Math 308 Final Exam Practice Problems

Math 308 Final Exam Practice Problems This review should not be used as your sole source for preparation for the exam You should also re-work all examples given in lecture and all suggested homework problems Consider the differential equation (3 x )y 3xy y = 0 (a) Seek power series solutions of the given differential equation about the point x 0 = 0 Find the recurrence relation (b) Find the first four terms in each of two solutions y and y (c) By evaluating the Wronskian W (y, y )(x 0 ), show that y and y form a fundamental set of solutions (d) If possible, find the general term in each solution Solve the linear system x + x x 3 = 5 x + x x 3 = 4 x x + 3x 3 = 8 by forming the augmented matrix and performing Gaussian elimination 3 Solve the linear system x + x 3x 3 = 3x x x 3 = x + 3x 5x 3 = 3 by forming the augmented matrix and performing Gaussian elimination 4 Solve the linear system x + x + x 3 = 3x x x 3 = 4 x + 5x + 5x 3 = by forming the augmented matrix and performing Gaussian elimination 5 Find the inverse of each matrix, if possible 3 (a) A = 6 4 (b) B = 4 8

6 Find the inverse of each matrix, if possible (a) A = 3 (b) B = 4 7 Solve the initial value problem x = ( 5 3 ) x, x(0) = Describe the behavior of the solution as t 8 Solve the initial value problem 5 x = x, x(0) = 3 Describe the behavior of the solution as t 9 Consider the system where α is a parameter α x = x, (a) Determine the eigenvalues in terms of α (b) Find the critical value(s) of α where the qualitative nature of the phase portrait for the system changes (c) Draw a phase portrait for a value of α slightly below, and for another value slightly above, each critical value 0 Solve the initial value problem 4 x = x, x(0) = 4 7 Describe the behavior of the solution as t Consider the initial value problem x = x, x(0) = 3 3 Find the matrix exponential function e At and use it to solve the initial value problem Describe the behavior of the solution as t

Consider the initial value problem 5 x = x, x(0) = 3 Find the matrix exponential function e At and use it to solve the initial value problem Describe the behavior of the solution as t 3 Find the general solution of the system 4 x = 3 x 4 Find the general solution of the system 0 x = 0 x 0 5 Use the Method of Undetermined Coefficients to solve the nonhomogeneous linear system x = x + e t 3 6 Use the Method of Variation of Parameters to solve the nonhomogeneous linear system 0 t x = x + 0 3

Solutions Solutions may contain errors or typos If you find an error or typo, please notify me at glahodny@mathtamuedu Consider the differential equation (3 x )y 3xy y = 0 (a) Seek power series solutions of the given differential equation about the point x 0 = 0 Find the recurrence relation Consider a solution in the form of a power series about x 0 = 0: y = a n x n n=0 and assume that the series converges in some interval x < ρ Differentiating term by term, we obtain y = y = na n x n, n= n(n )a n x n n= Substituting these series into the differential equation gives 3 3 (3 x ) n(n )a n x n 3x na n x n n= n(n )a n x n n= (n + )(n + )a n+ x n n=0 n= n= n(n )a n x n 3 na n x n n=0 Combining these four series, we obtain n= n(n )a n x n 3 na n x n n=0 [ ] 3(n + )(n + )an+ (n + ) a n x n = 0 n=0 a n x n = 0 n=0 a n x n = 0 n=0 a n x n = 0 For this equation to be satisfied for all x, the coefficient of each power of x must be zero Therefore, we obtain the recurrence relation 3(n + )a n+ (n + )a n = 0, n = 0,,, 3, n=0 4

(b) Find the first four terms in each of two solutions y and y The recurrence relation relates each coefficient to the second one before it Thus, the even-numbered coefficients (a 0, a, a 4, ) and the odd-numbered ones (a, a 3, a 5, ) are determined separately For the even-numbered coefficients, we have a = a 0 3, a 4 = 3a 3 4 = 3 a 0 3 4, a 6 = 5a 4 3 6 = 3 5 a 0 3 3 4 6, Similarly, for the odd-numbered coefficients, we have a 3 = a 3 3, a 5 = 4a 3 3 5 = 4 a 3 3 5, a 7 = 6a 5 3 7 = 4 6 a 3 3 3 5 7, Substituting these coefficients into the power series, we have Therefore, y = a 0 + a x + a 0! x + a 3! x3 + a 0 4! x4 + a 5! x5 + a 0 6! x6 + a = a 0 [ + 3 x + 3 3 4 x4 + 3 5 3 3 4 6 x6 + +a [ x + 3 3 x3 + 4 3 3 5 x5 + 4 6 3 3 3 5 7 x7 + y (x) = + x 6 + x4 4 + 5 43 x6 + y (x) = x + 9 x3 + 8 35 x5 + 6 945 x7 + 7! x7 + ] (c) By evaluating the Wronskian W (y, y )(x 0 ), show that y and y form a fundamental set of solutions ] It follows from the expansions in part (b) that y (0) = and y (0) = 0 differentiating the series for y (x) and y (x), we find that By y (x) = 3 x + 6 x + 5 7 x5 + y (x) = + 3 x + 8 7 x4 + Thus, at x = 0, we have y (0) = 0 and y (0) = Consequently, the Wronskian of y and y at x 0 = 0 is W (y, y )(0) = 0 0 = 0 So y and y form a fundamental set of solutions 5

