In the question Find a first integral of an ODE system, there is following system of equations:
$$ \left\{ \begin{array}{c} \dot x=2xy \\ \dot y=x+y^2 \\ \end{array} \right. $$
Second answer solved for $y$. I have question about this solution.
As far as I know $x$ and $y$ are some functions of let's say $t$. So I can rewrite system:
$$ \left\{ \begin{array}{c} x'(t)=2x(t)y(t) \\ y'(t)=x(t)+(y(t))^2 \\ \end{array} \right. $$
I will do the same as in the answer however I will keep track of variable $t$
$$\frac{y'(t)}{x'(t)}=\frac{x(t)+(y(t))^2}{2x(t)y(t)}$$
$$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{x(t)+(y(t))^2}{2x(t)y(t)}$$
$$\frac{dy}{dx}=\frac{x(t)+(y(t))^2}{2x(t)y(t)}$$
$$y'(x(t))=\frac{x(t)+(y(t))^2}{2x(t)y(t)}$$
$$2y(t)y'(x(t))=1+\frac{(y(t))^2}{x(t)}$$
Substitute $Y(t)=(y(t))^2$
$$\frac{dY}{dx}=\frac{\frac{dY}{dt}}{\frac{dx}{dt}}=2y(t)\frac{y'(t)}{x'(t)}=2y(t)\frac{dy}{dx}=2y(t)y'(x(t))$$
So:
$$\frac{dY}{dx}=1+\frac{Y(t)}{x(t)}$$
$$Y'(x(t))=1+\frac{Y(t)}{x(t)}$$
The author of solution claims that this is first order linear ODE, but on left side there is $Y'(x(t))$ and on right $Y(t)$ not $Y(x(t))$ or $Y'(t)$ on left. So I am not sure about this.
Author says that:
$$Y=x\ln|x|+cx$$
What i $Y$? Is it $Y(t)$ or $Y(x(t))$ or something else? I think it should be $Y(t)$ but then I think on the right side there should be $1+\frac{Y(x(t))}{x(t)}$ Did I make mistake in my notation?
The map $x\to Y(x)$ is such that $Y(x(t))=y^2(t)$. Then, after taking the derivative with respect to $t$ of both sides we find $$Y'(x(t))x'(t)=2y(t)y'(t)$$ and it follows that $$\frac{2y(t)y'(t)}{x'(t)}=1+\frac{(y(t))^2}{x(t)} \implies Y'(x(t))=1+\frac{Y(x(t))}{x(t)}.$$ By solving the linear ODE $Y'(x)=1+\frac{Y(x)}{x}$, we get $Y(x)=x\ln|x|+cx$ and therefore $$y^2(t)=x(t)\ln|x(t)|+cx(t)$$ or, after removing the parameter $t$, $$y^2=x\ln|x|+cx.$$
P.S. See also this Desmos page: the red curve (depending on the constant $c$) is the solution in the (normalized) vector field.