X m1 sin2x m2 cos2x for y thus equation so

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 0 yp = 2M cos(2x) 2M2 sin(2x) Lxe x + Le x yp = M1 1 sin(2x) + M2 cos(2x) + Lxe x 00 yp = 4M1 sin(2x) 4M2 cos(2x) + Lxe x 2Le x 0 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 dy + 12x 2 dx dx 1 2 y = 0. Use the Wronskian to verify...
View Full Document

This document was uploaded on 02/09/2014.

Ask a homework question - tutors are online