My differential equations textbook states to use the "elimination method" to crack this for $x$ and $y$. The final answer uses $t$ as the independent variable which both $x$ and $y$ are dependent on. I was able to work the problem but it seemed to involve a lot of work. How could I have done it faster?
The system consists of the following two linear ordinary differential equations written in linear differential operator forms along with their respective initial conditions:
$$\begin{align*} &(1) \: Dx = 2x + 3y + 2e^{2t} \qquad x(0) = -\frac{2}{3} &\\ &(2) \: Dy = x + 4y + 3e^{2t} \qquad y(0) = \frac{1}{3} & \end{align*}$$
The textbook states that the general solution for both $x$ and $y$ are:
$$\begin{align*} & x(t) = e^{5t} - \frac{5}{3}e^{2t} &\\ & y(t) = e^{5t} - \frac{2}{3}e^{2t} & \end{align*}$$
The following steps shows my work. I begin by solving for $y$ and $y'$ in (1):
$$\begin{align*} & \: 3y = x' -2x - 2e^{2t} & \\ & \: y = \frac{1}{3}x' - \frac{2}{3}x - \frac{2}{3}e^{2t} & \\ & \: y' = \frac{1}{3}x'' - \frac{2}{3}x' - \frac{4}{3}e^{2t} & \\ \end{align*}$$
I can now make substitutions (shown in parenthesis) for $y$ and $y'$ into equation (2) and simplify by collecting like terms and finally solve for $x$:
$$\begin{align*} & \: y' = x + 4y + 3e^{2t} &\\ & \: \Big(\frac{1}{3}x'' - \frac{2}{3}x' - \frac{4}{3}e^{2t}\Big) = x + 4\Big(\frac{1}{3}x' - \frac{2}{3}x - \frac{4}{3}e^{2t}\Big) + 3e^{2t} &\\ & \: \left[\frac{1}{3}x'' - \frac{2}{3}x' - \frac{4}{3}e^{2t} = x + \frac{4}{3}x' - \frac{8}{3}x - \frac{8}{3}e^{2t} + 3e^{2t} \right] 3x &\\ & \: x'' - 2x' - 4e^{2t} = 3x + 4x'' - 8x' - 8e^{2t} + 9e^{2t} &\\ & \: x'' - 6x' +5x = 5e^{2t} &\\ \end{align*}$$
The homogeneous part of the solution for $x_h$ couldn't be easier to find:
$$\lambda^2 - 6\lambda + 5 = 0 \qquad \lambda_1 = 1 \qquad \lambda_2 = 5 $$
$$\begin{align*} & x_h = c_1e^{t} +c_2e^{5t} &\\ \end{align*}$$
The particular part of the solution for $x$ is solved by taking an educated guess for $5e^{5t}$ and its derivatives and inserting into $x$:
$$x_p = Ce^{2t} \qquad x_p' = 2Ce^{2t} \qquad x_p'' = 4Ce^{2t} $$ $$x'' - 6x' +5x = 5e^{2t}$$ $$4Ce^{2t} - 6(2Ce^{2t}) + 5(Ce^{2t}) = 5e^{2t}$$ $$4C - 12C = 5C = 5 \qquad C = -\frac{5}{3}$$ $$x = x_h + x_p$$ $$x = c_1e^{t} + c_2e^{5t} - \frac{5}{3}e^{2t} \qquad $$ $$x' = c_1e^{t} + 5c_2e^{5t} - \frac{10}{3}e^{2t} \qquad $$
Now we are ready to solve for $y$ from equation (1). As before we do a few substitutions (also in parenthesis) but this time in $x$ and $x'$ using an intermediate previous step for $y$:
$$\begin{align*} & \: y = \frac{1}{3}x' - \frac{2}{3}x - \frac{2}{3}e^{2t} & \\ & \: y = \frac{1}{3}(c_1e^{t} + 5c_2e^{5t} - \frac{10}{3}e^{2t}) - \frac{2}{3}(c_1e^{t} + c_2e^{5t} - \frac{5}{3}e^{2t}) - \frac{2}{3}e^{2t} & \\ & \: y = -\frac{1}{3}c_1e^{t} + c_2e^{5t} -\ \frac{2}{3}e^{2t} & \\ \end{align*}$$
A specific solution is found by inserting initial conditions into $x$ and $y$ as follows:
$$x(0) = c_1 + c_2 - \frac{5}{3} = -\frac{2}{3}$$ $$y0) = -\frac{1}{3}c_1 + c_2 - \frac{2}{3} = \frac{1}{3}$$ $$c_1 = 0 \qquad c_2 = 1$$
My final solution matches the answer in text:
$$\begin{align*} & x(t) = e^{5t} - \frac{5}{3}e^{2t} &\\ & y(t) = e^{5t} - \frac{2}{3}e^{2t} & \end{align*}$$
Question: Are there less tedious or faster methods for solving these kinds of problems? If so how would it be done?
I do not know if this is really faster than what you did or even different from what you did very properly.
I extracted $y(t)$ for the first equation $$y(t)=\frac{1}{3} \left(x'(t)-2 x(t)-2 e^{2 t}\right)\tag 1$$ So, writing the second equation $$\frac{1}{3} \left(x''(t)-2 x'(t)-4 e^{2 t}\right)=\frac{4}{3} \left(x'(t)-2 x(t)-2 e^{2 t}\right)+x(t)+3 e^{2 t}$$ After simplifications, this leads to $$x''(t)-6 x'(t)+5 x(t)=5 e^{2 t}\tag 2$$ Solve, just as you did, $(2)$ and, knowing $x(t)$, go back to $(1)$ and get $y(t)$.
Edit
We can make this more general considering $$x'(t)=a x(t)+b y(t)+F(t)\tag 3$$ $$y'(t)=c x(t)+d y(t)+G(t)\tag 4$$ where $a,b,c,d$ are constants and $F(t), G(t)$ are arbitrary functions.
Extract $y(t)$ from $(3)$ $$y(t)=\frac{-a x(t)-F(t)+x'(t)}{b} \tag 5$$ Plug this in $(4)$ $$\frac{-a x'(t)-F'(t)+x''(t)}{b}=\frac{d \left(-a x(t)-F(t)+x'(t)\right)}{b}+c x(t)+G(t)$$ $$x''(t)-(a+d) x'(t)+(a d-bc) x(t)=b G(t)-d F(t)+F'(t)\tag 6$$ Solve $(6)$ and get $x(t)$ and reuse $(5)$ to get $y(t)$.