Answer to Question #216972 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

We know that the operator $D$ means a derivative $D=\frac{d}{dt}$ , but for now we will assume that this is just another letter and solve the specified system of equations by the "addition" method.

$$\left\{\begin{array}{l} \left.\left(D-2\right)x+Dy=10\sin2t\right|\times\left(D\right)\\[0.3cm] \left.Dx+\left(D+2\right)y=0\right|\times\left(D-2\right) \end{array}\right.\\[0.3cm] \left\{\begin{array}{l} D\left(D-2\right)x+D^2y=D\left(10\sin2t\right)\\[0.3cm] \text{Subtract the lower one from the upper equation}\\[0.3cm] D\left(D-2\right)x+\left(D^2-4\right)y=0 \end{array}\right.\\[0.3cm]\\[0.3cm] \left(\cancel{D^2}-\cancel{D^2}+4\right)y=D\left(10\sin2t\right)\\[0.3cm] 4y=\frac{d}{dt}\left(10\sin2t\right)=20\cos2t\to\boxed{y(t)=5\cos2t}\\[0.3cm] \text{To find the function} \quad x(t), \text{go back to the original system.}\\[0.3cm] \left\{\begin{array}{l} \left.\left(D-2\right)x+Dy=10\sin2t\right|\times\left(D+2\right)\\[0.3cm] \left.Dx+\left(D+2\right)y=0\right|\times\left(D\right) \end{array}\right.\\[0.3cm] \left\{\begin{array}{l} \left(D^2-4\right)x+D\left(D+2\right)y=\left(D+2\right)\left(10\sin2t\right)\\[0.3cm] \text{Subtract the lower one from the upper equation}\\[0.3cm] D^2x+D\left(D+2\right)y=0 \end{array}\right.\\[0.3cm]\\[0.3cm] \left(\cancel{D^2}-\cancel{D^2}-4\right)x=\left(D+2\right)\left(10\sin2t\right)\\[0.3cm] -4x=\left(\frac{d}{dt}+2\right)\left(10\sin2y\right)=20\cos2t+20\sin2t\to\\[0.3cm] \boxed{x(t)=-5\cos2t-5\sin2t}\\[0.3cm]$$

ANSWER

$$\left\{\begin{array}{l} x(t)=-5\cos2t-5\sin2t\\[0.3cm] y(y)=5\cos2t \end{array}\right.$$