My book gives the theorem:
Let $R$ be a rectangular region in the xy plane defined by $a \le x\le b$, $c \le y \le d$ that contains the point $(x_0, y_0)$ in its interior If $f(x, y)$ and $\dfrac {\delta f}{\delta y}$ are continuous on $R$, then there exists some interval $I_0 : (x_0-h, x_0+h), h>0$, contained in $[a, b]$, and a unique function $y(x)$, defined on $I_0$, that is a solution of the IVP.
I want to make sure i understood the gist of the theorem. What is a little bit confusing is that usually $f(x, y)$ denotes a function with 2 independent variables , but in this context I think they mean $\dfrac {dy}{dx}=f(x, y)$ so there is only one independent variable.
My understanding of how to apply the theorem goes like this:
Get the DE in a form of $\dfrac {dy}{dx}=f(x, y)$
Pretend $y$ is an independent variable and take the partial derivative with respect to $y$
The interval(s) where both $f(x, y)$ and $\dfrac {\delta f}{\delta y}$ are both continuous is the domain where an IVP will have a unique solution.
Is my understanding right? The reason I asked this question is because of the "pretend" part. If you think my understanding of partial derrivatives is flawed or incomplete please elaborate because I only know the bare basics of partial derivatives.
P.S. This is from my first differential equations class.
P.P.S. The only way it would make sense for me to "pretend" $y$ is an independent variable is if it was guaranteed that $y$ could take any value for at least one given value of $x$, which would imply that the DE must have an infinite number of solutions, unless the solution would be something like a vertical line.
