Question

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

Answer 1

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

Answer 2

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?

Answer 3

Add the equations, lol!

Answer 4

or find the determinant first?

Answer 5

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

Answer 6

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

Answer 7

This is 2013 where cellphones exist rofl