ordinary differential equationsstability-theory
I try to examine stability of non-homogeneous ODE system:
\begin{cases} Dy_{1} = y_{1}+2y_{2} +\frac{3}{x^4} \\ Dy_{2}= 3y_{1}+4y_{2}+ \frac{3}{x^4} \end{cases}
I tried to find solutions of such system and then examine whether solutions are stable, but I can't find them, is there any other method to determine stability of such system?
There is no asymptotic stability. Taken the system
$$ \cases{ \dot x_1 = x_2\\ \dot x_2 = -x_1^3 } $$
and multiplying by $x_1^3, x_2$ as
$$ \cases{ x_1^3\dot x_1 = x_1^3x_2\\ x_2\dot x_2 = -x_1^3x_2 } $$
and adding we have
$$ \frac 12x_1^4+x_2^2 = C $$
so those orbits characterize a center around the origin.
I'm not sure what you mean by "applying this method to the system... You have written down the linear system to be solved (apart from the correction mentioned by @Ian), now you just need to solve it. Your equations are $$ -3y_0 + 4y_1-y_2 = - 6h $$ $$ y_{i-1}-(2+h^2)y_i+y_{i+1} = 8h^2+3h^2 x_i(1-x_i), \quad i = 1, \cdots, n $$ $$ y_{n-1}-4y_n+3y_{n+1} = 2h (e-e^{-1}+3) $$
so, apart from the first and last lines, you have a tridiagonal system. Below you can visualize the numerical solution with $n=10$ (dots), compared to the exact solution $y(x)=-2 + e^{-t} + e^t - 3 t + 3 t^2$ (line).
It is also very instructive to observe the behaviour of the numerical solution when you use first order approximations of the derivatives to enforce the Neumann conditions, i.e. what happens if you replace the fisrt and last equations by $y_1-y_0=-3h$ and $y_{n+1}-y_n = h(e-e^{-1}+3)$.
Best Answer
First, solve the homogeneous equation:
$$Dy_1=y_1+2y_2$$ $$Dy_2=3y_1+4y_2$$
I am assuming you already know how to do this, so I will just write the solution here:
$$\left[\begin{matrix} y_1 \\ y_2\end{matrix}\right]=\left[\begin{matrix} \frac{1}{6}(-3-\sqrt{33})e^{(5-\sqrt{33})/2} & \frac{1}{6}(-3+\sqrt{33})e^{(5+\sqrt{33})/2} \\ e^{(5-\sqrt{33})/2} & e^{(5+\sqrt{33})/2}\end{matrix}\right]\left[\begin{matrix} c_1 \\ c_2\end{matrix}\right] \text{ where } c_1,c_2\in\mathbb{R}$$
Since there is one negative eigenvalue $\frac{5-\sqrt{33}}{2}$ and one positive eigenvalue $\frac{5+\sqrt{33}}{2}$, the origin is a saddle point and the system is unstable. Therefore, at the very least, you now know the stability of the ODE system.
Now, to actually solve the non-homogeneous equation, use variation of parameters by changing $c_1,c_2$ to $u_1(t),u_2(t)$:
$$\left[\begin{matrix} y_1 \\ y_2\end{matrix}\right]=\left[\begin{matrix} \frac{1}{6}(-3-\sqrt{33})e^{(5-\sqrt{33})/2} & \frac{1}{6}(-3+\sqrt{33})e^{(5+\sqrt{33})/2} \\ e^{(5-\sqrt{33})/2} & e^{(5+\sqrt{33})/2}\end{matrix}\right]\left[\begin{matrix} u_1(t) \\ u_2(t)\end{matrix}\right]$$ $$\left[\begin{matrix} \frac{1}{6}(-3-\sqrt{33})e^{(5-\sqrt{33})/2} & \frac{1}{6}(-3+\sqrt{33})e^{(5+\sqrt{33})/2} \\ e^{(5-\sqrt{33})/2} & e^{(5+\sqrt{33})/2}\end{matrix}\right]\left[\begin{matrix} u_1'(t) \\ u_2'(t)\end{matrix}\right]=\left[\begin{matrix} \frac{3}{x^4} \\ \frac{3}{x^4}\end{matrix}\right]$$
Here, you can use the second equation to solve for $u_1'(t),u_2'(t)$, integrate to solve for $u_1(t),u_2(t)$, and then plug into the first equation to solve for $y_1,y_2$. Good luck!