# Find the function M?

1. Oct 13, 2014

### Math10

1. The problem statement, all variables and given/known data
Find the function M such that M(x, y)dx+(x^2-y^2)dy=0 is exact.

2. Relevant equations
This is an exact equation.

3. The attempt at a solution
Here's my work:

integral of (x^2-y^2)dy=yx^2-y^3/3+h(x)=C
2xy+h'(x)=M(x, y)
But the answer in the book is M(x, y)=2xy+f(x). How do I get to the answer from there?

2. Oct 13, 2014

### zoki85

By putting h'(x)=f(x)

3. Oct 13, 2014

### gopher_p

So you have $M(x,y)=2xy+h'(x)$ and the book has $M(x,y)=2xy+f(x)$. Seems to me that your answer is basically the same as the book's; you just have different names for the the "extra" function of $x$.

4. Oct 13, 2014

### Staff: Mentor

They want a function M such that My = Nx.
IOW,
$$\frac{\partial M}{\partial y} = \frac{\partial (x^2 - y^2)}{\partial x} = 2x$$
If you integrate 2x with respect to y, you get 2xy + some arbitrary function of x. Taking the partial with respect to y of a function that involves only x is similar to taking the derivative of a constant.

5. Oct 13, 2014

### HallsofIvy

Let f(x,y) be any function of two variables and assume that x and y are themselves functions of some variable, t. Then we could write f as a function of the single variable t and, by the chain rule, $df/dt= (\partial f/\partial x) dx/dt+ (\partial f/\partial y)dy/dt$. In terms of "differentials" that is $df= (\partial f/\partial x)dx+ (\partial f/\partial y)dy$.

An expression of the form $M(x,y)dx+ N(x,y)dy$ is an "exact differential" if and only if there exist a function f such that $\partial f/\partial x= M(x, y)$ and $\partial f/\partial y= N(x, y)$. Of course, if that is true, we have $\partial M/\partial y= \partial^2 f/\partial x\partial y$ and $\partial N/\partial x= \partial^2 f/\partial y\partial x$. But those "mixed second partial derivatives" must be equal so to be an "exact differential" we must have $\partial M/\partial y= \partial N/\partial x$

Here, $N= x^2- y^2$ so that $\partial N/\partial x= 2x$. We must have $\partial M/\partial y= 2x$ so that, integrating with respect to y while holding x constant, we have $M= 2xy+ f(x)$. "f(x)" is the constant of integration- since we integrate with respect to y while holding x constant, that "constant of integration" can be any function of x.

6. Oct 14, 2014

### Math10

Thank you so much, @Mark44!