Question

solve the given system of differential equations:

dx/dt=2x+2y

dy/dt=x+3

Answer 1

solve the given system of differential equations:

$\begin{cases} \frac{dx}{dt}=2x+2y \\ \frac{dy}{dt}=x+3 \end{cases}$

Solution:

From the second equation:

$x=\frac{dy}{dt}-3$  (*)

$\frac{dx}{dt}=\frac{d^2y}{dt^2}$

Substitute $x$ and $\frac{dx}{dt}$ into the first equation:

$\frac{d^2y}{dt^2}=2(\frac{dy}{dt}-3 )+2y$

$\frac{d^2y}{dt^2}-2\frac{dy}{dt}-2y=-6$ (**)

First find the general solution of the corresponding homogeneous equation:

$\frac{d^2y_0}{dt^2}-2\frac{dy_0}{dt}-2y_0=0$

Let's compose and solve the characteristic equation:

$\lambda^2-2\lambda-2=0$

$\lambda_1=1-\sqrt3$ ,  $\lambda_2=1+\sqrt3$ .

$y_0=C_1e^{\lambda_1t}+C_2e^{\lambda_2t}=C_1e^{(1-\sqrt3)t}+C_2e^{(1+\sqrt3)t}$ .

Find a typical (specific) solution of the non-homogeneous equation in the form $y_p=A$

Derivatives of the solution: $\frac{dy_p}{dt}=0$ , $\frac{d^2y_p}{dt^2}=0$

Substitute those "solutions" into the (**):

$-2A=-6$

$A=3$

Add the typical and the complementary solutions to get the complete solution:

$y=y_0+y_p=C_1e^{(1-\sqrt3)t}+C_2e^{(1+\sqrt3)t}+3$ .

To find $x$ substitute $\frac{dy}{dt}$ into the (*):

$x=\frac{dy}{dt}-3 =C_1(1-\sqrt3)e^{(1-\sqrt3)t}+C_2(1+\sqrt3)e^{(1+\sqrt3)t}-3$

Answer: $\begin{cases} x =C_1(1-\sqrt3)e^{(1-\sqrt3)t}+C_2(1+\sqrt3)e^{(1+\sqrt3)t}-3 \\ y=C_1e^{(1-\sqrt3)t}+C_2e^{(1+\sqrt3)t}+3 \end{cases}$