Question

2) Solve the following system

𝑑𝑥

𝑑𝑡

= 𝑥 + 𝑦

𝑑𝑦

𝑑𝑡

= −2𝑥 − 𝑦 

Answer 1

$$\dfrac{dx}{dt}=x+y$$

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

The first equation of the system is equivalent to 

$$y=\frac{dx}{dt}-x$$

Then 

$$\frac{dy}{dt}=\frac{d^2x}{dt^2}-\frac{dx}{dt}$$

Put these in the second equation of the system. We have the following differential equation:

$$\frac{d^2x}{dt^2}-\frac{dx}{dt}=-2x-\frac{dx}{dt}+x$$

which is equivalent to

$$\frac{d^2x}{dt^2}+x=0$$

Its characteristic equation $k^2+1=0.$ 

Therefore, $x(t)=C_1\sin t+C_2\cos t$.

Then 

$$\frac{dx}{dt}=C_1\cos t-C_2\sin t$$

 and

$$y=\frac{dx}{dt}-x$$

$$=C_1\cos t-C_2\sin t-C_1\sin t-C_2\cos t$$

 .Consequently, the system has the general solution:

$\begin{cases} x(t)=C_1\sin t+C_2\cos t\\ y(t)=(-C_1-C_2)\sin t+(C_1-C_2))\cos t \end{cases}$