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I'm trying to understand exact differential equations by following this material. My question concerns the equation denoted by (1) and the step that follows. This post should have all relevant information, but check the link if anything is ambiguous.

Suppose we have some function Ψ(x,y).

Suppose we know that (∂Ψ)/(∂x) + (∂Ψ)/(∂y) * (dy)/(dx) = 0

Somehow, we step from the equation above to this equation here: d/(dx) * Ψ(x,y(x)) = 0

I do not see how the multi variate chain rule can be applied to the first equation, because it is not in the expected form. So how is that step made? Can you break it down into tiny steps? In particular, where does (dy)/(dx) from the first equation disappear? Also, why is y suddenly taking x as its parameter? Did something change during the step or was it presumed the entire time that y is a function of x?

2 Answers 2

1

What's being used is the concept of a total differential for a function $f$ of two variables $x$ and $y$:

$$df = \dfrac{\partial f}{\partial x} dx + \dfrac{\partial f}{\partial y} dy$$

If you divide by $dx$ you get

$$\dfrac{df}{dx} = \dfrac{\partial f}{\partial x} \dfrac{dx}{dx} + \dfrac{\partial f}{\partial y} \dfrac{dy}{dx}$$

which simplifies to

$$\dfrac{df}{dx} = \dfrac{\partial f}{\partial x} \cdot1 + \dfrac{\partial f}{\partial y} \dfrac{dy}{dx}$$

1

$y$ is a function of $x$. The notes are related to solving a differential equation of the form $f(x,y) + g(x,y)y'(x)=0.$ Note the $y'(x).$ $y$ is the function and $x$ is the independent variable.

So the chain rule gives $$\frac{d}{dx}\Phi(x,y(x)) = \partial_x\Phi + \partial_y\Phi \frac{dy}{dx}.$$

If you'd like to, you could write $\frac{dx}{dx}$ next to the first term.