Is M(x,y)dx + M(y,x)dy Always Exact?

[Saladsamurai]

Homework Statement

I thought that this was an interesting question.

(a) Show that $(x^3 + y)dx + (y^3 +x)dy = 0\qquad(1)$ is exact.
(b)More generally, is $M(x,y)dx + M(y,x)dy\qquad(2)$ exact? Explain.

Homework Equations

Test for exactness: [tex]\left(\frac{\partial{M}}{\partial{y}}\right)_x=\left(\frac{\partial{N}}{\partial{x}}\right)_y[/itex]

The Attempt at a Solution

(a) Applying the test is simple enough. 1 = 1. Exact.

(b) Now this is how I am thinking about (b). Please correct me if I am wrong. M is simply a rule. It is in the form M(x₁,x₂). It tells us how to operate on whatever is in the x₁ and x₂ spot. In equation (2), in the first term, x₁ = x and x₂ = y , and in the second term x₁ = y and x₂ = x. Wouldn't this imply that the equation will always be exact? Is there a way to show that it is or isn't?

Any thoughts?

 

[cronxeh]

Suppose there is a function z=f(x,y). Your M and N are partial derivatives of z w.r.t x and y

$$M = \left(\frac{\partial{z}}{\partial{x}}\right), N = \left(\frac{\partial{z}}{\partial{y}}\right)$$

By taking a derivative of M with respect to y, and taking derivative of N with respect to x, you are essentially testing for exactness. It is an exact differential equation if and only if:

$$\left(\frac{\partial{}}{\partial{y}}\right) \left(\frac{\partial{z}}{\partial{x}}\right) = \left(\frac{\partial{}}{\partial{x}}\right) \left(\frac{\partial{z}}{\partial{y}}\right)$$

Your objective is to find z.

I think to make it clear you need an example:

Suppose there are 2 functions, z1= 2*x^2 + y^2, z2 = x^2 + 2*y^2
M(z1) = 4x, N(z1) =2y M(z2) = 2x, N(z2) = 4y
My = 0, Nx = 0 My(z2) = 0, Nx(z2) = 0

So if I give you (4x)dx + (2y)dy = 0, your solution will be integral(M(z1))dx + integral(N(z1))dy =2x^2 + y^2 , and

if I give you (2x)dx + (4y)dy = 0, your solution will be integral(M(z2))dx + integral(N(z2))dy = x^2 + 2y^2.

This was only possible because both M and N were describing the same function z

 

[Saladsamurai]

Hi cronxeh I understand the test for exactness and how to find F(x,y) if it is exact. The question in part (b), is if equations of the form (2) are always exact. I am not sure how to apply what you have posted to part (b).

 

[cronxeh]

Hi cronxeh I understand the test for exactness and how to find F(x,y) if it is exact. The question in part (b), is if equations of the form (2) are always exact. I am not sure how to apply what you have posted to part (b).

So wait you saying part b is not a typo? You actually want to find whether M(x,y)dx + M(y,x)dy = c is always exact?

 

[arkajad]

You can try to see what happens by developing M(x,y) into Taylor series around (0,0) and try on $$M(x,y)=x^my^n$$. Once you are done, use $$(x_0,y_0)$$ instead of $$(0,0)$$. This may give you some insight and intuition to start with.

 

[Saladsamurai]

So wait you saying part b is not a typo? You actually want to find whether M(x,y)dx + M(y,x)dy = c is always exact?

Indeed it is not a typo.

 

[arkajad]

Notice that when you write

$$\frac{\partial M(x,y)}{\partial x}$$ it can be written as well as

$$\left(\frac{\partial M(s,t)}{\partial s}\right)_{s=x,t=y}$$

or

$$\left(\frac{\partial M(t,s)}{\partial t}\right)_{t=x,s=y}$$

It's all the same. Now, you can do it.

 

[D H]

(b)More generally, is $M(x,y)dx + M(y,x)dy\qquad(2)$ exact? Explain.

Hint (Big hint): Find a counterexample. It's not hard.