Exercises 0 More exercises are available in Elementary Differential Equations. If you have a problem to solve any of them, feel free to come to office hour. Problem Find a fundamental matrix of the given system of equations. Also find the fundamental matrix Φ(t satisfying Φ(0 = I.... x ( x ( 5 x (. Solution. First, we need to find the eigenvalues: ( λ( λ + 4 = 0, (4 λ λ = 0. (5 The eigenvalues are - and. If λ = : x y = x, (6 x y = y (7 y = x. (8 c ( (9 If λ = : x y = x, (0 x y = y ( x = y. ( c ( (
One possible fundamental matrix Ψ is: e t e Ψ = t e t e t (4 Now, we can compute Φ. We know the general solution: x = c e t e t. (5 We want to find x ( (t and x ( (t such that: ( x ( (0 = 0 ( x ( 0 (0 = (6 (7 We get: c = (8 c = 0 (9 Thus, we have: We also have: x ( = c = c = e t + 4 et e t + et (0 ( ( c = 0 ( c = (4 Thus, we get: We finally obtain: c = c = ( x ( = e t et 4 e t et Φ = e t + 4 et e t et e t + et 4 e t et (5 (6 (7 (8. First, we need to find the eigenvalues: ( λ( λ + = 0, (9 The eigenvalues are - and. λ = 0. (0
If λ = : x y = x ( x y = y ( y = x. ( c (4 If λ = : x y = x (5 x y = y (6 y = x. (7 c (8 A possible fundamental matrix is: e t e Ψ = t e t e t (9 Now, we can compute Φ. First, we need to compute x ( : c = (40 c = 0 (4 Thus, we have: Now, we need to compute x ( : x ( = c = c = e t + et e t + et (4 (4 (44 c = 0 (45 c = (46 Thus, we have: We finally obtain: c = c = ( x ( = e t et e t et Φ = e t + et e t et e t + et e t et (47 (48 (49 (50
. First, we need to find the eigenvalues: ( λ( λ + 5 = 0 (5 The eigenvalues are ±i. λ + = 0. (5 If λ = i: x 5y = ix (5 x y = iy (54 x = ( + iy (55 c + i cos t sin t c cos t sin t os t c sin t (56 (57 (58 One possible fundamental matrix Ψ is: ( cos t sin t sin t os t cos t sin t (59 Now, we can compute Φ. First, we need to compute x ( : c = (60 c = 0 (6 c = 0 (6 c = (6 Thus, we have: Now, we need to compute x ( : x ( = sin t os t sin t (64 c = 0 (65 c = (66 c = (67 c = (68 Thus, we have: We finally obtain: x ( = ( 5 sin t cos t sin t sin t os t 5 sin t Φ = sin t cos t sin t (69 (70 4
Problem Find the general solution of the system of equations.. 4 x (7 4 x (7. Solution. First, we need to compute the eigenvalues: ( λ( λ + 4 = 0 (7 λ λ + = 0 (74 The eigenvalue is λ = which is an eigenvalue of algebraic multiplicity two. If λ = : x 4y = x (75 x y = y (76 x = y. (77 c (78 We now have one solution of the system of ODE: To find the second solution, we need to solve: Therefore, we get: x ( = e t. (79 x 4y = x + (80 x y = y + (8 k x = y +. (8 ( + 0 Therefore, the general solution is: ( x = c e t te t + e 0 t (8 (84. First, we need to compute the eigenvalues: ( λ ( λ + 4 = 0 (85 λ + λ + = 0, (86 (λ + = 0 (87 The eigenvalue is λ = which is an eigenvalue of algebraic multiplicity two. 5
If λ = : x + y = x, (88 4 x y = y (89 We now have one solution of the ODE: x ( = x = y (90 c ( (9 ( e t (9 To find the second solution, we need to solve: Therefore, we get: x + y = x +, (9 4 x y = y + (94 x = y 4. (95 ( k + 4 0 Therefore, the general solution is: ( x = c e t te t + Problem Find the solution of the given initial value problem. 4 e t 0 (96 (97.. 4 x = x (98 4 7 ( x(0 = (99 x = x (00 x(0 = (0 6