(d) If possible, find the general term in each solution The results of part (b) suggest that a 0 = and, in general, if n = k, then a n = a k = a 0 3 5 (k ), k =,, 3, 3 k 4 (k) Similarly, a = and if n = k +, then Therefore, Solve the linear system a n = a k+ = y (x) = + y (x) = x + a 4 (k), k =,, 3, 3 k 3 5 (k + ) n= n= 3 5 (n ) 3 n 4 (n) xn, 4 (n) 3 n 3 5 (n + ) xn+ x + x x 3 = 5 x + x x 3 = 4 x x + 3x 3 = 8 by forming the augmented matrix and performing Gaussian elimination The augmented matrix for this linear system is 5 4 3 8 Using Gaussian elimination, we have 5 4 3 8 Thus, the solution is (,, ) 4 5 0 4 4 0 3 0 4 0 0 0 3 0 0 0 0 0 0 0 0 6 R R R 3 + R (R R ) R + R R 3 R (R R 3 )

3 Solve the linear system x + x 3x 3 = 3x x x 3 = x + 3x 5x 3 = 3 by forming the augmented matrix and performing Gaussian elimination The augmented matrix for this linear system is 3 3 3 5 3 Using Gaussian elimination, we have 3 3 3 5 3 3 0 7 7 7 0 0 0 0 0 0 0 0 0 0 0 0 R 3R R 3 R R + R 3 7 R (R 3 + R ) Therefore, x 3 = α is a free variable Using the second equation, we have Using the first equation, we have x = x 3 = α x = x 3 = α Thus, the solution set is {(α, α, α) α R} 7

4 Solve the linear system x + x + x 3 = 3x x x 3 = 4 x + 5x + 5x 3 = by forming the augmented matrix and performing Gaussian elimination The augmented matrix for this linear system is 3 4 5 5 Using Gaussian elimination, we have 3 4 5 5 0 4 4 0 4 4 0 4 4 0 0 0 Since 0, the system is inconsistent There is no solution R 3R R 3 R (R 3 + R ) 5 Find the inverse of each matrix, if possible 3 (a) A = 6 The determinant of A is det(a) = 6 ( 6) = 0 Therefore, A is nonsingular with inverse A = ( = 6 6 3 (b) B = 4 4 8 4 ) The determinant of B is det(b) = 6 6 = 0 Therefore, B is singular and B does not exist 8

6 Find the inverse of each matrix, if possible (a) A = The determinant of A is = = 9 0 Therefore, A is nonsingular and A can be found using Gaussian elimination 0 0 0 0 0 0 0 3 3 0 R R R 0 0 0 0 3 0 3 R 0 0 ( 0 0 R ) 3 3 3 0 0 0 R 3 3 3 3 0 0 0 3 3 0 0 R + R 3 3 3 0 0 R 0 R 3 3 3 0 0 0 3 3 0 0 ( R 3 3 3 ) 0 0 0 3 3 Therefore, we have obtained 3 (b) B = 4 A = 0 3 3 3 3 3 3 0 3 The determinant of B is 3 4 = 3 4 Therefore, B is singular and B does not exist 4 = 0 9

7 Solve the initial value problem x = ( 5 3 ) x, x(0) = Describe the behavior of the solution as t To find the eigenvalues of the coefficient matrix, let 5 λ 3 λ = λ 6λ + 8 = (λ )(λ 4) = 0 The eigenvalues are λ = and λ = 4 If λ =, then 3 v = 0 3 It follows that 3v = v and so v =, 3 T If λ = 4, then v = 0 3 3 It follows that v = v and so v =, T The general solution is v v x(t) = c e t + c 3 e 4t Applying the initial condition, we obtain c + c x(0) = = 3c + c It follows that c = 3/ and c = 7/ Therefore, the solution of the initial value problem is x(t) = 3 e t + 7 e 4t 3 As t, the solution trajectory diverges from the origin along the line x = x 0

