How to solve a first-order linear system of differential equations

Question

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!

Answer 1

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 $x.$

$y' = 2x + 4y\\ x'' = x' - 2x- 4y\\ y = x-x'\\ x'' = x' - 2x- 4x + 4x'\\ x'' - 5x' + 6x = 0$

Answer 2

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.

$$x=Ae^{3t}+Be^{2t}.$$

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*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 $Ae^{3t}$,

$$(D-2)x=A(3-2)e^{3t}=Ce^{3t}.$$

Answer 3

$$\begin{align*} \frac{dx}{dt} & = x-y \\ \frac{dy}{dt} & = 2x + 4y \end{align*}$$

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...