Is a differential equation involving a multivariate function and exactly one of its partial derivatives an ODE?

I haven't taken any class in differential equations, so most of what I know about them is from small lectures in other classes. Please forgive my naivete.

According to someone in this this post, the answer is yes: a differential equation involving a multivariate function and exactly one of its partial derivatives is indeed an ODE. However, I can't understand why.

If we have $$f(x,y) = y * \frac{df(x,y)}{dx}$$, don't the various values we can plug into independent variable $$y$$ affect the output of both $$f(x,y)$$ and $$\frac{df(x,y)}{dx}$$? Therefore, since we have multiple independent variables to worry about, we should treat this equation more as a PDE than ODE.

Please feel free to correct my way of thinking as I am a greenhorn.

• This is an ODE with a parameter, not a PDE really. The method of solution will be to just treat each fixed $y$ separately and solve the equation in each case as an ODE.
– Ian
Jul 10, 2021 at 2:54

$$f(x,y)=y\frac{df}{dx}$$ If $$y$$ is not a function of $$x$$ this is an ODE in which $$y$$ is a parameter. The solution is : $$f=Ce^{x/y}$$ $$C$$ is an arbitrary constant wrt $$x$$. Nothing prevents $$C$$ to be function of $$y$$ or function of any other variable insofar those variables are not function of $$x$$. Thus in order to obtain all solutions : $$\boxed{f=C(y)e^{x/y}}$$ where $$C(y)$$ is an arbitrary function.
Now consider the original equation as a PDE : $$f(x,y)=y\frac{\partial f}{\partial x}$$ $$y\frac{\partial f}{\partial x}+0\frac{\partial f}{\partial y}=f$$ The Charpit-Lagrange characteristic ODEs are : $$\frac{dx}{y}=\frac{dy}{0}=\frac{df}{f}$$ This implies the first characteristic equation : $$y=c_1$$ A second characteristic equation comes from solving $$\frac{dx}{c_1}=\frac{df}{f}$$ $$e^{-x/c_1}f=c_2$$ The general solution of the PDE on implicit form $$c_2=F(c_1)$$ is : $$e^{-x/c_1}f=F(c_1)=e^{-x/y}f=F(y)$$ with arbitrary function $$F$$. $$\boxed{f(x,y)=F(y)e^{x/y}}$$ The result is consistant with the above result since $$C(y)$$ and $$F(y)$$ are both arbitrary functions.
That is a interesting question that may diverge according to the author. Based on Erwin Kreyszig (vide Sec. 12.1), he states that a equation such as $$u_{xx}(x,y) - u(x,y) = 0$$ is a PDE that can be solves as an ODE since there is no $$y$$-derivative involved. Therefore, we can be solve it as $$u^{''} - u = 0$$. The solution is $$u(x,y) = A(y) e^{x} + B(y) e^{-x},$$ which is the very same solution that we have for an ODE, except by that $$A$$ and $$B$$ may depend on $$y$$ (in a ODE, it would be a constant). Regardless, you can treat it as an ODE to solve it.