If I'm given a system of equation of the form $$\begin{cases} \frac{dx}{dt}= ax+by \\ \frac{dx}{dt}= cx+ey\end{cases}$$

I get the general solution finding the eigenvalues and eigenvectors of the matrix $\left( \begin{array}{ccc} a & b\\ c & e \\ \end{array} \right)$, let's say that I found the eigenvalues $\lambda_1,\lambda_2$ and eigenvectors $v_1,v_2$. Then the general solution have the form $(x,y)=c_1e^{t\lambda_1}+c_2e^{t\lambda_2}$.

What happens if I don't have "enough" eigenvectors to write a solution like the former?

This is what happened when I tried to find the solution for the system

$$\begin{cases} \frac{dx}{dt}= 2x-y \\ \frac{dx}{dt}= x+4y\end{cases}$$

Here the eigenvalues will be given by the roots of $p(\lambda)=(2-\lambda)(4-\lambda)+1=(\lambda-3)^2$. Now if I can only find one eigen vector associated to the double eigenvalue: $$\left( \begin{array}{ccc} 2-3 & -1\\ 1 & 4-1 \\ \end{array} \right)\to\left( \begin{array}{ccc} 1 & 1\\ 0 & 0 \\ \end{array} \right)\implies \left( \begin{array}{ccc} x \\ y \\ \end{array} \right)=\operatorname{gen}\Bigg\{\left( \begin{array}{ccc} 1\\ -1 \end{array} \right) \Bigg\}$$

Then the solution will be simply $\left( \begin{array}{ccc} x \\ y \\ \end{array} \right)= c_1e^{3t}\left( \begin{array}{ccc} 1 \\ -1 \\ \end{array} \right)$ ? or do I need another eigenvector to write the general solution?

  • 1
    $\begingroup$ shouldn't be there a $\frac{dy}{dt}$ somewhere? $\endgroup$ – tired Nov 20 '14 at 15:21
  • $\begingroup$ When you don't have enough eigenvectors, you take generalized eigenvectors. See this answer. Edit: The question isn't a duplicate, but my answer answers your question. $\endgroup$ – Git Gud Nov 20 '14 at 15:30

You need to find a vector $\vec{\rho}$ such that $(A-\lambda I)\vec{\rho} = \vec{\eta}.$
Then, the general solution will be $$c_1e^{\lambda t}\vec{\eta} + c_2(te^{\lambda t}\vec{\eta} + e^{\lambda t}\vec{\rho}).$$ DERIVATION:
Suppose the a solution was of the form $te^{\lambda t}\vec{\eta}$, since that's what you do with second order non-homogeoneous equations. Then, $$\vec{\eta}e^{\lambda t}+\lambda te^{\lambda t}\vec{\eta} = Ate^{\lambda t}\vec{\eta}.$$However, this implies that $$\vec{\eta} = 0.$$But eigenvectors cannot be zero, so this is wrong. Now the problem was that there was a lone term with an exponential in it so let's see if we can correct that. Let's try the guess $$te^{\lambda t}\vec{\eta} + e^{\lambda t}\vec{\rho}.$$Then, $$e^{\lambda t}\vec{\eta}+\lambda te^{\lambda t}\vec{\eta}+\lambda e^{\lambda t}\vec{\rho} = Ate^{\lambda t}\vec{\eta} + Ae^{\lambda t}\vec{\rho}$$ $$\vec{\eta}+\lambda t\vec{\eta}+\lambda\vec{\rho}=At\vec{\eta}+A\vec{\rho}$$ $$\vec{\eta}+\lambda\vec{\rho} = A\vec{\rho}$$ $$\vec{\eta} = A\vec{\rho}-\lambda I\vec{\rho}$$ $$(A-\lambda I)\vec{\rho} = \vec{\eta}.$$

  • $\begingroup$ What if no such vector $\rho$ exists? $\endgroup$ – Pancake_Senpai Jan 8 '19 at 14:47

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