# For what values of $r$ does $y=e^{rx}$ satisfy $y'' + 5y' - 6y = 0$?

For what values of $$r$$ does $$y=e^{rx}$$ satisfy $$y'' + 5y' - 6y = 0$$?

### Attempt:

$$y' = [e^{rx}] (r)$$

$$y''= r^2e^{rx}$$

• Why'd you stop there? Plug them in! Nov 13 '13 at 23:24
• The function $x\mapsto e^{rx}$ is a solution to the ODE $y''+5y'-6y=\bf 0$, if, and only if, $r^2+5r-6=0$. Nov 13 '13 at 23:25
• and then you find $r_1=1, r_2=-6$ and $y(x)=Ae^{x}+Be^{-6x}$. Nov 13 '13 at 23:28
• @GitGud when you plugged it in, why did the e^rx just dissapear? Nov 14 '13 at 0:10
• @GitGud oh wait nvm it gets factored out and can never equal zero, makes sense thank you all! Nov 14 '13 at 0:11

If you plug them in, you obtain : $$r^2+5r-6=0$$ Solving this equation you get $r=1$ or $r=-6$.
That means that the general solution of the suggested ODE is : $$y(x)=ae^t + be^{-6t}, (a,b) \in \Bbb R^2$$