differential equations, diagonalizable matrix

Solution 1:

Given $n\in \mathbb N, A\in \mathbb R^{n\times n}$, a non trivial interval $I, t_0\in I, y_0\in \mathbb R^n$ and $b\colon I\to \mathbb R^n$ a continuous function, consider the initial value problem $$y'+Ay=b, y(t_0)=y_0.$$

Let $f\colon I\to\mathbb R^n$ be a differentiable function.

Fact: For all $t\in \mathbb R$, $e^{At}$ is invertible and $\left(e^{At}\right)^{-1}=e^{-At}$.

There exists $C\in \mathbb R^n$ such that for all $t\in I$ the following holds: $$\begin{align} f'(t)+Af(t)=b(t)&\iff e^{At}(f'(t)+Af(t))=e^{At}b(t)\\ &\iff e^{At}f'(t)+e^{At}Af(t)=e^{At}b(t)\\ &\iff e^{At}f'(t)+Ae^{At}f(t)=e^{At}b(t)\\ &\iff \int \limits _{t_0}^te^{As}f'(s)+Ae^{As}f(s)\mathrm ds=\int \limits_{t_0}^te^{As}b(s)\mathrm ds+C\\ &\iff e^{At}f(t)=\int \limits_{t_0}^te^{As}b(s)\mathrm ds+C\\ &\iff f(t)=e^{-At}\int \limits_{t_0}^te^{As}b(s)\mathrm ds+e^{-At}C. \end{align}$$

Taking into account $f(t_0)=y_0$ after some simple calculations it follows that $C=e^{At_0}y_0$.

This explicitly finds a solution and it shows it is unique.

All you need to do is compute matrices exponentials. No numerical results are needed if you can find the antiderivative on the RHS and certainly that's not a problem when $b$ is the null function.

As for question $1$ it indeed the same if you find complex eigenpairs, but in this case, if you want real solutions, you need to take real and imaginary parts to get them.