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MTH5201 Fall 2015
HW Assignment 1: Prob. Set 2.6, #9
The due date for this assignment is Wed. Aug. 26.
9. (a) Find a second-order homogeneous linear ODE of which the functions cos 5x, sin 5x
are solutions. (b) Show linear independence by the Wronskian. (c) Solve the initial
value problem y (0) = 3, y 0 (0) = 5:
(a) Let y1 (x) = cos 5x, y2 (x) = sin 5x. Then these are solutions of the second-order
homogeneous linear ODE
y 00 + 25y = 0; x 2 R;
since
y100 (x) + 25y1 (x) = ( 5 sin 5x)0 + 25 cos 5x = 25 cos 5x + 25 cos 5x = 0;
y200 (x) + 25y2 (x) = (5 cos 5x)0 + 25 sin 5x = 25 sin 5x + 25 sin 5x = 0
for all x 2 R.
(b) The Wronskian of these solutions is
W (y1 ; y2 ) (x) = y1 (x) y20 (x)
y10 (x) y2 (x) = 5 cos2 5x + 5 sin2 5x = 5 6= 0
for all x 2 R. Therefore by Theorem 2 in Sec. 2.6 of the book this implies the
solutions y1 , y2 are linearly independent on R.
(c) The coe¢ cients of this ODE are continuous on R with y1 , y2 are linearly independent on R and so by Theorem 4 in Sec. 2.6 the general solution of the ODE
is
y (x) = c1 y1 (x) + c2 y2 (x) = c1 cos 5x + c2 sin 5x
for arbitrary constants c1 , c2 . Thus, to solve the IVP we …nd
3 = y (0) = c1 cos 0 + c2 sin 0 = c1 ;
5 = y 0 (0) = 5c1 sin 0 + 5c2 cos 0 = 5c2 :
Thus,
c1 = 3; c2 =
1
and so the solution to the IVP is
y (x) = 3 cos 5x
1
sin 5x: