Question

Find the general solution to the given system:

dx/dt=x+2y

dy/dt=2x+y

Answer 1

Let us find the general solution to the given system:

$\begin{cases} \frac{dx}{dt}=x+2y\\ \frac{dy}{dt}=2x+y \end{cases}$

The first equation of the system is equivalent to $y=\frac{1}{2}(\frac{dx}{dt}-x)$. Then $\frac{dy}{dt}=\frac{1}{2}(\frac{d^2x}{dt^2}-\frac{dx}{dt})$. Put these in the second equation of the system. We have the following differential equation:

$\frac{1}{2}(\frac{d^2x}{dt^2}-\frac{dx}{dt})=2x+\frac{1}{2}(\frac{dx}{dt}-x)$

which is equivalent to

$\frac{1}{2}\frac{d^2x}{dt^2}-\frac{dx}{dt}-\frac{3}{2}x=0$

and to

$\frac{d^2x}{dt^2}-2\frac{dx}{dt}-3x=0$

Its characteristic equation $k^2-2k-3=0$ is equivalent to $(k+1)(k-3)=0$, and thus has the solutions $k_1=-1$ and $k_2=3.$

Therefore, $x(t)=C_1e^{-t}+C_2e^{3t}$.

Then $\frac{dx}{dt}=-C_1e^{-t}+3C_2e^{3t}$ and

$y=\frac{1}{2}(\frac{dx}{dt}-x)=\frac{1}{2}(-C_1e^{-t}+3C_2e^{3t}-(C_1e^{-t}+C_2e^{3t}))=-C_1e^{-t}+C_2e^{3t}$ .

Consequently, the system has the general solution:

$\begin{cases} x(t)=C_1e^{-t}+C_2e^{3t}\\ y(t)=-C_1e^{-t}+C_2e^{3t} \end{cases}$