# Exact differential equations

## Homework Statement

Let M(x,y) = yf(xy) and N(x,y) = xg(xy), where f(v) and g(v) are functions of a single real variable v, defined and continuously differentiable for all real values of v. Under what conditions on f and g is the form Mdx + Ndy exact for all values of x, y in the plane? In that case, find the function u(x,y) such that exact equation holds, and use that information to infer the general solution to the equation y' = -(M(x,y)/N(x,y)).

## The Attempt at a Solution

Okay, so the way I see it, I have to use the product rule and the chain rule to take the derivatives, so

$$\frac{\partial M}{\partial y} = xyf_y(xy) + f(xy)$$

and

$$\frac{\partial N}{\partial x} = xyg_x(xy) + g(xy)$$.

If the form is exact, then we have

$$f(xy) + xyf_y(xy) = g(xy) + xyg_x(xy)$$

so maybe a good condition is something like

$$\frac{f(xy) - g(xy)}{f_y(xy) - g_x(xy)} = -xy$$

provided the denominator is nonzero. But then, how do use this to find u(x,y)? I need to integrate M(x,y) with respect to x, but f(xy) depends on x. I have no clue what f is, and I don't see how this condition will help at all. What am I missing or doing wrong? Thanks for any help.

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HallsofIvy
Homework Helper
Two comments: First, since f and g are functions of a single variable, "$f_y(xy)$" and "$g_x(xy)$" are not correct notation. It should be " f '(xy)" and "g '(xy)". Second, rather than that fraction, I would be inclined to write the formula (which is correct) as $g(xy)- f(xy)= xy(f '(xy)- g '(xy))$.

Now, solve the same way you normally would. Since yf(xy)dx+ xg(xy)dy is exact, there exist a fuction, U(x,y), such that
$$dU= \frac{\partial U}{\partial x}dx+ \frac{\partial U}{\partial y}dy= yf(xy)dx+ xg(xy)dy$$
that means that
$$\frac{\partial U}{\partial x}= yf(xy)$$
and so, integrating with respect to x while treating y as a constant,
$$U(x,y)= F(xy)+ W(y)$$
where F is an anti-derivative of f and W is an arbitrary differentiable function (the "constant" of integration).

Then
$$\frac{\partial U}{\partial y}= xf(xy)+ W'(y)= xg(xy)$$
and
$$W'(y)= x(g(xy)- f(xy))$$
In order that that be possible, $x(g(xy)- f(xy))$ must be a function of y only.

Thanks so much for responding. I'm confused a little about the conclusion. If x(g(xy) - f(xy)) only depends on y, what does that mean for the solution? From the condition on exactness, I guess it means that x^2y(f'(xy) - g'(xy)) only depends on y, but I don't see how that helps. Sorry for my confusion.

I was wondering about the derivatives issue as well. It makes sense that it should be f'(xy) instead of f_y(xy), since there's only one variable, but I thought it would be the case that I first evaluate f(v) with v = xy, then take the partial derivative with respect to y. So, for example, if f(v) = v^2, then f(xy) = x^2y^2. Now f'(v) = 2v, so f'(xy) = 2xy. But ∂/∂y (x^2y^2) = 2x^2y, which is different. Maybe more explicitly,

$$\frac{\partial}{\partial y} yf(xy) = f(xy) + y \left ( \frac{\partial}{\partial y} f(xy) \right )$$

right? Now at this point, shouldn't I evaluate f at xy and take the partial derivative with respect to that expression? It seems wrong to just differentiate f if I'm not differentiating with respect to the variable xy.

So right now, it seems that I can only reduce this to an equation

u(x,y) = F(xy) + G(xy) + C(y) + D(x)

where C and D are constants of integration since I don't know anything about the functions f and g.

$$\frac{\partial}{\partial y} f(xy) \right = f'(xy) \cdot x.$$