3 Systems of differential equations The five vectors vl, v2, v3, v4, and v5 are easily seen to be linearly independent. Hence, every solution x(t) is of the form x(t)=c 1 1 0 0 0 0 1 0 0 1 +c4 0 +cz 0 +c3 0 +c5et5t 0 0 0 1 -1 -1 -1 -1 1 2 3 4 5 EXERCISES In each of Problems 1-6 find all solutions of the given differential equation. 1. x=(~ . x= ( -2 -4 j)x 2 . -3) 1 X 3. 2 x=(! 20 i)x s. x= ( -7~ 05 ~)x -14 -I~)x x= ( -1b -2 I -I 2 3 u-[~ 10I46 2II59 4230'~] 4. 0 In each of Problems 7-12, solve the given initial-value problem. x=(! Dx, x(O)=n) s. x=( -i -nx, 1 -I) 9. x={! 3 -I x, x(O)=( =D 3 -I -I2 10. x=O r)x, x(O)= ( ~;) 10 7. x(O)=(~) -3 u.x=(~ -I -I -2~)x, x(O)= ( -~) 12. I x(O)=( -7!) x=(-! 2I -qx, -3 13. (a) Show that e"Av,t0 constant, is a solution of x=Ax if Av=A.v. (b) Solve the initial-value problern x=( (see Exercise 12). 340 -! ~ =Dx, x(I)=(_D 3.9 Complex roots 14. Three solutions of the equation x= Ax are Find the eigenvalues and eigenvectors of A. 15. Show that the eigenvalues of A- 1 are the reciprocals of the eigenvalues of A. 16. Show that the eigenvalues of An are the nth power of the eigenvalues of A. 17. Show that ;\\.=0 is an eigenvalue of A if detA=O. 18. Show, by example, that the eigenvalues of A + B are not necessarily the sum of an eigenvalue of A and an eigenvalue of B. 19. Show, by example, that the eigenvalues of AB are not necessarily the product of an eigenvalue of A with an eigenvalue of B. 20. Show that the matrices A and T- 1AT have the same characteristic polynomial. 21. Suppose that either B-I or A-I exists. Prove that AB and BA have the same eigenvalues. Hint: Use Exercise 20. (This result is true even if neither B- 1 or A-I exist; however, it is more difficult to prove then.) 3.9 Complex roots If X= a + i is a complex eigenvalue of A with eigenvector v=v 1 + iv2 , then x( t) = eMv is a complex-valued solution of the differential equation x=Ax. (I) This complex-valued solution gives rise to two real-valued solutions, as we now show. Lemma 1. Let x(t)=y(t)+ iz(t) be a complex-valued so/ution of (1). Then, both y(t) and z(t) are rea/-valued so/utions of (1). PROOF. If x(t)=y(t)+ iz(t) is a complex-valued solution of (1), then y( t) + iz( t) = A(y( t) + iz( t)) = Ay( t) + iAz( t). (2) Equating real and imaginary parts of (2) gives y( t) = Ay( t) and i:( t) = Az( t). Consequently, both y(t)=Re{x(t)} and z(t)=Im{x(t)} are real-va1ued solutions of (1). 0 The complex-valued function x( t) = e