- #1

jostpuur

- 2,112

- 18

[tex]

f'(x) = -\frac{xf(x)}{Af(x)+ Bx^2}

[/tex]

with some positive initial value [itex]f(x)>0[/itex], and with positive constants [itex]A,B>0[/itex].

First question: Does an explicit formula exist? I couldn't figure it out.

Second question:

I see that [itex]f(x)>0\implies f'(x)<0[/itex], and on the other hand a constant [itex]f=0[/itex] is a solution for all [itex]x>0[/itex]. So clearly [itex]0< f(x)<f(0)[/itex] will hold for all [itex]0<x<\infty[/itex]. Therefore with large [itex]x[/itex] we have [itex]Af(x)+Bx^2\approx Bx^2[/itex] and

[tex]

f'(x)\approx -\frac{f(x)}{Bx}

[/tex]

which implies that in some sense

[tex]

f(x) \sim x^{-\frac{1}{B}}

[/tex]

will probably hold. The second question is that how do you prove something rigor with this approximation. The approximations

[tex]

f(x) = O\big(x^{-\frac{1}{B}}\big)

[/tex]

and

[tex]

f(x)= Cx^{-\frac{1}{B}}+ O\big(x^{-\frac{1}{B}-1}\big)

[/tex]

probably hold, but how do you prove them?