monfatiellymeer monfatiellymeer
  • 21-11-2016
  • Mathematics
contestada

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

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antonsandiego
antonsandiego antonsandiego
  • 22-11-2016
I am pretty sure that the soluton I share below is truly correct one : 
[tex]y = e^(rx) y' = re^(rx) y'' = r^2e^(rx)[/tex]
Then go to :[tex]2y'' + y' - y = 0 2r^2 e^(rx) + r e^(rx) - e^(rx) =0[/tex]
The next step you have to do is dividing all the items : 
[tex]2r^2 + r - 1 = 0[/tex]
Andddd here you are! that's the answer - r = -1, r = 1/2. Hope it's clear :) Regards.
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