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For $y'=\frac{1}{x}\begin{pmatrix}0 &1 \\ 2 & -1\end{pmatrix}y$ the solutions are given by $y_1=\begin{pmatrix}x \\ x\end{pmatrix},y_2=\begin{pmatrix}x^{-2} \\ -2x^{-2}\end{pmatrix}$ with $x\in \Bbb R,y\in \Bbb R^2$

I would like to know how they got this solutions.

I tried the following:$$det\frac{1}{x}\begin{pmatrix}0-\lambda &1 \\ 2 & -(1+\lambda)\end{pmatrix}=\frac{1}{x}(\lambda^2+\lambda-2)$$ $\Rightarrow \lambda_{1,2}=-\frac{1}{2}+/-\frac{3}{2}\Rightarrow\lambda_1=-2 , \lambda_2=1$

Than $$ker\begin{pmatrix}-1&1 \\ 2 & -2\end{pmatrix}\Rightarrow v_1=\begin{pmatrix}1 \\ 1\end{pmatrix}\Rightarrow y_1=e^x\begin{pmatrix}1 \\ 1\end{pmatrix}$$

also $$ker\begin{pmatrix}2&1 \\ 2 & 1\end{pmatrix}\Rightarrow v_1=\begin{pmatrix}1 \\ -2\end{pmatrix}\Rightarrow y_2=e^{-2x}\begin{pmatrix}1 \\ -2\end{pmatrix}$$

I guess this way is wrong

Edit:

$y_1'=\frac{1}{x}y_2 $ and $y_2'=\frac{1}{x}(2y_1-y_2)$

  • 2
    Those eigenvalues and subsequent statements are valid for linear systems. Your equations are nonlinear – 2017-02-07
  • 0
    @Moo: Yes, at the beginning in the matrix yes, in the end I didn't used it – 2017-02-07
  • 2
    And you need to clarify the dimensions of $x$ and $y$ – 2017-02-07
  • 0
    @polfosol: yes, and for non-linear ? I see that $y_1'=\frac{y_2}{x}$ – 2017-02-07

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Let $x = e^u$. Then we have $$ \def\d#1#2{\frac{d#1}{d#2}} \d yu = \d yx \cdot \d xu = y' \cdot x $$ Hence, your equation reads $$ \d yu = \begin{pmatrix} 0 & 1 \\ 2 & -1 \end{pmatrix}y $$ Now solve this linear system as usual and resubstitue $e^u$ by $x$, giving the stated solutions.