8 Solve the initial value problem 5 x = x, x(0) = 3 Describe the behavior of the solution as t To find the eigenvalues of the coefficient matrix, let λ 5 3 λ = λ + λ + = 0 The roots of this quadratic are λ = ( ± 4 8) = ± i The eigenvalues are λ = + i and λ = i If λ = + i, then i 5 v = 0 i It follows that v = ( + i)v and so v = + i, T Splitting the eigenvector into its real and imaginary parts, we obtain v = a + ib = + i 0 Therefore, two real-valued linearly independent solutions are u (t) = e t cos t e t sin t, 0 u (t) = e t sin t + e t cos t 0 The general solution is cos t sin t x(t) = c e t cos t Applying the initial condition, we obtain c + c x(0) = = It follows that c = and c = c v sin t + cos t + c e t sin t Therefore, the solution of the initial value problem is cos t 3 sin t x(t) = e t cos t sin t As t, the solution trajectory spirals towards the origin

9 Consider the system where α is a parameter α x = x, (a) Determine the eigenvalues in terms of α To find the eigenvalues of the coefficient matrix, let λ α λ = λ + λ + α + = 0 The eigenvalues are λ = ( ± 4 4(α + )) = ( ± α) = ± α (b) Find the critical value(s) of α where the qualitative nature of the phase portrait for the system changes The eigenvalues are complex conjugates if α > 0 and are real otherwise Thus, α = 0 is a critical value, where the eigenvalues change from real to complex, or vice versa For < α < 0, both eigenvalues are real and negative, so all trajectories approach the origin which is an asymptotically stable node For α <, both eigenvalues are real (one positive and one negative), so the trajectories diverge from the origin which is a saddle point (c) Draw a phase portrait for a value of α slightly below, and for another value slightly above, each critical value The phase portraits illustrating the behavior of solution trajectories for different values of α are shown below α = α = 05 α = 05 5 5 5 05 05 05 x 0 x 0 x 0-05 -05-05 - - - -5-5 -5 - - - 0 x - - - 0 x - - - 0 x (Note: On the exam, you will not be required to sketch a direction field or phase portrait)

0 Solve the initial value problem 4 x = x, x(0) = 4 7 Describe the behavior of the solution as t 3 To find the eigenvalues of the coefficient matrix, let λ 4 4 7 λ = λ + 6λ + 9 = (λ + 3) = 0 There is a double eigenvalue λ = λ = 3 If λ = 3, then 4 4 v = 0 4 4 It follows that v = v and so v =, T To find a generalized eigenvector w of the coefficient matrix A corresponding to the eigenvalue λ = 3, let (A + 3I)w = v That is, 4 4 /4 4 4 0 0 0 It follows that w = w + /4 and so w = /4, 0 T Thus, the general solution is [ ( )] x(t) = c e 3t + c te 3t + e 3t 40 Applying the initial condition, we obtain ( c + x(0) = c ) 4 = It follows that c = and c = 4 c v 3 Therefore, the solution of the initial value problem is 3 + 4t x(t) = e 3t + 4t As t, the solution trajectory approaches the origin 3

Consider the initial value problem x = x, x(0) = 3 Find the matrix exponential function e At and use it to solve the initial value problem Describe the behavior of the solution as t To find the eigenvalues of the coefficient matrix, let λ 3 λ = λ = (λ + )(λ ) = 0 The eigenvalues are λ = and λ = If λ =, then 3 v = 0 3 It follows that 3v = v and so v =, 3 T If λ =, then v = 0 3 3 It follows that v = v and so v =, T Thus, a fundamental matrix is v v e t e X(t) = t 3e t At the initial time t 0 = 0, we have ( X(0) =, X (0) = 3 e t 3 The matrix exponential function is given by e At = X(t)X (0) = e t + 3 et e t et 3 e t + 3 et 3 e t et Therefore, the solution of the initial value problem is x(t) = e At x 0 = 7 e t 3 e t 3 As t, the solution trajectory diverges from the origin along the line x = x ) 4

Consider the initial value problem 5 x = x, x(0) = 3 Find the matrix exponential function e At and use it to solve the initial value problem Describe the behavior of the solution as t To find the eigenvalues of the coefficient matrix, let λ 5 λ = λ + = 0 The eigenvalues are λ = i and λ = i If λ = i, then i 5 v = 0 i It follows that v = ( + i)v and so v = + i, T Splitting the eigenvector into its real and imaginary parts, we obtain v = a + ib = + i 0 Therefore, two real-valued linearly independent solutions are u (t) = cos t sin t, 0 u (t) = sin t + cos t 0 Thus, a fundamental matrix is cos t sin t sin t + cos t X(t) = cos t sin t At the initial time t 0 = 0, we have X(0) =, X (0) = 0 v 0 The matrix exponential function is given by sin t + cos t 5 sin t e At = X(t)X (0) = sin t cos t sin t Therefore, the solution of the initial value problem is 3 cos t + sin t x(t) = e At x 0 = sin t + cos t 5

