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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.

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  • $\begingroup$ 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. $\endgroup$
    – Ian
    Jul 10, 2021 at 2:54

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$$f(x,y)=y\frac{df}{dx}$$ 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.

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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.

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