Answer to Question #217807 in Differential Equations for Lwazi Boyce

Question

Solve for x and y in the following set of simultaneous differential equations by using D-operator methods: (D-2)x + Dy = 10sin2t

Dx + (D+2)y = 0

Answer 1

Let's write the given system of ODE in the matrix form

$\begin{pmatrix} D-2 & D\\ D & D+2 \end{pmatrix} \begin{pmatrix} x\\ y \end{pmatrix} = \begin{pmatrix} 10 \sin 2t\\ 0 \end{pmatrix}$

Where $D$ is the differential operator $D=\frac{d}{dt}$.

The determinant of this system is $(D-2)(D+2)=D^2-4$ is invertible, therefore the matrix of the system is invertible too.

For arbitrary invertible matrix 2x2 we have

$\begin{pmatrix} a & b\\ c & d \end{pmatrix}^{-1} = \frac{1}{ad-bc}\begin{pmatrix} d & -c\\ -b & a \end{pmatrix}$

Therefore,

$\begin{pmatrix} D-2 & D\\ D & D+2 \end{pmatrix}^{-1}=-\frac{1}{4}\begin{pmatrix} D+2 & -D\\ -D & D-2 \end{pmatrix}$

and

$\begin{pmatrix} x\\ y \end{pmatrix} =\begin{pmatrix} D-2 & D\\ D & D+2 \end{pmatrix}^{-1}\begin{pmatrix} 10 \sin 2t\\ 0 \end{pmatrix} =-\frac{1}{4}\begin{pmatrix} D+2 & -D\\ -D & D-2 \end{pmatrix}\begin{pmatrix} 10 \sin 2t\\ 0 \end{pmatrix}$

$x=-\frac{1}{4}(D+2)10 \sin 2t=-\frac{1}{4}(20\cos2t+20\sin 2t)=-5\cos2t-5\sin 2t$

$y=-\frac{1}{4}D(10\sin 2t)=-5 \cos 2t$

Answer. $x=-5\cos2t-5\sin 2t$, $y=-5 \cos 2t$.