# Eigenvalues of $\ddot {y} +\dot{y}-6y = 0$

I want to write $\ddot {y} +\dot{y}-6y = 0$ in the form

$$\dot{x} = \cdots$$

$$\dot{y} = \cdots$$

then create a matrix from which I can calculate $\lambda_1,~\lambda_2$.

But I keep getting the wrong eigenvalues if I let $x = \dot{y}$.

• Just solve the characteristic equation. – Rodrigo de Azevedo Aug 7 '17 at 23:11
• "But I keep getting the wrong eigenvalues if I let x=y˙" This is the part we ought to see if this was to become an interesting question. – Did Oct 14 '17 at 13:48
• @Did How did you even find this? I posted this about ten weeks ago! – Alt-Rock Oct 14 '17 at 13:57

We let:

$$x_1 = y \\ x_2 = y'$$

Next

$$x_1' = y' = x_2 \\ x_2' = y'' = -y' + 6 y = - x_2 + 6 x_1$$

We can now write

$$X' = \begin{bmatrix} 0 & 1 \\ 6 & -1 \end{bmatrix}x$$

Can you take it from here?

Note: as an alternate approach, we can find the roots of the characteristic equation $\lambda^2 + \lambda - 6 = 0 \implies \lambda_1 = -3, \lambda_2 = 2.$