So I was scrolling through Youtube when I saw this video by Maths505 where he solves the equation$$y'(x)=y^{-1}(x)$$which I thought that I might be able to solve. Here is my attempt:
I first tried to think of some trivial solutions to this problem. I did this by taking the $y(x)$ of both sides to get that$$y(y'(x))=x$$although this didn't really help.
I then remembered this Wikipedia article that stated
Laisant proved that if $F$ is an antiderivative of $f$, then the antiderivatives of $f^{-1}$ are:$$\int f^{-1}(y)dy=yf^{-1}(y)-F\circ f^{-1}(y)+c$$
This means that we can take the antiderivative of our original equation with respect to $x$ to get$$y(x)+c_0=xy^{-1}(x)-Y(y^{-1}(x))+c_1$$where$$Y(x)=\int y(x)dx$$although I then remembered that since we determined that $y(y'(x))=x$, then $y'(x)$ must be the inverse function of $y(x)$, meaning that we could rewrite this as$$y=xy'-y+b,\quad b=c_1-c_0$$which then we can do this:$$y=x\dfrac{dy}{dx}-y+b\\\implies\dfrac{2y-b}x=\dfrac{dy}{dx}\\\implies\dfrac{dx}x=\dfrac{dy}{2y-b}\\\implies\ln|x|+c_2=\dfrac12\ln|2y-b|+c_3\\\implies2\ln|x|+a=\ln|2y+b|,\quad 2(c_3-c_2)=a\\\implies gx^2=2y+b,\quad e^a=g\\\implies y(x)=\dfrac{gx^2-b}2$$and taking $g=1,b=0$ gets us the function$$y(x)=\dfrac{x^2}2$$although it seems I did something wrong, because $y(y'(x))$ gets $\dfrac{x^2}2$ instead of $x$, which is what I want.
So my question is: What did I do wrong? I'm not sure how I got the incorrect solution.
