- #1

Mark Brewer

- 38

- 4

## 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(x

^{2}- y

^{2})

## The Attempt at a Solution

partial derivative du/dx = (-2x)sin(x

^{2}- y

^{2})dx

partial derivative du/dy = (2y)sin (x

^{2}- y

^{2})dy

P = (-2x)sin(x

^{2}- y

^{2})dx ; Q = (2y)sin (x

^{2}- y

^{2})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 (x

^{2}- y

^{2}))((-2x)sin(x

^{2}- y

^{2}) - (2y)sin (x

^{2}- y

^{2})) = 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(x

^{2}- y

^{2}))

and

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

^{2}- y

^{2}))

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.