. Solution. First, we need to compute the eigenvalues: ( λ( 7 λ + 6 = 0, (0 λ + 6λ + 9 = 0, (0 (λ + = 0. (04 The eigenvalue is λ = which is an eigenvalue of algebraic multiplicity two. If λ = : x 4y = x, (05 4x 7y = y, (06 x = y (07 c (08 We now have one solution of the ODE: x ( = ( e t (09 To find the second solution, we need to solve: x 4y = x + (0 4x 7y = y + ( x = y + 4. ( Therefore, we get: k ( + 4 0 Therefore, the general solution is: ( ( x = c e t te t + 4 0 Now we can use the I.C., to determine c and c : ( (4 c 4 = (5 c = (6 Thus, we finally have: ( (( ( x = e t + 4 te t + 4 0 (7. First, we need to compute the eigenvalues: ( λ( λ + 9 = 0, (8 4 λ λ + = 0, (9 4 ( λ = 0. (0 The eigenvalue is λ = which is an eigenvalue of algebraic multiplicity two. 7
If λ = : x + y = x, ( x y = y ( y = x. ( c ( (4 We now have one solution of the ODE: x ( = e t. (5 To find the second solution, we need to solve: x + y = x +, (6 x y = y (7 y = x +. (8 Therefore, we get: 0 k + Therefore, the general solution is: ( x = c e t te t 0 + e t (9 (0 Now we can use the I.C., to determine c and c : c = ( c + c = ( Thus, we finally get: 4 Problem 4 ( x = e t + (( te t 0 + e t ( Find the general solution of the given system of equations... x + e t 4 t x + 8 4 t t (4 (5 t > 0 8
4. Solution. First, we need to solve the homogeneous problem: x = x (6 We need to compute the eigenvalues: ( λ( λ + = 0 (7 The eigenvalues are λ = and λ =. λ = 0 (8 If λ = : x y = x (9 x y = y (40 x = y (4 c (4 We have one solution of the corresponding homogeneous ODE: c e t (4 If λ = : x y = x (44 x y = y (45 y = x (46 c (47 We have one more solution of the corresponding homogeneous ODE: c e t (48 The general solution of the corresponding homogeneous ODE is: x c = c e t e t (49 The nonhomogeneous term is: ( e t e t 0 = + t 0 t Therefore, we look for a particular solution of the form: (50 x p = ate t + be t t + d, (5 x p = a(t + e t + be t. (5 9
Substituting in the ODE, we get: ate t + (a + be t = A ( ate t + be t t + d ( e t + 0 + 0 t (5 where We get the system of equations: A =. (54 Aa = a (55 ( Ab = a + b (56 0 ( 0 Ac = (57 Ad = c (58 We know that the solution of equation (55 is: a = α α (59 We can know solve equation (56: b b = α + b (60 b b = α + b (6 b = b + α (6 b + α b = α + b (6 α = b = b + (64 (65 Thus, we have: ( ( b = k + 0 (66 We can choose the value of k. Choosing different value of k will only change the value of c. So we can pick k = 0. We can now solve equation (57: c c = 0 (67 c c = (68 c = (69 c = (70 We can now solve equation (58: d d = (7 d d = (7 0
d = 0 (7 d = (74 We finally get: x = c e t e t + ( te t + ( ( ( 0 + t 0 (75. First, we need to solve the homogeneous problem: x 4 = 8 4 (76 We need to compute the eigenvalues: (4 λ( 4 λ + 6 = 0, (77 The eigenvalue is λ = 0. λ = 0. (78 If λ = 0: 4x y = 0 (79 8x 4y = 0 (80 y = x (8 c (8 We now have one solution of the ODE: To find the second solution, we need to solve: x ( = (8 4x y = (84 8x 4y = (85 y = x Therefore, we get: ( 0 k Therefore, the solution of the corresponding homogeneous problem is: ( x c = c t ( 0 (86 (87 (88 To find the particular solution, we use the method of variation of parameters. We need to solve the system of equations: t u t t u = t (89 u + u t = t, (90 ( u + t u = t (9
u = t u t, (9 t u t + u t u = t (9 u = t 4t t (94 u = 4t + t (95 We finally get: ( x = c t u = t + 4t ln t (96 u = t t (97 ( ( 0 + ( t 4t ln t + ( t t ( t ( 0 (98