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.


1 Answer 1


The equation can be written as $y'(x)=f(x,y(x))$. I think this makes things a little bit more clear.

Your confussion comes from the fact that we use the same letter $y$ to denote the unknown function $y(x)$ and the second variable of the function $f$.

  • $\begingroup$ Hmm yes but $y$ is not really an independent variable so how can we take the partial derivative with respect to $y$? Or does it take on the properties of an independent variable because there are an infinite number of solutions? $\endgroup$
    – Ovi
    Oct 7, 2014 at 16:33
  • $\begingroup$ $f$ is a function of two independent variables: the first, usually called $x$, and the second, usually called $y$. You can restate the theorem in the following terms: ... $f$ has continuous partial derivative with respect to the second variable... $\endgroup$ Oct 7, 2014 at 19:32
  • $\begingroup$ but $y$ represents a solution, $y(x)$ so how can it be independent when it depends on $x$? $\endgroup$
    – Ovi
    Oct 7, 2014 at 20:18
  • $\begingroup$ $f$ is a continuous function of two variables: $f(v_1,v_2)$. $\partial f/\partial v_2$ is continuous. The solution $y(x)$ verifies the equality $y'(x)=f(x,y(x))$. $\endgroup$ Oct 7, 2014 at 20:53

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