Answer to Question #271852 in Differential Equations for Wannyan

Question

dx/dt=x-4y

dy/dt=x+y

Answer 1

$Let\\ \frac{d}{dt}=D.\\ \text{The system of equation is given by,}\\ Dx=x-4y\\ \implies(D-1)x+4y=0\space ------(1)\\ And\\ Dy=x+y\\ \implies -x+(D-1)y=0\space ------(2)\\ \text{First Multiply (2) with (D-1) and add(1),we get}\\ [4+(D-1)^2]y=0\\ (D^2-2D+5)y=0 \text{Auxiliary equation,}\\ m^2-2m+5=0\\ m=\frac{2\pm \sqrt{-16}}{2}\\ =1\pm 2i\\ \text{The solution is given by,}\\ y(t)=e^t(Acos(2t)+Bsin(2t))\space ------(3)\\ \text{Putting the value of y in (2), we get}\\ -x+(D-1)(e^t(Acos(2t)+Bsin(2t))=0\\ x=e^t(-2Asin(2t)+2Bcon(2t))+e^t(Acos(2t)+Bsin(2t))-e^t(Acos(2t)+Bsin(2t))\\ x(t)=e^t(-2Asin(2t)+2Bcon(2t))\space ------(4)\\ \text{Thus, (3) and (4) are the required solution.}$