I am trying to find the roots of the differential equation $y''+7y'+6y=0$.
I assume the following:
$$ y=e^{rx} \\ y'=r*e^{rx} \\ y''=r^2*e^{rx}. $$ Then I substitute that into the dif. equation: $$ r^2*e^{rx}+7*(r*e^{rx})+2*e^{rx}=0 \\ e^{rx}*(r^2+7r+2)=0. $$ And I end up with the following quadratic equation which I need to find the roots of: $$ r^2+7r+2=0. $$
It's impossible to factor, so I tried to use the quadratic formula but end up with wrong results for r: $$ r=((-7+\sqrt{7^2-4*1*2})/(2*1))=-0.298 \\ r=((-7-\sqrt{7^2-4*1*2})/(2*1))=-6.702. $$ I used the same method for all other linear 2nd order homogeneous differential equations where I got correct results. But this time, I messed up.
What did I do wrong?!
Your original equation is: $$y''+7y'+6y=0$$
After you substitute $e^{rx}$ you write:
$$r^2e^{rx}+7re^{rx}+\color{red}{2}e^{rx}=0$$
The $\color{red}{2}$ should be a $\color{green}{6}.$