MATH 2090  Answers to the Review Problems for Exam 4  Fall 2001

1. For each of the matrices A below, find a real-valued fundamental set of solutions (explain why your solutions form a fundamental set), a fundamental matrix, and the general solution of the system x⃗′ = A x⃗. Determine if the matrix A is defective or diagonalizable in each case. If A is diagonalizable, find a diagonal matrix D and an invertible matrix S so that S^{−1}AS = D. Also, solve the initial value problem in part (a).

(a) A = [ \begin{matrix}1 & 2 \\ 8 & 1 \end{matrix} ] , x⃗(0) = [ \begin{matrix}3 \\ 2 \end{matrix} ] (b) A = [ \begin{matrix}4 & −5 \\ 1 & 2 \end{matrix} ] (c) A = [ \begin{matrix}7 & −4 \\ 4 & −1 \end{matrix} ] (d) A = [ \begin{matrix}2 & −1 & 3 \\ 3 & 1 & 0 \\ 2 & −1 & 3 \end{matrix} ] (e) A = [\begin{matrix}2 & −3 & 2 \\ −1 & 0 & 2 \\ −1 & −3 & 5 \end{matrix} ]

(a) fund. set: \{x⃗_{1} = e^{5t} [ \begin{matrix}1 \\ 2 \end{matrix} ] , x⃗_{2} = e^{−3t} [\begin{matrix}−1 \\ 2 \end{matrix} ]\}, W [x⃗_{1}, x⃗_{2}] = 4e^{2t} ≠ 0, fund. matrix: [ \begin{matrix}e^{5t} & −e^{−3t} \\ 2e^{5t} & 2e^{−3t} \end{matrix} ]

A is diagonalizable. S = [ \begin{matrix}1 & −1 \\ 2 & 2 \end{matrix} ], D = [ \begin{matrix}5 & 0 \\ 0 & −3 \end{matrix} ]  x⃗ = 2e^{5t} [ \begin{matrix}1 \\ 2 \end{matrix} ] − e^{−3t} [\begin{matrix}−1 \\ 2 \end{matrix} ] solves the IVP

(b) fund. set: \{x⃗_{1} = [ \begin{matrix}e^{3t} cos 2t − 2e^{3t} sin 2t \\ e^{3t} cos 2t \end{matrix} ] , x⃗_{2} = [ \begin{matrix}e^{3t} sin 2t + 2e^{3t} cos 2t \\ e^{3t} sin 2t \end{matrix} ]\}, W [x⃗_{1}, x⃗_{2}] = −2e^{6t} ≠ 0,

fund. matrix: [ \begin{matrix}e^{3t} cos 2t − 2e^{3t} sin 2t & e^{3t} sin 2t + 2e^{3t} cos 2t \\ e^{3t} cos 2 & e^{3t} sin 2t \end{matrix} ]

A is diagonalizable. S = [ \begin{matrix}1 + 2i & 1 − 2i \\ 1 & 1 \end{matrix} ], D = [ \begin{matrix}3 + 2i & 0 \\ 0 & 3 − 2i \end{matrix} ]

(c) fund. set: \{x⃗_{1} = e^{3t} [ \begin{matrix}1 \\ 1 \end{matrix} ] , x⃗_{2} = te^{3t} [ \begin{matrix}1 \\ 1 \end{matrix} ] + e^{3t} [ \begin{matrix}1/4 \\ 0 \end{matrix} ]\}, W [x⃗_{1}, x⃗_{2}] = −e^{6t}/4 ≠ 0,

fund. matrix: [ \begin{matrix}e^{3t} & te^{3t} + e^{3t}/4 \\ e^{3t} & te^{3t} \end{matrix} ] A is defective. The eigenvalue λ = 3 has algebraic multiplicity 2 and geometric multiplicity 1.

