Consider the linear system of differential equations: \begin{align}\frac{dx}{dt}= \ \ \ \ \ x \ + \ y \\ \frac{dy}{dt}=-2x+4y \end{align} (a) Solve the system and find the general solution. (b) Sketch the phase portrait.

To solve the system I have found the eigen values of the asociated matrix which are $ \lambda=2,3$.

Edit: I now need only part (b) because I already solved part (a). The general solution is given by $A e^{2t} \begin{pmatrix}1 \\ 1 \end{pmatrix}+B e^{3t} \begin{pmatrix} 1 \\ 2 \end{pmatrix}$

Is there any help for me? Thank you.

  • 1
    $\begingroup$ The zero vector is always an eigenvector. But in order to calculate an eigenvector you have to solve $A\begin{bmatrix}x\\y\end{bmatrix} = \lambda \begin{bmatrix}x\\y\end{bmatrix}$ instead of $A\begin{bmatrix}x\\y\end{bmatrix}=0$. $\endgroup$ – Wauzl May 30 '17 at 8:51
  • $\begingroup$ Also your matrix should be $\begin{bmatrix}1&1\\-2&4\end{bmatrix}$ $\endgroup$ – Wauzl May 30 '17 at 8:53
  • $\begingroup$ I get eigenvalues of 2 and 3. An eigenvalue of -1 doesn't make the determinant of A-lambda*I equal zero! $\endgroup$ – Andrew Gray May 30 '17 at 8:54
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    $\begingroup$ The phase portrait simply describes the regions of the $(x,y)$-plane where $x'=0$, where $y'=0$, where $x'>0$ and $y'>0$, where $x'>0$ and $y'<0$, where $x'<0$ and $y'>0$, and where $x'<0$ and $y'<0$. Why is it difficult to determine these regions and to draw them all on a single figure? $\endgroup$ – Did May 30 '17 at 9:30
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    $\begingroup$ @Wauzl, The zero vector $\vec0$ always satisfies $A\vec0 = \lambda\vec0$ but some texts (like the one I taught from last semester) require eigenvectors to be not the zero vector, so I'd be careful with saying "the zero vector is always an eigenvector." AFAIK it's actually standard to require an eigenvector to be not $\vec0$. $\endgroup$ – tilper May 30 '17 at 13:48

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