 2) + 8 r 2 +4(r +2i)(r 2i) r +2i; the results for r 2 2i are found b complex conjugation of the answer (i.e. change the sign of i everwhere). We have 2 2i 2 x 4 2 2i x i + i We know from the theor that when the matrix has a complex eigenvalue r ff ± ifi with eigenvector u a ± ib then the two linearl independent solutions to x Ax are x e fft cos fita e fft sin fitb, x 2 e fft sin fita+ e fft cos fitb. Here this gives, with r +2i x cos 2t sin 2t x 2 sin 2t + cos 2t so that the general solution is given b cos 2t x gen (t) C + C2 cos 2t + sin 2t sin 2t sin 2t cos 2t 3

4 3 Problem Find a fundamental matrix for the sstem x Ax for the matrix 2 A 8 Solution Form the characteristic polnomial P (r) det (A ri) det r 2 8 r (r +) 2 +6(r 4i)(r +4i) r +4i; the results for r 2 4i are found b complex conjugation of the answer (i.e. change the sign of i everwhere). We have 4i 2 x 8 4i x 2i + i 2 We know from the theor that when the matrix has a complex eigenvalue r ff ± ifi with eigenvector u a ± ib then the two linearl independent solutions to x Ax are x e fft cos fita e fft sin fitb, x 2 e fft sin fita+ e fft cos fitb. Here this gives, with r +4i x cos 2t sin 2t x 2 sin 2t + cos 2t 2 2 so that the Fundamental Matrix is given b cos 2t sin 2t Φ(t) 2 sin 2t 2 cos 2t 4

5 4 Problem 9.6.3a Find the solution to the IVP 3 x x ; x() 2 Solution Form the characteristic polnomial P (r) det (A 3 ri) det r 2 r x (r + )(r +3)+2r 2 +4r +5(r +2) 2 + r 2+i; the eigenvector is found from 3+2 i x 2 +2 i x + i i where the first equation was used to find the components of the eigenvector. Following problem (), we find an independent set as (with r 2+i) x e 2t cos t cos t e 2t sin t e 2t sin t cos t x 2 e 2t sin t + e 2t cos t e 2t sin t cos t sin t so that the general solution is given b cos x gen (t) C e 2t t + C2e sin t cos 2t t sin t cos t sin t A fundamental matrix for the sstem is then cos Φ(t) e 2t t sin t cos t 5 sin t cos t sin t ;

6 the general solution is given in terms of the Fundamental matrix b the equivalent form x gen (t) Φ(t) Solving for the IC Φ(()) C C C C Finall, the solution to the IVP is x(t) Φ(t) (x 2 (t) x (t)) e t cos t + sin t 2 sin t 6

7 5 Problem 9.6.3d modified Find the solution to the IVP 3 x x ; x(ß2) 5 Solution Form the characteristic polnomial P (r) det (A 3 ri) det r 5 r (r + )(r 3)+3r 2 2r +2(r ) 2 + r +i; the eigenvector is found from 3 i x 5 i x 2+i 2 + i 5 5 where the second equation was used to find the components of the eigenvector. Following problem (), we find an independent set as (with r +i) 2 x e t cos t 2 cos t sin e t sin t e t t 5 5 cos t x 2 e t sin t e t cos t so that the general solution is given b 2 cos t sin x gen (t) C e t t 5 cos t e t cos t + 2 sin t 5 sin t + e t cos t + 2 sin t 5 sin t A fundamental matrix for the sstem is then 2 cos t sin Φ(t) e t t cos t + 2 sin t 5 cos t 5 sin t 7 ;

8 the general solution is given in terms of the Fundamental matrix b the equivalent form x gen (t) Φ(t) Solving for the IC Φ((ß2)) 2 e ß2 5 C C C e ß2 5 Finall, the solution to the IVP is C 5 2 x(t) e ß2 5 Φ(t) 2 e ß2 5 2 e ß2 (2x (t)+x 2 (t)) 5 cos e t ß2 t sin t + 2 cos t 8