Exact Differential Equations for Level Curves of u(x,y) = cos(x^2-y^2)

[Mark Brewer]

Homework Statement

Given u(x.y), find the exact differential equation du = 0. What sort of curves are the solution curves u(x,y) = constant? (These are called the level curves of u).

u = cos(x² - y²)

The Attempt at a Solution

partial derivative du/dx = (-2x)sin(x² - y²)dx
partial derivative du/dy = (2y)sin (x² - y²)dy

P = (-2x)sin(x² - y²)dx ; Q = (2y)sin (x² - y²)dy

I found that they're not exact differential equations

so, I'm using integration factors.

1/Q(P dx - Q dy) = R(x)

(1/(2y)sin (x² - y²))((-2x)sin(x² - y²) - (2y)sin (x² - y²)) = R(x)

I get, R(x) = -x/y

Then,

F = exp^(R(x))dx

I get, F = exp^((-x^2)/(2y))

Then,

M = FP and N = FQ

so,

M = (exp^((-x^2)/(2y)))((-2x)sin(x² - y²))

and

N = (exp^((-x^2)/(2y)))((2y)sin (x² - y²))

I then took the partial derivative of M in respect to y, and this is where I am getting stuck.

I also have to take the partial derivative of N in respect to x.

Any help would be much appreciated.

 

[Chestermiller]

At constant u,
$$du=\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy=0$$
What does this tell you about dy/dx?

Chet

 

[Mark Brewer]

At constant u,
$$du=\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy=0$$
What does this tell you about dy/dx?

Chet

I believe it tells me that the partial derivatives are continuous

 

[Mark Brewer]

I believe it tells me that the partial derivatives are continuous

it may also tell me that the partials in respect to x and y are equal to u and 0...?

 

[Muhannad]

Checking for exactness,
P = (-2x)sin(x2 - y2)dx, Q = (2y)sin (x2 - y2)dy

dP/dy = (4xy)cos(x^2- y^2)
dQ/dx = (4xy)cos(x^2 - y^2)

dP/dy = dQ/dx, they are exact!
So, don't use the "integration factors", just find the general solution.