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AMath350.W13.A5.Sol

# AMath350.W13.A5.Sol - AMath 350 Assignment#5 Winter 2013...

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AMath 350 Assignment #5 Winter 2013 Solutions 1. Solve y 0 + 3 y = 5 e - 3 x + 6 xe 3 x : By inspection, the complementary function is y h = Ce - 3 x . For a particular solution, the 5 e - 3 x term would normally require a term of the form Ae - 3 x , but since this matches the complementary function, we’ll need to multiply it by x . Meanwhile, the 6 xe 3 x term will require a trial function of the form ( Ax + B ) e 3 x , by the usual rule for products. So, we try y p = Axe - 3 x + ( Bx + C ) e 3 x = ) y 0 p = Ae - 3 x - 3 Axe - 3 x + Be 3 x + 3 ( Bx + C ) e 3 x Plugging these into the ODE, we obtain the equation Ae - 3 x - 3 Axe - 3 x + 3 Bxe 3 x + ( B + 3 C ) e 3 x + 3 Axe - 3 x + 3 Bxe 3 x + 3 Ce 3 x = 5 e - 3 x + 6 xe 3 x = ) A = 5 , 6 B = 6 , B + 6 C = 0 = ) A = 5 , B = 1 , C = - 1 6 = ) y = Ce - 3 x + 5 xe - 3 x + x - 1 6 e 3 x . 2. a) y 00 - 2 y 0 + y = 1 x e x Solution: The associated homogeneous equation has characteristic equa- tion m 2 - 2 m + m = 0 , i.e. m = 1 , repeated, so y h = C 1 e x + C 2 xe x . Applying the variation of parameters method, we seek a particular solu- tion in the form y p = u ( x ) e x + v ( x ) xe x . As shown in the Course Notes, this requires us to solve the system u 0 e x + v 0 xe x = 0 (1) u 0 e x + v 0 ( e x + xe x ) = 1 x e x . (2) Subtracting (1) from (2) yields v 0 e x = 1 x e x , so v 0 = 1 x , and hence v = ln x . Meanwhile, (1) gives us u 0 e x = - v 0 xe x = e x , so u 0 = 1 , and u = x . Therefore y p = xe x + xe x ln x. When we write out the general solution, we see that the first of these terms will be absorbed by the C 2 term; the solution is y = C 1 e x + C 2 xe x + xe x ln x. b) y 00 + y = sin 2 x Solution: The complementary function is y h = C 1 cos x + C 2 sin x , so we look for a particular solution in the form y p = u ( x ) cos x + v ( x ) sin x.

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AMATH 350 Assignment #5 Solutions - Fall 2011 Page 2 The unknown functions must satisfy the system of equations u 0 cos x + v 0
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