(d) fund. set: \{x⃗_{1} = [\begin{matrix}−3 \\ 9 \\ 5 \end{matrix} ] , x⃗_{2} = e^{2t} [ \begin{matrix}1 \\ 3 \\ 1 \end{matrix} ] , x⃗_{3} = e^{4t} [ \begin{matrix}1 \\ 1 \\ 1 \end{matrix} ]\}, W [x⃗_{1}, x⃗_{2}, x⃗_{3}] = −4e^{6t} ≠ 0,

fund. matrix: [\begin{matrix}−3 & e^{2t} & e^{4t} \\ 9 & 3e^{2t} & e^{4t} \\ 5 & e^{2t} & e^{4t} \end{matrix} ] A is diagonalizable. S = [\begin{matrix}−3 & 1 & 1 \\ 9 & 3 & 1 \\ 5 & 1 & 1 \end{matrix} ] D = [ \begin{matrix}0 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{matrix} ]

(e) fund. set: \{x⃗_{1} = e^{t} [ \begin{matrix}1 \\ 1 \\ 1 \end{matrix} ] , x⃗_{2} = e^{3t} [\begin{matrix}−3 \\ 1 \\ 0 \end{matrix} ] , x⃗_{3} = e^{3t} [ \begin{matrix}2 \\ 0 \\ 1 \end{matrix} ]\}, W [x⃗_{1}, x⃗_{2}, x⃗_{3}] = 2e^{7t} ≠ 0,

fund. matrix: [ \begin{matrix}e^{t} & −3e^{3t} & 2e^{3t} \\ e^{t} & e^{3t} & 0 \\ e^{t} & 0 & e^{3t} \end{matrix} ] A is diagonalizable. S = [ \begin{matrix}1 & −3 & 2 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{matrix} ] D = [ \begin{matrix}1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{matrix} ]

2. The differential equation x^{3}y′′′ + x^{2}y′′ − 2xy′ + 2y = 0, x > 0 has three solutions of the form y(x) = x^{r}. Find these solutions, show that they are linearly independent on the interval (0, ∞), and find the general solution of this differential equation.

The functions y_{1} = x, y_{2} = x^{2}, and y_{3} = x^{−1} are solutions of this DE.

The Wronskian W [x, x^{2}, x^{−1}] = 6/x is non-zero on I = (0, ∞), so {y_{1}, y_{2}, y_{3}} is linearly independent on I.

The general solution of the DE is y = c_{1}x + c_{2}x^{2} + c_{3}x^{−1}

MATH 2090  Answers to the Review Problems for Exam 4  Fall 2001

3. For each of the non-homogeneous linear differential equations P (D)y = F (x) below, find the complementary solution, find the annihilator of the function F (x), and determine the form of a particular solution. In parts (a) and (b), also find the general solution of the differential equation, and solve the initial value problem in part (a).

(a) (D − 1)(D − 2)(D − 3)y = 6e^{4x}, y(0) = 4, y′(0) = 10, y′′(0) = 30 (b) y′′′ + 3y′′ + 3y′ + y = 2e^{−x} + 3e^{2x}

(c) (D^{2} − 2D + 2)(D − 3)^{2}(D + 2)y = x^{2} − e^{x} cos x + 3e^{3x}

(a) complementary solution: y_{c} = c_{1}e^{x} + c_{2}e^{2x} + c_{3}e^{3x} D − 4 annihilates of 6e^{4x} trial solution: y_{p} = Ae^{4x} general solution: y = c_{1}e^{x} + c_{2}e^{2x} + c_{3}e^{3x} + e^{4x} solution of the IVP: y = e^{x} + e^{2x} + e^{3x} + e^{4x}

(b) complementary solution: y_{c} = c_{1}e^{−x} + c_{2}xe^{−x} + c_{3}x^{2}e^{−x} (D + 1)(D − 2) annihilates 2e^{−x} + 3e^{2x} trial solution: y_{p} = Ax^{3}e^{−x} + Be^{2x} general solution: y = c_{1}e^{−x} + c_{2}xe^{−x} + c_{3}x^{2}e^{−x} + \frac{1}{3}x^{3}e^{−x} + \frac{1}{9}e^{2x}

(c) complementary solution: y_{c} = c_{1}e^{x} cos x + c_{2}e^{x} sin x + c_{3}e^{3x} + c_{4}xe^{3x} + c_{5}e^{−2x} D^{3}(D^{2} − 2D + 2)(D − 3) annihilates x^{2} − e^{x} cos x + 3e^{3x} trial solution: y_{p} = A + Bx + Cx^{2} + Dxe^{x} cos x + Exe^{x} sin x + F x^{2}e^{3x}

4. Suppose λ is an eigenvalue of the matrix A, and that v⃗ is a corresponding eigenvector.

(a) Show that λ is an eigenvalue of the matrix A^{T}.

