HW#1 - Math 293, HW 1 Solutions 1.1:5,15,43,46 Wednesday,...

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Math 293, HW 1 Solutions 1.1:5,15,43,46 Wednesday, Aug 28th, 2007 6). Given the ODE y ′′ + 4 y + 4 y = 0 verify that y 1 = e 2 x and y 2 = xe 2 x are solutions. We begin with y 1 . Differentiate the proposed solution twice y 1 = e 2 x y 1 = - 2 e 2 x y ′′ 1 = 4 e 2 x and substitute into the ODE y ′′ + 4 y + 4 y = 0 (4 e 2 x ) + 4 ( - 2 e 2 x ) + 4 ( e 2 x ) = 0 (4 - 4 2 + 4) e 2 x = 0 0 = 0 Next we look at y 2 . Again, differentiate the proposed solution twice y 2 = xe 2 x y 2 = - 2 xe 2 x + e 2 x y ′′ 2 = 4 xe 2 x - 4 e 2 x and substitute into the ODE y ′′ + 4 y + 4 y = 0 (4 xe 2 x - 4 e 2 x ) + 4 ( - 2 xe 2 x + e 2 x ) + 4 ( xe 2 x ) = 0 (4 xe 2 x - 4 * 2 xe 2 x + 4 xe 2 x ) + ( - 4 e 2 x + 4 e 2 x ) = 0 (4 - 8 + 4) xe 2 x + ( - 4 + 4) e 2 x = 0 0 = 0 15. Substitute y = e rx into the ODE y ′′ + y - 2 y = 0 to find the appropriate value(s) for
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HW#1 - Math 293, HW 1 Solutions 1.1:5,15,43,46 Wednesday,...

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