SOLUTIONS TO PRACTICE EXAM 3, SPRING 2004

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1 8034 SOLUTIONS TO PRACTICE EXAM 3, SPRING 004 Problem Let A be a real matrix and consider the linear system of first order differential equations, y y (t) Ay(t), y(t) (t) y (t) Let α be a real number, let β be a nonzero real number, and Mlet, M be matrices with real entries Suppose that the general solution of the linear system is, e y(t) (k M + k M ) αt cos(βt), e αt sin(βt) where k, k are arbitrary real numbers (a) Prove that M and M each satisfy the following equation, α β AM i M i D, D β α Solution: By assumption, e αt cos(βt) d e αt cos(βt) AM i e αt M sin(βt) i dt e αt sin(βt) And, d e αt cos(βt) αe αt cos(βt) βe αt sin(βt) α β e αt cos(βt) dt e αt sin(βt) αe αt sin(βt) + βe αt sin(βt) β α e αt sin(βt) Therefore, for each real number t, e (AM i M i D) αt cos(βt) e αt 0 sin(βt) But for t 0 and t π/(β), the vectors give a basis for R Therefore AM i M i D 0 (b) Consider the linear system of differential equations, z (t) A z z(t), z(t) (t) z (t) Use (a) to show that for every pair of real numbers k, k, the following function is a solution of the linear system, [ ] z(t) (k M + k M ) Solution: Because AM i M i D, also Now, A M i A(AM i ) A(M i D) (AM i )D (M i D)D M i D D α β α β α β αβ β α β α αβ α β Date : Spring 004

2 And, d α β αβ dt αβ α β Thus, d e (α e M (α i D β )t A β )t M i dt M i Therefore, for each pair of real numbers k, k, d e A (k (α β )t M + k M ), dt (k M + k M ) ie, is a solution of z (t) A z(t) z(t) (k M + k M ), Problem Consider the following inhomogeneous nd order linear differential equation, y y, y(0) y 0, y (0) v 0 Denote by Y (s) the Laplace transform, Y (s) L[y(t)] e st y(t)dt 0 (a) Find an expression for Y (s) as a sum of ratios of polynomials s in Solution: By rules of the Laplace transform, L[y (t)] sy (s) y 0 and L[y (t)] s Y (s) sy 0 v 0 Therefore, (s Y (s) sy 0 v 0 ) Y (s) L [y y] L[] s Gathering terms and simplifying, (s )(s + )Y (s) (s )Y (s) v 0 + sy 0 + s Therefore, s y 0 + sv 0 + Y (s) (s + )s(s ) (b) Determine the partial fraction expansion Y of (s) Solution: Because each factor in the denominator is a linear factor with multiplicity, the Heaviside cover up method determines all the coefficients, s y 0 + sv 0 + y 0 v 0 + s + + ( ) + y 0 + v 0 + (s + )s(s ) s s (c) Determine y(t) by computing the inverse Laplace transform Y of (s) Solution: The inverse Laplace transform of /(s a) is e at Therefore, y 0 + v 0 + y(t) L [Y (s)] y 0 v 0 + e t + e t + (y 0 + ) cosh(t) + v 0 sinh(t)

3 Problem 3 The general skew symmetric real matrix is, 0 b A, b 0 where b is a real number Prove that the eigenvalues Aof of the form λ ±iµ for some real number µ Determine µ and find all values of b such that there is a single repeated eigenvalue Solution: The trace is Trace(A) 0, and the determinant is det(a) 0( b ) b Therefore the characteristic polynomial is p A (λ) λ Trace(A)λ + det(a) λ + b Therefore the eigenvalues of A are ±ib There is a repeated eigenvalue b iff 0 There is a more involved proof that for every positive integer n, for every skew symmetric real n n matrix A, every eigenvalue of A is purely imaginary The idea is that Con n there is a Hermitian inner product, which assigns to each pair of vectors, z w z, w, the complex number, Observe this has the properties, Problem 4 Let λ be a real number and let A be the following 3 3 matrix, A λ 0 0 λ 0 0 λ z n w n z, w z w + + z n w n z + z, w z, w + z, w, z, w + w z, w + z, w, λz, w λ z, w, z, λw λ z, w, w, z z, w, z, z 0, if z 0 Because A is a real skew symmetric matrix, for every pair of vectors z, w the following equation holds, Az, w z, Aw Suppose that λ C is an eigenvalue and let z be a (nonzero) λ eigenvalue Then, λ z, z λz, z Az, z z, Az z, λz λ z, z Because z is nonzero, z, z is nonzero Therefore λ λ, which implies that λ is a pure imaginary number Let a, a, be real numbers Consider the following initial value problem, y(t) Ay(t), y(0) 3 a a

4 Denote by Y(s) the Laplace transform of y(t), ie, Y (s) Y(s) Y (s), Y i (s) L[y i (t)], i,, 3 Y 3 (s) (a) Express both L[y (t)] and L[Ay(t)] in terms of Y(s) Solution: First of all, a L[y (t)] sy(s) a Secondly, L[Ay(t)] AY(s) (b) Using part (a), find an equation that Y(s) satisfies, and iteratively solve the equation Yfor 3 (s), Y (s) and Y (s), in that order Solution: By part (a), a sy(s) a AY(s) Written out, this is equivalent to the system of 3 equations, (s λ)y (s) a + Y (s) (s λ)y (s) a + Y 3 (s) (s λ)y3 (s) Solving this iteratively, and, Y 3 (s), s λ a a Y (s) + Y 3 (s) + s λ s λ s λ (s λ), a a a Y (s) + Y (s) + + s λ s λ s λ (s λ) (s λ) 3 (c) Determine y(t) by applying the inverse Laplace transform Yto (s), Y (s) and Y 3 (s) Solution: The relevant inverse Laplace transforms are, L [/(s λ)] e λt, L [/(s λ) ] te λt, L [/(s λ) 3 ] t e λt Therefore, y t λt (t) a e λt + a te λt + e, λt te λt y (t) a e +, y λt 3 (t) e In matrix form, this is, t t y(t) a e λt 0 + a e λt + e λt t 0 0 4

5 Problem 5 For each of the following matrices A, compute the following, (i) Trace(A), (ii) det(a), (iii) the characteristic polynomial p A (λ) det(λi A), (iv) the eigenvalues of A (both real and complex), and (v) for each eigenvalue λ a basis for the space λ eigenvectors of (a) The matrix with real entries, 0 A 0 Hint: See Problem 3 Solution: In Problem 3, we computed Trace(A) 0, A) det(, p A (λ) λ +, and the eigenvalues are λ ± ±i For the eigenvalue λ + i, denote an eigenvector by, v v +, + v +, Then v +, iv +,, eg, v +,, v +, i Therefore an eigenvector for λ + i is, v + i Similarly, an eigenvector for λ i is, v i (b) The 3 3 matrix with real entries, 3 A Solution: Because this is an upper triangular matrix, clearly Trace(A) , det(a) , and p A (λ) (λ 3)(λ 5)(λ 3) λ 3 λ + 39λ 45 The eigenvalues are λ 5 and λ 3 (the eigenvalue 3 has multiplicity ) For the eigenvalue λ 5, the eigenvectors are the nonzero nullvectors of the matrix, A 5I Either by using row operations to put this matrix in row echelon form, or by inspection, a basis for the nullspace is, v 0 5

6 For the eigenvalue λ 3, the eigenvectors are the nonzero nullvectors of the matrix, 0 A 3I In this case, a basis for the nullspace is, v 0 0 6