# Exact Equation - theory

An exact equation has the form

$$M(x,y) + N(x,y) \: y^{\prime} = 0$$

where

$$M(x,y) = \frac{\partial \psi}{\partial x} (x,y)$$

and

$$N(x,y) = \frac{\partial \psi}{\partial y} (x,y) \mbox{.}$$

If $$y=\phi (x)$$ and $$\psi (x,y) = c$$, then

$$M(x,y) + N(x,y) \: y^{\prime} = \frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{dy}{dx} = \frac{d}{dx} \psi \left[ x, \phi (x) \right] = 0 \mbox{.}$$

I can't understand this:

$$\frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{dy}{dx} = \frac{d}{dx} \psi \left[ x, \phi (x) \right]\mbox{.}$$

Any help is highly appreciated.

Related Introductory Physics Homework News on Phys.org

#### OlderDan

Homework Helper
thiago_j said:
I can't understand this:

$$\frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{dy}{dx} = \frac{d}{dx} \psi \left[ x, \phi (x) \right]\mbox{.}$$
From the definition of an "exact differential" or "total differential" for a function f(x,y) whose integral is path independent

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

http://mathworld.wolfram.com/ExactDifferential.html

If you divide by dx you have

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

and since you have y as a function of x, there is a derivative of y wrt x

#### James R

Homework Helper
Gold Member
$$\frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{dy}{dx} = \frac{d}{dx} \psi \left[ x, \phi (x) \right]\mbox{.}$$

Well, $\psi$ is a function of x and y, so

$$\frac{d\psi}{dx} = \frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y}\frac{\partial y}{\partial x}$$

Now put $y=\phi (x)$, and that's all there is to it.

Thank you, guys! Thus, we have

$$M(x,y) + N(x,y) \: y^{\prime} = 0$$

$$d\psi = M(x,y) \: dx + N(x,y) \: dy = 0$$

$$\frac{d\psi}{dx} = M(x,y) \: + N(x,y) \: \frac{dy}{dx} = 0$$

$$\frac{d\psi}{dx} = \frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{dy}{dx} = 0 \mbox{.}$$

M

#### MalleusScientiarum

I'm not exactly sure of how pragmatic it is to convert an ordinary differential equation into a partial differential equation. Maybe I miss something?

#### HallsofIvy

Homework Helper
No one changed an ordinary differential equation into a partial differential equation. They just used partial derivatives to show show that the sum of two differentials can, in some circumstances, be written as a single differential- which makes the problem very easy.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving