To solve a linear second order differential equation we first find the complementary function and then the particular integral. To find the complementary function we must find the roots of the auxiliary equation m2+ 5m + 6 = 0 which are m = -3 and m = -2 and which can be found by factorising. This means that the complementary function of the differential equation is xc = A e-3t+B e-2t where A and B are constants.As 4 e-t is to the right of the equals sign, the particular integral must be of the form xp =C e-t where C is a constant. We then differentiate the particular integral twice and substitute into the original differential equation. Differentiating gives dxp/dt = -Ce-t and d2xp/dt2 = Ce-t. Substituting these into the differential equation gives 2C e-t= 4e-t, therefore C =2 and so xp=2e-t. Finally we add the complementary function to the particular integral to reach the general solution x = A e-3t+B e-2t+ 2e-t.
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