# Solution of $(x^2+f^2(x))f'(x)=1,\forall x\in [1,+\infty)$

A question of interest that arose during my doing a seemingly easy exercise is: The exercise considered a function $f:[1,+\infty)\to\mathbb{R}$ differentiable with $f(1)=1$ and $f'(x)\left( x^2+f^2(x)\right) =1,\forall x\in [1,+\infty)$.

Anyway, the exercise's questions were easy for me but the last one's conclusion is interesting: $1<f(x)<2,\forall x\in (1,+\infty)$.

So, as a general study of the function I have shown that f is monotonically increasing, concave downwards and $\lim_\limits{x\to +\infty}{f'(x)}=0$. My question here is: Can we find a closed type for f? If yes, what is it?

• The equation you have can be converted to Bernoulli's equation by division by $y^2$, where y is f(x). May 2, 2016 at 16:49
• I see no solution of that differential equation here...
– user171110
May 13, 2016 at 9:52
• tutorial.math.lamar.edu/Classes/DE/Bernoulli.aspx this shows how to solve theses types of equations. May 14, 2016 at 11:06

I can't find the exact solution but I will show the inequality:
$1<f(x)<2,\forall x\in (1,+\infty).$

Since $f$ is strictly increasing in the domain of interest, we have $f(x) > 1,\; \;\forall x > 1,$ which justifies the lower bound of $f$. Consequently,
$f'(x) = 1/(x^2+f^2(x)) < 1/(x^2+1),\quad\forall x > 1.$

Now we set $g(x) = \arctan x +1-\pi/4,\; w(x)=f(x)-g(x).$ It is easy to see that:
$w(1) = 0,\quad w'(x) < 0,\quad \forall x > 1.$

Hence $w(x) < 0 ,\quad \forall x> 1.$ Namely,
$f(x) < g(x)=\arctan x +1-\pi/4<\pi/2+1-\pi/4=1+\pi/4<2,\quad\forall x\in (1,+,\infty).$

Remark. Though $g(x)$ does not solve the euqaiont, if we plug $g(x)$ into the left hand side of the differential equation, it does not differ too much from the right hand side $1$ I think.