Question

dx dt=2x-4y 

dx dt=2x+4y

Answer 1

$$\dfrac{dx}{dt}=2x-4y$$ $$\dfrac{dy}{dt}=2x+4y$$

$$Dx-2x+4y=0$$ $$Dy-2x-4y=0$$

$$(D-2)x+4y=0$$ $$-2x+(D-4)y=0$$

Multiply first equation by $2$ and second equarion by $D-2$

$$2(D-2)x+8y=0$$ $$-2(D-2)x+(D-2)(D-4)y=0$$

Add two equations

$$-2(D-2)x+(D-2)(D-4)y$$

$$+(-2(D-2)x+(D-2)(D-4)y)=0$$

Simplify

$$D^2-6D+16y=0$$

Auxiliary equation

$$r^2-6r+16=0$$

$$r=3\pm i\sqrt{7}$$

$$y(t)=c_1e^{3t}\cos(\sqrt{7}t)+c_2e^{3t}\sin(\sqrt{7}t)$$

Then

$$\dfrac{dy}{dt}=3c_1e^{3t}\cos(\sqrt{7}t)+3c_2e^{3t}\sin(\sqrt{7}t)$$

$$-\sqrt{7}c_1e^{3t}\sin(\sqrt{7}t)+\sqrt{7}c_2e^{3t}\cos(\sqrt{7}t)$$

Substitute

$$\dfrac{dy}{dt}=2x+4y$$

$$x=-2y+\dfrac{1}{2}\dfrac{dy}{dt}$$

$$x=-2c_1e^{3t}\cos(\sqrt{7}t)-2c_2e^{3t}\sin(\sqrt{7}t)$$

$$+\dfrac{3}{2}c_1e^{3t}\cos(\sqrt{7}t)+\dfrac{3}{2}c_2e^{3t}\sin(\sqrt{7}t)$$

$$-\dfrac{\sqrt{7}}{2}c_1e^{3t}\sin(\sqrt{7}t)+\dfrac{\sqrt{7}}{2}c_2e^{3t}\cos(\sqrt{7}t)$$

$$x(t)=(-\dfrac{1}{2}c_1+\dfrac{\sqrt{7}}{2}c_2)e^{3t}\cos(\sqrt{7}t)$$

$$+(-\dfrac{1}{2}c_2-\dfrac{\sqrt{7}}{2}c_1)e^{3t}\sin(\sqrt{7}t)$$

$$y(t)=c_1e^{3t}\cos(\sqrt{7}t)+c_2e^{3t}\sin(\sqrt{7}t)$$