Page 99 - Schaum's Outline of Differential Equations

82 LINEAR DIFFERENTIAL EQUATIONS: THEORY OF SOLUTIONS [CHAR 8

8.55. Prove directly that the set given in Problem 8.49 is linearly dependent.

8.56. Using the results of Problem 8.42, construct the general solution of y" -4y = 0.

8.57. Using the results of Problem 8.43, construct the general solution of y" - 5y' + 6y = 0.
8.58. What can one say about the general solution of y" + 16y = 0 if two particular solutions are known to be y 1 = sin 4x
and y 2 = cos 4x1
8
8.59. What can one say about the general solution of y" — Sy' = 0 if two particular solutions are known to be y^ = e * and
3-2=1?
8.60. What can one say about the general solution of y" + y' = 0 if two particular solutions are known to be y 1 = 8 and
y 2= 1?
8.61. What can one say about the general solution of y'" - y" = 0 if two particular solutions are known to be y± = x and
x
y 2 = e ?
8.62. What can one say about the general solution of y"' + y" + y' + y = 0 if three particular solutions are known to be the
functions given in Problem 8.49?

8.63. What can one say about the general solution of y'" — 2y" — y' + 2y = 0if three particular solutions are known to be
the functions given in Problem 8.48?

8.64. What can one say about the general solution of cfy/dx 5 = 0 if three particular solutions are known to be the functions
given in Problem 8.47?

8.65. Find the general solution of y" + y = x , if one solution is y = x - 2, and if two solutions of y" + y = 0 are sin x and
2
2
cos x.
2
2
8.66. Find the general solution of y" — y = x , if one solution is y = —x — 2, and if two solutions of y" — y = 0 are e*
and 3e".
8.67. Find the general solution of y'" — y" — y' + y = 5, if one solution is y = 5, and if three solutions of y'" — y"
x
— y' + y = 0 are e", e~ , and xe".
2
8.68. The initial-value problem y' - (2lx)y = 0; y(0) = 0 has two solutions y = 0 and y = x . Why doesn' t this result violate
Theorem 8.1?

8.69. Does Theorem 8.1 apply to the initial-value problem y' - (2/x)y = 0; y(i) = 3?
8.70. The initial-value problem xy' — 2y = 0; y(0) = 0 has two solutions y = 0 and y = x . Why doesn't this result violate
2
Theorem 8.1?