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


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

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

  • 3
    $\begingroup$ Why'd you stop there? Plug them in! $\endgroup$
    – dfeuer
    Nov 13 '13 at 23:24
  • 2
    $\begingroup$ The function $x\mapsto e^{rx}$ is a solution to the ODE $y''+5y'-6y=\bf 0$, if, and only if, $r^2+5r-6=0$. $\endgroup$
    – Git Gud
    Nov 13 '13 at 23:25
  • 1
    $\begingroup$ and then you find $r_1=1, r_2=-6$ and $y(x)=Ae^{x}+Be^{-6x}$. $\endgroup$
    – alexjo
    Nov 13 '13 at 23:28
  • $\begingroup$ @GitGud when you plugged it in, why did the e^rx just dissapear? $\endgroup$
    – Jessica
    Nov 14 '13 at 0:10
  • $\begingroup$ @GitGud oh wait nvm it gets factored out and can never equal zero, makes sense thank you all! $\endgroup$
    – Jessica
    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 : $$y(x)=ae^t + be^{-6t}, (a,b) \in \Bbb R^2$$


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