Given a curve $C=\{(x,f(x)\in \mathbb{R}\times\mathbb{R}\mid x\in(r_1,r_2)\}$ with has the following property.(f(x) is $C^3$-function)
At any point $(a,f(a))\in C$ if we change coordinate system by defining tangent line at this point to be X-axis and normal line at this point to be Y-axis ,then this curve can be represented in new coordinate system $C=\{(X,Y)\mid Y=F(X)\}$ so that $F'''(0)=0$
Show that $(1+(f'(x))^2)f'''(x)=3f'(x)f''(x)^2$ and solve this differential equation.
I can show that any point $(x,f(x))$ in normal XY-plane will be represented by $((x-a)cos\theta+(f(x)-f(a))sin\theta,-(x-a)sin\theta+(f(x)-f(a))cos\theta)$ when $tan\theta=f'(a)$in this new coordinate system but still cannot find $F'''(X)$ in terms of $x,f'(x),...$
for the second part let g(x)=f'(x) so this ODE turn into second order DE but still do not found any techniques to deal with this.
thanks for your help
Firstly, we consider the behavior of the function on the new coordinate. It is obvious that $F(0)=0$, also $F'(0)=0$ since the $x$ coordinate is just the tangent line. So to compute $F'''(x)$, we need $\lim_{x\to 0}3\cdot\frac{F(x)-F(-x)}{x^3}$, this can be verified via Taylor expansion on new coordinate.$\newcommand{\e}{\epsilon}$
Secondly, we move the point $(a,f(a))$ to $(0,0)$, the point $(a+\e,f(a+\e))$ is translated to $(\e,f(a+\e)-f(a))$. Consider the distance with the line $y=f'(a)x$ in the translated coordinate, this distance is assigned a sign corresponding to the $y$ axis in the new coordinate system, see the picture
$$y_1=\frac{f(a+\e)-f(a)-f'(a)\e}{\sqrt{1+f'(a)^2}},\\x_1=\frac{f'(a)(f(a+\e)-f(a))+\e}{\sqrt{1+f'(a)^2}}=\sqrt{1+(f'(a))^2}\e+\frac{f''(a)f'(a)}{2\sqrt{1+(f'(a))^2}}\e^2+o(\e^2)$$
Thus when $x_1\to -x_1$, the $\e\to \e_1=-\e-\frac{f''(a)f'(a)}{1+f'(a)^2}\e^2+o(\e^2)$
$$\lim_{x_1\to 0}\frac{y_1(\e)-y_1(-\e-\frac{f''(a)f'(a)}{1+f'(a)^2}\e^2)}{x_1^3}=0\Rightarrow \frac{f''(a)\e^2/2!+f'''(a)\e^3/3!-f''(a)(\e_1)^2/2!-f'''(a)(\e_1)^3/3!}{\left(\sqrt{1+(f'(a))^2}\e\right)^3}=0$$
Expand and remove the higher order term, we get
$$\frac{1}{3}f'''(a)-\frac{2}{2}\left(\frac{f''(a)^2f'(a)}{1+f'(a)^2}\right)=0.$$
Which is exactly $$(1+(f'(x))^2)f'''(x)=3f'(x)f''(x)^2$$
Note that this implies $$\left(\frac{1+(f'(x))^2}{f''(x)}\right)'=-f'(x)\Rightarrow \frac{1+(f'(x))^2}{f''(x)}= -f(x)+C\Rightarrow (f(x)f'(x))'=C f''(x)-1.$$
Hence $f(x)f'(x)=C_1 f'(x)-x+C_2$ and $((f(x)-C_1)^2)'=-2x+2C_2$. The final answer is
$$(f(x)-C_1)^2=-x^2+2C_2 x+C_3$$