This is not a duplicate, don't flag me please
I thought about a smooth function $f$ which has this propety: in every point the slope of the function is equal to the distance from zero or mathematicly: $f'(x)=\sqrt{f^2(x)+x^2}$
I tried to find it so I defined new function $g$ like that:
$ g(0,y):=f(0)$, and in every other point the slope is $\lfloor f(x) \rfloor$
It easy to see that $g$ is continuous and $g(y\lfloor \frac xy \rfloor,y)=f(0)(y+1)^{\lfloor \frac xy \rfloor}$
and $g'(x)=g(y\lfloor \frac xy \rfloor)\le g(x)$
and $g(0)=f(0)$
so:
$\forall_{y\in\mathbb{R}}f(x)\ge g(x,y)\ge (1+y)^{\lfloor \frac xy \rfloor}\Rightarrow f(x)\ge g(x,-1)\ge \lim_{y\rightarrow -1}(1+y)^{\lfloor -x\rfloor}f(0)=\lim_{y\rightarrow 0}\frac{f(0)}{y^{\lfloor x\rfloor}}=\infty$
So if $f(0)\neq 0$ the function is not continuous so $f(0)=0$
Then with induction we see $\frac n4 \notin \mathbb{N} \Rightarrow f^{(n)}(0)=0$ and $b_n=\sum_{k=1}^{n-1}\frac {(n-2)!}{(k-1)!(n-k-1)!} b_kb_{n-k}$
($b_n:=\frac {f^{(n)}(0)}{n!}$)
So because of the propeties of maclaurin series $f(x)=\sum_{n=1}^\infty \frac {b_{4n}x^{4n}}{(4n)!}$
We now define $a_n:=b_{4n}$
$f^{(4)}(x)=2\Rightarrow a_1=2$
We can also define $y:=\sum_{n=1}^\infty \frac {a_nx^n}{(4n-1)!}$
Then we apply sum from 2: $\sum_{n=2}^\infty \frac{a_nx^x}{(4n-1)!}(4n-1)=\sum_{n=2}^\infty \sum_{k=1}^{n-1} \frac {a_ka_{n-k}x^kx^{n-k}}{(4k-1)!(4(n-k)-1)!}$
This is actualy $4xy'-y-\frac{a_1x}{2}=y^2$ so because $a_1=2$ we know: $4xy'=x+y+y^2$this is Riccati equation so: $\frac y{4x}=-\frac{u'}u\Rightarrow u''+\frac 3{4x} u'+\frac 1{16x} u=0\Rightarrow u=C_1x^{\frac 18}J_{-\frac 14}(\frac {\sqrt x}2) +C_2x^{\frac 18}J_{\frac 14}(\frac {\sqrt x}2)$
We know that $\frac{y}{4x}$ defined at zero so:
$u=C_1x^{\frac 18}J_{-\frac 14}(\frac {\sqrt x}2)\Rightarrow y=4x\left( -\frac {x^{\frac 18}J_{-\frac 14}(\frac {\sqrt x}2)'}{x^{\frac 18}J_{-\frac 14}(\frac {\sqrt x}2)}\right)$
Since $J_v'(x)=\frac {J_{v-1}(x)-J_{v+1}(x)}2$:
$y(x)=\frac{-\frac 12 x^{\frac 18}J_{-\frac14}(\frac {\sqrt x}2)-\frac 12 x^{\frac 58}J_{-\frac 54}(\frac {\sqrt x}2)+\frac 12 x^{\frac 58}J_{\frac 34}(\frac {\sqrt x}2)}{x^{\frac 18}J_{-\frac 14}(\frac {\sqrt x}2)}\Rightarrow y(x^4)=\frac{-\frac 12 x^{\frac 12}J_{-\frac14}(\frac {x^2}2)-\frac 12 x^{\frac 52}J_{-\frac 54}(\frac {x^2}2)+\frac 12 x^{\frac 52}J_{\frac 34}(\frac {x^2}2)}{x^{\frac 12}J_{-\frac 14}(\frac {x^2}2)}$
We know that $f(x)=\int \frac {y(x^4)}x dx=$ because each term in the meclaurin series of $y(x^4)$ is bigger them $f$'s like the index of that term and to multiply each term we devide by $x$ and integrate. so:
$f(x)=\int \frac{-\frac 12 x^{\frac 12}J_{-\frac14}(\frac {x^2}2)-\frac 12 x^{\frac 52}J_{-\frac 54}(\frac {x^2}2)+\frac 12 x^{\frac 52}J_{\frac 34}(\frac {x^2}2)}{x^{\frac 32}J_{-\frac 14}(\frac {x^2}2)}dx=\, \int \frac {x(J_{\frac 34}(\frac {x^2}2)-J_{-\frac 54}(\frac {x^2}2))}{2J_{-\frac 14}(\frac {x^2}2)}-\frac 1{2x}dx=\\\int \frac {x(J_{\frac 34}(\frac {x^2}2)-J_{-\frac 54}(\frac {x^2}2))}{2J_{-\frac 14}(\frac {x^2}2)}dx-\frac{\ln (x)}2$
Since $J_v'(x)=\frac {J_{v-1}(x)-J_{v+1}(x)}2$:
$f(x)+\ln{(\sqrt x)}=\frac {\ln (x)}2 +f(x)=-\int \frac x{J_{-\frac 14}(\frac {x^2}2)} \frac {dJ_{-\frac 14}(\frac {x^2}2)}{d\frac {x^2}2}dx=-\int \frac{dJ_{-\frac 14}(\frac {x^2}2)}{J_{-\frac 14}(\frac {x^2}2)}=-\ln \left( J_{-\frac 14}(\frac {x^2}2)\right) +c$
So:
$f(x)=-\ln \left( J_{-\frac 14}(\frac {x^2}2)\right)-\ln (\sqrt x)+c=\, -\ln \left( \sqrt x J_{-\frac 14}(\frac {x^2}2)\right)+c$
But that function is undefined at zero so there is no such function which is smooth
Where is my mistake??
Is that true??