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AMath350.W13.A4.Sol

# X m1 sin2x m2 cos2x for y thus equation so

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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 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...
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