Find the general solution of the second order differential equation y''(t)+y(t) = 5exp(2t)

The general solution yGS(t) to a differential equation in the form ay''(t)+by'(t)+cy(t)=f(t) is the sum of the complementary function yCF(t) and a particular solution yPS(t).First we find the complementary function, which is the general solution to the equation y''(t)+y(t)=0, so the right hand side is zero. The characteristic equation associated with this equation is r2+1=0, which has solutions r=±i. Hence yCF(t) = Acos(t)+Bsin(t) where A and B are arbitrary constants.Next we find a particular solution. We guess based on the right hand side that a particular solution has the form yPS(t)=Ce2t, where C is a constant to be determined.Differentiating twice we find that yPS''(t)=4Ce2t, so yPS''(t)+yPS(t) = 4Ce2t + Ce2t = 5Ce2t.But since y''(t)+y(t)=5e2t, this means that 5Ce2t must equal 5e2t, so we see that C=1, so yPS(t)=e2t.Therefore yGS(t) = yCF(t) + yPS(t) = Acos(t) + Bsin(t) + e2t.

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