Question

(D-1)x + Dy = 0

Dx + 2Dy = 4$e^2t$

Answer 1

Let us solve the system:

$\begin{cases} (D-1)x + Dy = 0\\ Dx + 2Dy = 4e^{2t} \end{cases}$

$\begin{cases} Dy = -(D-1)x \\ Dx + 2Dy = 4e^{2t} \end{cases}$

$\begin{cases} Dy = -(D-1)x \\ Dx - 2(D-1)x = 4e^{2t} \end{cases}$

$\begin{cases} Dy = -(D-1)x \\ -(D-2)x = 4e^{2t} \end{cases}$

$\begin{cases} Dy = -(D-1)x \\ (D-2)xe^{-2t} = -4 \end{cases}$

$\begin{cases} Dy = -(D-1)x \\ D(xe^{-2t}) = -4 \end{cases}$

$\begin{cases} Dy = -(D-1)x \\ xe^{-2t} = -4t+C_1 \end{cases}$

$\begin{cases} Dy = -(D-1)x \\ x = -4te^{2t}+C_1e^{2t} \end{cases}$

$\begin{cases} Dy = 4e^{2t}+8te^{2t}-2C_1e^{2t}-4te^{2t}+C_1e^{2t}= 4e^{2t}+4te^{2t}-C_1e^{2t}\\ x = -4te^{2t}+C_1e^{2t} \end{cases}$

We conclude that the solution is of the following form:

$\begin{cases} y =e^{2t}+2te^{2t}-\frac{1}{2}C_1e^{2t}+C_2\\ x = -4te^{2t}+C_1e^{2t} \end{cases}$