What is the relationship between exact equations and partial derivatives?

  • Thread starter DivGradCurl
  • Start date
  • Tags
    Theory
In summary, an exact equation has the form M(x,y) + N(x,y) \: y^{\prime} = 0, where M(x,y) and N(x,y) are partial derivatives of a function \psi with respect to x and y, respectively. If y=\phi (x) and \psi (x,y) = c, then the equation can be simplified to \frac{d\psi}{dx} = 0. This is achieved by dividing both sides by dx and using the fact that \psi is a function of x and y, so its total derivative is equal to its partial derivative with respect to x plus the partial derivative with respect to y multiplied by the derivative of y with respect to x
  • #1
DivGradCurl
372
0
An exact equation has the form

[tex]M(x,y) + N(x,y) \: y^{\prime} = 0[/tex]

where

[tex]M(x,y) = \frac{\partial \psi}{\partial x} (x,y)[/tex]

and

[tex]N(x,y) = \frac{\partial \psi}{\partial y} (x,y) \mbox{.}[/tex]

If [tex]y=\phi (x)[/tex] and [tex]\psi (x,y) = c[/tex], then

[tex]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{.}[/tex]

I can't understand this:

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

Any help is highly appreciated.
 
Physics news on Phys.org
  • #2
thiago_j said:
I can't understand this:

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

[tex] df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy [/tex]

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

If you divide by dx you have

[tex] \frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx} [/tex]

and since you have y as a function of x, there is a derivative of y wrt x
 
  • #3
[tex]\frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{dy}{dx} = \frac{d}{dx} \psi \left[ x, \phi (x) \right]\mbox{.}[/tex]

Well, [itex]\psi[/itex] is a function of x and y, so

[tex]\frac{d\psi}{dx} = \frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y}\frac{\partial y}{\partial x}[/tex]

Now put [itex]y=\phi (x)[/itex], and that's all there is to it.
 
  • #4
Thank you, guys! Thus, we have

[tex]M(x,y) + N(x,y) \: y^{\prime} = 0[/tex]

[tex]d\psi = M(x,y) \: dx + N(x,y) \: dy = 0[/tex]

[tex]\frac{d\psi}{dx} = M(x,y) \: + N(x,y) \: \frac{dy}{dx} = 0 [/tex]

[tex]\frac{d\psi}{dx} = \frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{dy}{dx} = 0 \mbox{.}[/tex]
 
  • #5
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?
 
  • #6
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.
 

1. What is an exact equation in mathematics?

An exact equation in mathematics is a type of differential equation where the coefficients of the variables and their derivatives are continuous and differentiable. This means that the equation can be solved using direct integration rather than using methods such as separation of variables.

2. How is an exact equation different from an inexact equation?

An inexact equation is a type of differential equation where the coefficients of the variables and their derivatives are not continuous or differentiable. This means that the equation cannot be solved using direct integration and requires other methods such as separation of variables. In contrast, an exact equation can be solved using direct integration.

3. What are the steps to solve an exact equation?

The steps to solve an exact equation are as follows:

  1. Check if the equation is exact by verifying if it satisfies the exactness condition.
  2. If the equation is not exact, then make it exact by multiplying both sides by an integrating factor.
  3. Integrate both sides of the equation with respect to one of the variables.
  4. Add a constant of integration.
  5. Use the given initial conditions to solve for the constant.

4. Can an exact equation have multiple solutions?

Yes, an exact equation can have multiple solutions. This is because the constant of integration that is added after integrating can have different values, resulting in different solutions. However, the solutions will all satisfy the original equation.

5. How is an exact equation used in real-life applications?

Exact equations are used in various fields such as physics, engineering, and economics to model real-life situations. For example, in physics, exact equations are used to describe the behavior of physical systems such as fluid flow or electrical circuits. In engineering, they are used to design and analyze systems such as control systems or mechanical systems. In economics, exact equations are used to model economic systems and predict outcomes.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
808
Replies
8
Views
232
  • Introductory Physics Homework Help
Replies
21
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
289
  • Introductory Physics Homework Help
Replies
21
Views
2K
Replies
8
Views
1K
Replies
3
Views
401
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Advanced Physics Homework Help
Replies
10
Views
573
Back
Top