(b) Show that λ^{2} is an eigenvalue of the matrix A^{2}, and that v⃗ is a corresponding eigenvector.

(c) If A is non-singular, show that 1/λ is an eigenvalue of A^{−1}, and that v⃗ is a corresponding eigenvector.

(a) Recall that the determinant of a matrix is equal to the determinant of its transpose.

Since det[A − λI] = det[(A − λI)^{T}] = det[A^{T} − (λI)^{T}] = det[A^{T} − λI],

the matrices A and A^{T} have the same eigenvalues.

(b) If Av⃗ = λv⃗, then A^{2}v⃗ = A · Av⃗ = Aλv⃗ = λAv⃗ = λλv⃗ = λ^{2}v⃗.

Thus λ^{2} is an eigenvalue of A^{2} and v⃗ is a corresponding eigenvector.

(c) Note that if A is non-singular, then the eigenvalues of A are non-zero (why?).

If Av⃗ = λv⃗ and A is non-singular, then v⃗ = A^{−1}Av⃗ = A^{−1}λv⃗ = λA^{−1}v⃗. So \frac{1}{λ}v⃗ = A^{−1}v⃗.

Thus 1/λ is an eigenvalue of A^{−1} and v⃗ is a corresponding eigenvector.

5. Let X(t) = [ x⃗_{1}(t) x⃗_{2}(t) ⋯ x⃗_{n}(t) ] be a fundamental matrix for the system x⃗′ = A(t) x⃗ on the interval I.

(a) Show that X′(t) = A(t) X(t).

(b) Show that the general solution of the system can be written as x⃗(t) = X(t) c⃗, where c⃗ is a constant vector.

(c) If t_{0} is in I, show that the solution of the IVP x⃗′ = A(t) x⃗, x⃗(t_{0}) = α⃗ can be written as x⃗(t) = X(t)X^{−1}(t_{0})α⃗.

(a) Each column of X(t) is a solution, so x⃗_{1}′ = A x⃗_{1}, x⃗_{2}′ = A x⃗_{2}, …, x⃗_{n}′ = A x⃗_{n}.

Thus X′ = [ x⃗_{1}′ x⃗_{2}′ ⋯ x⃗_{n}′] = [ A x⃗_{1} A x⃗_{2} ⋯ A x⃗_{n}] = A [ x⃗_{1} x⃗_{2} ⋯ x⃗_{n}] = A X

(b) The columns of X form a fundamental set of solutions. So the general solution is x⃗(t) = c_{1} x⃗_{1}(t) + c_{2} x⃗_{2}(t) + ⋯ + c_{n} x⃗_{n}(t) = X(t) c⃗, where c⃗ = [ c_{1} c_{2} … c_{n}]^{T}

(c) From part (b), the general solution of the system of DE is x⃗(t) = X(t) c⃗.

To solve the IVP, we must find c⃗ so that x⃗(t_{0}) = X(t_{0}) c⃗ = α⃗.

Since the columns of X(t) form a fundamental set of solutions and t_{0} is in I, the Wronskian W [ x⃗_{1}, x⃗_{2}, …, x⃗_{n}] (t_{0}) is different from zero. In other words, det X(t_{0}) ≠ 0. So X(t_{0}) is non-singular, and the solution of the system of equations X(t_{0}) c⃗ = α⃗ is c⃗ = X^{−1}(t_{0}) α⃗. Thus the solution of the IVP is x⃗(t) = X(t) c⃗ = X(t)X^{−1}(t_{0}) α⃗.