$\displaystyle f'(x)=x(1-f(x))$
Evaluate $\displaystyle \int_0^{\infty} x(1-f(x))$
f(0)= 10
I actually see how you got that now
f'(x)=x(1-f(x))
dy/dx=x(1-y) where y=f(x)
$\displaystyle \int \frac{1}{1-y} dy = \int x dx$
$\displaystyle -ln|1-y| = \frac{1}{2} x^2 + C$
$\displaystyle -ln|1-f(x)| = \frac{1}{2} x^2 + C$
then using the initial condition f(0) = 10 I can find C
$\displaystyle -ln|1-f(0)| = \frac{1}{2} (0)^2 + C$
$\displaystyle -ln|9|= C$
thanks for the help i think i can finish the problem from here but if i get stuck i will post it.