Question

For what values of r does the function y= e^rx satisfy the differential equation 2y"+y'-y=0?

Answer 1

I am pretty sure that the soluton I share below is truly correct one : 
$y = e^(rx) y' = re^(rx) y'' = r^2e^(rx)$
Then go to :$2y'' + y' - y = 0 2r^2 e^(rx) + r e^(rx) - e^(rx) =0$
The next step you have to do is dividing all the items : 
$2r^2 + r - 1 = 0$
Andddd here you are! that's the answer - r = -1, r = 1/2. Hope it's clear :) Regards.