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

The general solution y_GS(t) to a differential equation in the form ay''(t)+by'(t)+cy(t)=f(t) is the sum of the complementary function y_CF(t) and a particular solution y_PS(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 y_CF(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 y_PS(t)=Ce^2t, where C is a constant to be determined.Differentiating twice we find that y_PS''(t)=4Ce^2t, so y_PS''(t)+y_PS(t) = 4Ce^2t + Ce^2t = 5Ce^2t.But since y''(t)+y(t)=5e^2t, this means that 5Ce^2t must equal 5e^2t, so we see that C=1, so y_PS(t)=e^2t.Therefore y_GS(t) = y_CF(t) + y_PS(t) = Acos(t) + Bsin(t) + e^2t.