This is the system: $$\begin{cases} \dot{x}=x+2y+e^{-t}\\ \dot{y}=2x+y+1 \end{cases}$$

Now I first solve the homogeneous one, without the vector $(e^{-t},1)$, so I have to find the eigenvalues of the matrix

$$\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$$ which are $(1-\lambda)^2-4=\lambda^2-2\lambda-3$ and $\lambda_1=3$, $\lambda_2=-1$. By the first one I obtain the eigenvector $u=(1,1)$ for example, for the second one I obtain the matrix $$\begin{pmatrix} 0 & 2\\ 2 & 0 \end{pmatrix}$$ that has null eigenvector. Now, I find the generalized eigenvector: $(A-\lambda I_d)u_1=u$ and obtain $$\begin{cases} 2u_1^1=1\\ 2u_2^2=1 \end{cases}$$ I chose the eigenvector $u_1=(\frac{1}{2},0)$ is it correct?

I think no, because going on in the resolution of the equation I have to write the $A$ matrix of the system as a sum $A=P+N$ where $P=Sdiag[\lambda_i]S^{-1}$, $S=\begin{pmatrix} 1 & \frac{1}{2}\\ 1 & 0\end{pmatrix}$ and $N$ is nihilipotent of order 2 and then: $$e^{At}=e^{Pt}e^{Nt}=Sdiag[e^{\lambda_i}]S^{-1}(I_d+Nt)$$ I usually do like this when the eigenvalues of the matrix have multiplicity, is it correct to solve like this in this case? And if not, what is the correct method? Thanks and sorry for bad anglisk

  • 1
    $\begingroup$ For $\lambda=-1$ the matrix is $A+E=\begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix}$, not $\begin{pmatrix} 0& 2 \\ 2 & 0 \end{pmatrix}$ $\endgroup$
    – AVK
    Commented Jul 13, 2018 at 13:25

1 Answer 1


You start off well and then get lost a little bit towards the end. Let me try to clear up the confusion.

The first step towards a solution, as you correctly note, is to transform the system of ODEs into a matrix equation. Concretely, you'll find

$$\mathbf{x}(t)=A \mathbf{x}(t)+\mathbf{g}(t)$$

with $\mathbf{g}(t)=(e^t,1)^T$, $\mathbf{x}(t)=(x(t),y(t))^T$ and the matrix $A$ has the form $$A=\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}.$$

The standard ansatz $\mathbf{x}(t)=\mathbf{\mu} e^{r t}$ shows you that solutions to the homogeneous equation (without the $\mathbf{g}(t)$ part) willl have to satisfy the following two properties:

  1. $r$ needs to be an eigenvalue of $A$
  2. $\mathbf{\mu}=(\mu_x,\mu_y)^T$ has to be an eigenvector to A

Eigenvalues and eigenvectors are easily found, as you wrote above, you'll find $\begin{pmatrix}1 \\1 \end{pmatrix}$ for EV 3 and $\begin{pmatrix}-1 \\1 \end{pmatrix}$ for EV -1. Once you've got this, your ready to write down the general solution to the homogeneous system, which has the form

$$\mathbf{x}_c(t)=C_1 e^{3 t}\begin{pmatrix}1 \\1 \end{pmatrix} +C_2 e^{- t} \begin{pmatrix}-1 \\1 \end{pmatrix}.$$ In the same vein as you'd treat an nonhomogeneous ODE, the full solution will be the sum of the above plus a particular solution to the nonhomogeneous system, $\mathbf{x}(t)=\mathbf{x}_c(t)+\mathbf{x}_p(t)$. To that end, introduce the fundamental matrix $M$, whose rows are the two solutions to the above eigenvector problem:

$$M(t)=\begin{pmatrix} e^{3t} & -e^{-t} \\ e^{3t} & e^{-t} \end{pmatrix}.$$

For the particular solution, we make the ansatz $\mathbf{x}_p(t)=M(t)\mathbf{u}(t)$, with some yet-to-be-determined vector $\mathbf{u}(t)$. Inserting this into the very first equation yields $\dot{\mathbf{u}}(t) =M(t)^{-1} \mathbf{g}(t)$, solutions to which are obtained by integration.

I'll stop at this point to give you a chance to put the final pieces together yourself, but don't hesitate if you have additional questions!


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