Question

Solve the following system of ODE

x'+x-y = e^2t

Answer 1

$\textsf{Case 1.}\\ \textsf{If} \, y(t) = x(t) \\ x' + x - y = e^{2t}\\ \displaystyle\frac{\mathrm{d}x}{\mathrm{d}t} + x - y = e^{2t}\\ \textsf{If}\hspace{0.1cm} y = x = e^{mt},\\ \frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\mathrm{d}x}{\mathrm{d}t} = me^{mt}\\ \implies me^{mt} + e^{mt} - e^{mt}= e^{2t}\\ me^{mt} = e^{2t}\\ me^{m} = e^2 \implies me^m = e^2\\ m\,\textsf{can only be solved for using}\\\textsf{numerical methods, such}\\\textsf{as the Lambert}W\,\textsf{function.}\\ m = W_n(e^2), \hspace{0.1cm} \textsf{where}\,W_n(z) \\\textsf{is the Lambert}W \,\textsf{function}.\\ m = W_n(e^2) \hspace{0.1cm}\textsf{is the solution}\\\textsf{to the equation.}\,\textsf{where}\, n \in \mathbb{Z}\\ \therefore y = x = e^{W_n(e^2)} \hspace{0.1cm}\textsf{is a solution to the first}\\\textsf{order linear ODE.}\, \forall n \in \mathbb{Z}. \\ \textsf{Case 2.}\\ \textsf{If} \, y(t) \,\textsf{is a multiple of}\, x(t) \\ x' + x - y = e^{2t}\\ \displaystyle\frac{\mathrm{d}x}{\mathrm{d}t} + x - y = e^{2t}\\ \textsf{If}\hspace{0.1cm} x = e^{mt}, y = me^{mt} \frac{\mathrm{d}x}{\mathrm{d}t} = me^{mt}\\ \implies me^{mt} + e^{mt} - me^{mt}= e^{2t}\\ e^{mt} = e^{2t} \implies m = 2\\ m = 2\hspace{0.1cm}\textsf{is the only solution}\\\textsf{to the equation.}\\ \therefore x= e^{2t}, y = 2e^{2t} \hspace{0.1cm}\textsf{is a solution to the first}\\\textsf{order linear ODE.}\\ \textsf{We now have two solutions for}\, x \, \& \, y.$