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 ,buts inceth ismatchesthecomp lementaryfunct ion ,we l lneedto multiply it by x . Meanwhile, the 6 xe 3 x term will require a trial function of the form ( Ax + B ) e 3 x ,bytheusua lru leforproducts .So ,wetry y p = Axe - 3 x +( Bx + C ) e 3 x = ) y 0 p = Ae - 3 x - 3 Axe - 3 x + Be 3 x +3( + C ) e 3 x Plugging these into the ODE, we obtain the equation Ae - 3 x - 3 Axe - 3 x Bxe 3 x B C ) e 3 x + 3 Axe - 3 x 3 x 3 x e - 3 x xe 3 x = ) A , 6 B =6 ,B C =0 = ) A =1 ,C = - 1 6 = ) y = - 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 ,i .e . m ,repea ted ,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 (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 v 0 = 1 x ,andhence v =ln x . Meanwhile, (1) gives us u 0 e x = - v 0 xe x = e x ,s o u 0 ,and u = x . Therefore y p = xe x + xe x ln x. When we write out the general solution, we see that the ±rst 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 =s in 2 x Solution: The complementary function is y h = C 1 cos x + C 2 sin x ,sowe 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 sin x =0 (3) - u 0 sin x + v 0 cos x =s in 2 x. (4) Dividing (3) by
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AMath350.W13.A5.Sol - AMath 350 Assignment#5 Winter 2013...

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