Homework 4 Solution 1 Sol.

So we have A = 1 4: Thus a particular solution is y p = 1 4 xe x, and so the general solution is y = y c + y p = C 1ex + C 2e 3x + 1 4 xex: 10. y00 4y0+ 4y0= e2x Sol. The characteristic equation m2 4m + 4 = (m 2)2 = 0 has a root m = 2 with multiplicity 2. The complementary solution is y c = C 1e 2x + C 2xe 2x: ................
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