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Unformatted text preview: 2x).
For ) ( ): y ( Thus
equation, so multiply by x:
For F2 (x): yp2 (x) = Le x . This is already a solution of the associated homogeneous
yp by M yp2(x) = Lxe cos(2x)
equation, so multiply = x: 1 sin(2x) + M2 x . Thus + Lxe x
yp = 2M cos(2x) 2M2 sin(2x) Lxe x + Le x
yp = M1 1
sin(2x) + M2 cos(2x) + Lxe x
4M1 sin(2x) 4M2 cos(2x) + Lxe x 2Le x
yp = 2M1 cos(2x) 2M2 sin(2x) Lxe x + Le x
00 x x AMATH 350 Page 4 Assignment #4 Solutions- Winter 2013 (3.8) is both easier to use and more powerful, though; observing that e x
is continuous for all x, ln (x + 1) is continuous for all x > 1, and our
initial condition is at x0 = 0, we can conclude that a unique solution exists
on the entire interval ( 1, 1).
(If we were to apply the E/U Theorem for ﬁrst-order equations (2.4) we
would only be able to conclude that the solution existed on some interval
containing 0 and not extending to -1).
5. The functions y1 = x, y2 = x3 , and y3 = x4 are all solutions to the equation
x3 d3 y
dx3 5 x2 d2 y
dx 1 2 y = 0. Use the Wronskian to verify...
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This document was uploaded on 02/09/2014.
- Winter '14