I am looking for general solutions for the linear sODE's $$\textbf{x}'(t) = A\textbf{x}(t)$$ with $t \geq 0$ and $A \in \mathbb{R}^{n \times n}$

Let focus on just real eigenvalues and eigenvectors. For the case of $n=2$, where we have one eigenvalue, $\lambda \in \mathbb{R}$, such that $am(\lambda)=2, gm(\lambda)=1$ I know that the general solution is $$\textbf{x}(t) = e^{\lambda t}(c_1 \textbf{v} + c_2\textbf{w}) + te^{\lambda t}(c_2 \textbf{w})$$ where $c_1,c_2 \in \mathbb{R}$, $\textbf{v}$ is the eigenvector corresponding to $\lambda$ and $\textbf{w}$ is a generalized eigenvector of $A$.

But what would be the general solution be in the case $n = 3$ with one eigenvalue, $\lambda \in \mathbb{R}$, such that $am(\lambda)=3, gm(\lambda)=1$ or in the case of two eigenvalues $\lambda_1,\lambda_2$ with $am(\lambda_1)=2, gm(\lambda_1) = 1, am(\lambda_2)=1, gm(\lambda_2) = 1$?

So, all in all, how would one find the general solution to such systems of linear differential equations?

Full answers are appreciated, but I prefer some hints to find the solution myself.

Thanks in advance!

Bonus: The same question but then with difference equations.


This process is explained in numerous ODE textbooks and lecture notes; why reinvent the wheel?

  1. Find the eigenvalues (you are assuming they are real).
  2. For each eigenvalue $\lambda$, find a basis of the eigenspace, i.e., of the null space of $A-\lambda I$.
  3. Each basis element $v$ yields a solution $e^{\lambda t}v $.
  4. If the dimension of the eigenspace is equal to the algebraic multiplicity of $\lambda$, we are done with $\lambda$.
  5. Otherwise look for generalized eigenvectors on top of each $v$ as in step 3. These come from equations $(A-\lambda I)w_1=v$, $(A-\lambda I)w_2=w_1$, etc.
  6. The generalized eigenvectors contribute solutions $$e^{\lambda t} (tv+w_1),\quad e^{\lambda t} \left(\frac{t^2}2 v+tw_1+w_2\right), \dots$$ where the coefficients are of the form $t^k/k!$
  7. The number of solutions obtained in this way is $n$. Since they are linearly independent by construction, their linear combinations yield the general solution of the system.
| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.