Question

For what values of r does the function y = e^{rx} satisfy the differential equation 7y'' + 48y' − 7y = 0?

Answer 1

\[y = e^{rx}, y' = re^{rx}, y'' = r^2e^{rx}\]\[7(r^2e^{rx})+48(re^{rx})-7(e^{rx})=0\]\[7e^{rx}(r^2+7r-1) = 0\]Am I doing it correctly so far?

Answer 2

Is the middle coefficient 48 or 49 ?

Answer 3

Other than that you seem to be on right track.

Notice that e^x is never 0.
So the other factor must be 0 for the overall product to be 0.

Answer 4

opps, the middle coefficient is 48, but I got the problem now. :)