monfatiellymeer monfatiellymeer

21-11-2016
Mathematics

Answered

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

...?

Verified Answer :

antonsandiego antonsandiego

22-11-2016

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.

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