How to solve a first-order linear system of differential equations I am not quite sure how to start this problem, I haven't seen anything like this before.
Solve the following system of ODEs :
\begin{align*}
\frac{dx}{dt} & = x-y \\
\frac{dy}{dt} & = 2x + 4y
\end{align*}
Thanks!
 A: There are two ways you can go....
one is to create a matrix.
$\mathbf x = \begin{bmatrix} x\\y\end{bmatrix}$
$\mathbf x' = \begin{bmatrix} 1 &-1\\2&4\end{bmatrix} \mathbf x$
And solve it just like you solve any other linear diff eq.
$\mathbf x' = A\mathbf x\\
\mathbf x = e^{At}\mathbf x(0)\\
A = P^{-1}DP\\
e^{At} = P^{-1}e^{Dt}P\\
$
The other is to differentiate one of the lines and make a second order diff eq.
$x' = x-y\\
x'' = x' - y'$
Then use the given equations to get everything it terms of $x.$
$y' = 2x + 4y\\
x'' = x' - 2x- 4y\\
y = x-x'\\
x'' = x' - 2x- 4x + 4x'\\
x'' - 5x' + 6x = 0$
A: You can work this out by introducing the differential operator $D:=\dfrac d{dt}$ and rewrite the system as
$$\begin{cases}(D-1)x+y=0,\\2x-(D-4)y=0.\end{cases}$$
Then applying  factors $D-k$ on the left and summing the equations, you can eliminate an unknown or another as with an ordinary linear system.
$$\begin{cases}(D-4)(D-1)x+2x=0,\\
2y+(D-1)(D-4)y=0.\end{cases}$$
In both cases, you get a second order equation
$$(D^2-5D+6)x=(D-3)(D-2)x=0$$ and similar for $y$.
Even better, the solution of the latter is simply the sum of the solutions of $(D-3)x=0$ and $(D-2)x=0$*, i.e.
$$x=Ae^{3t}+Be^{2t}.$$

*Indeed, let $z=(D-2)x$. We have
$$(D-3)z=0$$ which has the solution 
$$z=Ce^{3t}.$$
Then
$$(D-2)x=Ae^{3t}$$ has for solution that of the homogeneous part, $Be^{2t}$ plus a particular solution of the non homogeneous equation. With the ansatz $Ae^{3t}$,
$$(D-2)x=A(3-2)e^{3t}=Ce^{3t}.$$
A: $$\begin{align*}
\frac{dx}{dt} & = x-y \\
\frac{dy}{dt} & = 2x + 4y
\end{align*}$$
You can also add both equation and get
$$x'+y'=3(x+y) \implies \int \frac {d(x+y)}{x+y}=3\int dt$$
$$ \ln(x+y)=3t+C \implies x=Ke^{3t} -y$$
Then you have that
$$x'=x-y \implies x'=2x-Ke^{3t}$$
Which is easy to solve...