3 Find the general solution of the system 4 x = 3 x To find the eigenvalues of the coefficient matrix, let λ 4 3 λ = λ 3 + λ + 5λ 6 λ The eigenvalues are λ =, λ =, and λ 3 = 3 If λ =, then 0 4 0 3 0 0 = (λ )(λ + )(3 λ) = 0 0 0 0 4 0 0 0 0 0 It follows that v 3 = α, v = 4α, v = α, and so v =, 4, T If λ =, then 3 4 0 3 4 0 0 0 0 0 0 0 0 0 0 It follows that v 3 = α, v = α, v = α, and so v =,, T If λ 3 = 3, then 4 0 3 0 4 0 0 0 0 0 0 0 0 0 It follows that v 3 = α, v = α, v = α, and so v 3 =,, T Therefore, the general solution is x(t) = c e t 4 + c e t + c 3 e 3t 6

4 Find the general solution of the system 0 x = 0 x 0 To find the eigenvalues of the coefficient matrix, let λ λ λ = (λ + ) ( λ) = 0 The eigenvalues are λ = and λ = λ 3 = If λ =, then 0 0 0 0 0 0 0 0 0 0 0 It follows that v 3 = α, v = α, v = α, and so v =,, T If λ =, then 0 0 0 0 0 0 0 0 0 0 0 0 It follows that v = α, v 3 = β, v = α β Two linearly independent eigenvectors can be found by taking α = 0 and β = 0, respectively That is, v =, 0, T and v 3 =,, 0 T Therefore, the general solution is x(t) = c e t + c e t 0 + c 3 e t 0 (Note: Since the coefficient matrix is a real symmetric matrix, we are able to find linearly independent eigenvectors for the repeated eigenvalue λ = ) 7

5 Use the Method of Undetermined Coefficients to solve the nonhomogeneous linear system x = x + e t 3 To find the eigenvalues of the coefficient matrix, let λ 3 λ = λ = 0 The eigenvalues are λ = and λ = If λ =, then v = 0 3 3 It follows that v = v and so v =, T It λ =, then 3 v = 0 3 It follows that 3v = v and so v =, 3 T Therefore, the homogeneous solution is x h (t) = c e t + c e t 3 Consider a particular solution of the form v v x p (t) = ate t + be t Substituting the particular solution into the differential equation and collecting terms, we obtain Aa = a Ab = (a + b) It follows that a is an eigenvector of A corresponding to the eigenvalue λ = Thus, a = α, α T, where α is a free variable Therefore, the second equation can be rewritten as α (A I)b = α + The augmented matrix for this linear system is ( α α 3 3 α + 0 0 4 α ) 8

The system can be solved if and only if α =, in which case a =, T augmented matrix is 0 0 0 and the It follows that b = β +, β T, where β is a free variable Choosing β = 0, we have a = b = 0 Thus, a particular solution of the nonhomogeneous system is x p (t) = te t + e t 0 Therefore, the general solution of the nonhomogeneous system is x(t) = c e t + c e t + te t + e t 3 0 9

6 Use the Method of Variation of Parameters to solve the nonhomogeneous linear system 0 t x = x + 0 To find the eigenvalues of the coefficient matrix, let λ λ = λ + = 0 The eigenvalues are λ = i and λ = i If λ = i, then i v = 0 i It follows that v = iv and so v = i, T Splitting the eigenvector into its real and imaginary parts, we obtain 0 v = + i 0 Two linearly independent solutions of the homogeneous system are 0 u (t) = cos t sin t, 0 0 u (t) = sin t + cos t 0 Therefore, a fundamental matrix for the homogeneous system is sin t cos t X(t) = cos t sin t v The inverse of this matrix is sin t cos t X (t) = cos t sin t Thus, X (t)g(t) dt = = = sin t cos t t dt cos t sin t t sin t + cos t t dt cos t + sin t t cos t t sin t cos t + sin t t sin t + t cos t sin t cos t 0

Therefore, a particular solution of the nonhomogeneous sytem is x p (t) = X(t) X (t)g(t) dt sin t cos t t = cos t t sin t cos t + sin t cos t sin t t sin t + t cos t sin t cos t t = t The general solution of the nonhomogeneous equation is sin t cos t x(t) = c + c cos t + sin t ( t t )