Use the auxiliary equation k2-5k+6=0. Solving this gives roots k=2 and k=3, which are real and distinct roots. This means that the complementary function is of the form y=Ae^(k1x)+Be^(k2x), where k1 and k2 are roots of the auxiliary equation and A and B are real constants. Therefore the complementary function for this differential equation is y=Ae2t+Be3t.
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