If f'(x)=[f(x)]^2, and f(0)=1, then f(6)=(1)/(n) for some integer n. What is n?

Question

if
`f'(x)=[f(x)]^2`, and
`f(0)=1`, then
`f(6)=(1)/(n)` for some integer
`n`. What is
`n`?

Answer 1

The ODE is separable: We can write

`f'(x)=f(x)^2\implies(\mathrm df)/(\mathrm dx)=f^2\implies(\mathrm df)/(f^2)=\mathrm dx`

Integrating both sides gives

`-\frac1f=x+C`

so that

`f(0)=1\implies-1=C`

and so

`-\frac1f=x-1\implies f(x)=\frac1{1-x}`

Then
`f(6)=-\frac15`, making
`n=-5`.