Question

If f'(x)=f(x) and f(3)=e^2, then f(4)=me^n for some integers m and n. What are m and n?

m = ______.
n = ______.

Answer 1

m=1 and n=3

Explanation:

One way of doing this is solving the differential equation of f(x).

We have that
`f'(x)=f(x)` so
`f'(x)/f(x)=1`, that is,
`(\ln(f(x))'=1`. Integrating in both sides respect to x,
`\ln(f(x))=x+C` for some constant C. Therefore
`f(x)=e^(x+C)=ke^x` for all x, and for some constant
`k=e^C`. In particular, taking x=3,
`f(3)=ke^3=e^2`. From this,
`k=e^(-1))`, then
`f(x)=e^(x-1)`. Choosing x=4, we have that
`f(4)=e^3` so the integers are m=1 and n=3.