Consider the family of functions C1e^x + C2e^-x. FOr what values of x is there a function y in this family such that y(x)=1 y'(x)= 2

Question

usukidoll · Answer

This is what I got so far...
c1e^x+c2e^-x = y(x)
c1e^x-c2e^-x = y'(x)

usukidoll · Answer

c1e^x+c2e^-x = 1
c1e^x-c2e^-x = 2

do I put this in augmented matrix form and then find the values from there?

anonymous · Answer

Add the equations, lol!

usukidoll · Answer

or find the determinant first?

chaise · Answer

Sub in y(x)=1 and y'(x)=2 and solve simultaneously for x. c1 and c2 are homogenous I think

usukidoll · Answer

alright...ohh sorry man I'm using a book from 1962 and they use weird words x.X

usukidoll · Answer

This is 2013 where cellphones exist rofl

usukidoll · Answer

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