Deriving y & Solving y'' + 4y' - 5y = 0 with C

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It's been a while, but how would you go about deriving

y = C e^-^5^x

and plug it into this equation

y'' + 4y' - 5y = 0

i can do this without C the answer would be 0, but what do I do with C there?
 
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You just differentiate it as usual, leaving the C in front of the terms, and then plug the first and second derivatives into the equation.
 
alright thought so, so I should get the same answer if C wasn't there, which is 0

25Ce^-^5^x+4(-5Ce^-^5^x)-5Ce^-^5^x = 0
 
Yes, that's correct
 
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