Question

if $f'(x)=[f(x)]^2$, and $f(0)=1$, then $f(6)=\frac{1}{n}$ for some integer $n$. What is $n$?

Answer 1

The ODE is separable: We can write

$f'(x)=f(x)^2\implies\dfrac{\mathrm df}{\mathrm dx}=f^2\implies\dfrac{\mathrm df}{f^2}=\mathrm dx$

Integrating both sides gives

$-\dfrac1f=x+C$

so that

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

and so

$-\dfrac1f=x-1\implies f(x)=\dfrac1{1-x}$

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