Question

Find the general solution to

y'' − y'− 2y = 2e^3x

Answer 1

$\text{The general solution to the given differential equation is }\\ y = y_p(x) + y_c(x)\\ \text{where $y_p(x)$ represents particular solution and $y_c(x)$ represents the complimentary }\\ \text{solution.}\\ \text{The characteristic equation is given by }y^2-y-2\\ \text{Hence y = -1 and y = 2, therefore the complimentary solution is given by}\\ y_c = C_1e^{-x}+C_2e^{2x}\\ y_p(t) = Ae^{3x}, y'_p(t) = 3Ae^{3x}, y''_p(t) = 9Ae^{3x}\\ \text{Substituting the above into the given differential equation, we have}\\ 9Ae^{3x}-3Ae^{3x}-2Ae^{3x}=2Ae^{3x}\\ \implies A = \frac{1}{2}\\ \therefore y_p(x)= \frac{1}{2}e^{3x}\\ \implies y = C_1e^{-x}+C_2e^{2x}+\frac{1}{2}e^{3x}$