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

[Dr Game]

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?

 

[cristo]

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.

 

[Dr Game]

alright thought so, so I should get the same answer if C wasn't there, which is 0

$$25Ce^-^5^x+4(-5Ce^-^5^x)-5Ce^-^5^x = 0$$

 

[cristo]

Yes, that's correct