Solve the system of ordinary differential of second order non-homogeneous

Solution to the System of Ordinary Differential Equations

We solve the following system of ODEs:

$\ddot{x} - \dot{x} + x = 3$

$\ddot{y} + \dot{y} - y = 0$

The differential equation for $x(t)$ is:

$\ddot{x} - \dot{x} + x = 3.$

First, find the complementary solution by solving the homogeneous equation:

$\ddot{x} - \dot{x} + x = 0.$

The characteristic equation is:

$r^2 - r + 1 = 0.$

The roots of this equation are:

$r = \frac{1 \pm \sqrt{-3}}{2} = \frac{1}{2} \pm i\frac{\sqrt{3}}{2}.$

The complementary solution is therefore:

$x_c(t) = e^{t/2}\left(C_1 \cos\left(\frac{\sqrt{3}}{2}t\right) + C_2 \sin\left(\frac{\sqrt{3}}{2}t\right)\right).$

For the particular solution, assume $x_p = A$ . Substituting into the equation, we get:

$A = 3.$

Thus, the general solution for $x(t)$ is:

$x(t) = e^{t/2}\left(C_1 \cos\left(\frac{\sqrt{3}}{2}t\right) + C_2 \sin\left(\frac{\sqrt{3}}{2}t\right)\right) + 3.$

The differential equation for $y(t)$ is:

$\ddot{y} + \dot{y} - y = 0.$

The characteristic equation is:

$r^2 + r - 1 = 0.$

The roots of this equation are:

$r = \frac{-1 \pm \sqrt{5}}{2}.$

Thus, the general solution for $y(t)$ is:

$y(t) = C_3 e^{\frac{-1 + \sqrt{5}}{2}t} + C_4 e^{\frac{-1 - \sqrt{5}}{2}t}.$

The solutions to the system of ODEs are:

$x(t) = e^{t/2}\left(C_1 \cos\left(\frac{\sqrt{3}}{2}t\right) + C_2 \sin\left(\frac{\sqrt{3}}{2}t\right)\right) + 3,$

$y(t) = C_3 e^{\frac{-1 + \sqrt{5}}{2}t} + C_4 e^{\frac{-1 - \sqrt{5}}{2}t},$

where $C_1, C_2, C_3, C_4$ are constants determined by initial conditions.