Evaluating $\int xy \ d(xy)$

Question

In the "Integrating Factors" section of an old introductory differential equations book, this example problem is presented: $$\frac{x \ dy-y \ dx}{x^2}=xy \ (x \ dy+y \ dx)$$ $$d \left( \frac{y}{x}\right)=xy \ d(xy) \quad (*)$$ The book then simply says, "Integrate," and shows the answer: $\frac{y}{x}=\frac{1}{2} \ x^2y^2+k$.

I'm puzzled by the right side of equation (*) though. Integrating it would produce an integral of the form $\int xy \ d(xy)$, which is new to me. I'm familiar with differentials and iterated integrals, but I've never seen an integral with two variables simultaneously in a $d( \ )$ expression with variables preceding the $d( \ )$.

The left-hand side of (*) is easy: an integral of a differentiated function is simply the function itself (plus a constant).

But for the right side, can I just treat $(xy)$ as some sort of unit and integrate, since it's followed by $d(xy)$? $$\int xy \ d(xy)= \frac{1}{2}(xy)^2=\frac{1}{2}x^2y^2$$ That just seems too simplistic, especially since two variables are being multiplied together. I'd expect it to require something akin to integration by parts.

Yet, if I do the following (which seems more legitimate), I get a wrong answer for the right-hand side: $$\int xy \ d(xy)= \int xy \ (x \ dy+y \ dx) = \int (x^2y \ dy \ + \ xy^2dx)$$ $$=x^2\frac{y^2}{2}+y^2\frac{x^2}{2}=\frac{2x^2y^2}{2}=x^2y^2 \ + \ k$$ I don't think the discrepancy can be resolved simply by saying the constants are different. Where am I going wrong?

Answer 1

One of the wonderful things about differentials (as opposed to, say, partial derivatives) is that they don't care about independent/dependent variables, interact extremely well with algebraic manipulations, and so forth.

For example, you know that $\mathrm{d}\left( \frac{1}{2} t^2 \right) = t \, \mathrm{d}t$, and if we have $t = xy$ then it immediately follows that $\mathrm{d}\left( \frac{1}{2} (xy)^2 \right) = xy \, \mathrm{d} xy$. Thus, $\frac{1}{2}(xy)^2$ is an antiderivative of $xy \, \mathrm{d}xy$.

And this fact is true always; when $x$ and $y$ are: independent, dependent on each other, dependent on additional variables, or even constant!

Now, an important point that you've overlooked is, just as there is a difference between the differential $\mathrm{d}$ and partial derivatives, there is a difference between the 'antidifferential' and the partial antiderivatives.

When you calculated $\int x^2 y\,\mathrm{d}y = \frac{1}{2} x^2y^2$, you computed a partial antiderivative. That is, you antidifferentiated subject to the constrant that $x$ is held constant. This is no good, since you were trying to invert the differential, not a partial derivative!

The same happens if you consider definite integration instead of indefinite integration — the appropriate integral would be a path integral $\int_\gamma x^2 y \, \mathrm{d} y $, where $\gamma$ is a path restricted to the one-dimensional space of allowed values of $(x,y)$. Arbitrary paths in the plane are disallowed — such as the vertical paths that would be analogous to the partial antiderivative in $y$.

---

Now, partial antidifferentiation can be used to compute an antidifferential. If $x$ and $y$ are independent variables and $z$ a scalar depending on them, then

$$ \mathrm{d} z = \frac{\partial z}{\partial x} \mathrm{d} x +\frac{\partial z}{\partial y} \mathrm{d} y $$

(where, as usual, $\partial/\partial x$ means to hold $y$ constant and vice versa)

So, if an antidifferential exists, we can get information from partial antiderivatives; e.g. $$ \int \frac{\partial z}{\partial x} \mathrm{d} x = z + c(y) $$ (where, as usual, $\int \ldots \mathrm{d}x$ means the partial antiderivative while holding $y$ constant) Then, if the solution is not obvious, take the differential again (or just take the partial derivative in $y$) to get a differential equation you can solve for $c(y)$.

For example, in the given problem, if we define $z$ be such that $$ \mathrm{d}z = x^2 y \, \mathrm{d}y + x y^2 \, \mathrm{d}x $$ then in the generic domain where $x$ and $y$ are independent, the partial antiderivatives in $y$ and $x$ you computed imply, respectively, $$ z = \frac{x^2 y^2}{2} + C_1(x) $$ $$ z = \frac{x^2 y^2}{2} + C_2(y) $$ at which point the general solution is clear: $$ z = \frac{x^2 y^2}{2} + C $$ And as before, since the equation $$ \mathrm{d}\left(\frac{x^2 y^2}{2} + C\right) = x^2 y \, \mathrm{d}y + x y^2 \, \mathrm{d}x $$ holds when $x$ and $y$ are independent, it holds always.

Answer 2

I guess you could do this:

1. Make the substitution $u = xy$.
2. Then the integral becomes $\int u\, du$.

That should be easy enough to solve, then just "de-substitute" u as xy.

After doing some thinking, I decided to substitute $d(xy)$ as $$(y + xy')dx. \tag{1}$$ Then, substituting (1) back into the original integral, we get $$\int xy^2 + x^2yy'.$$ Any hints on how to evaluate this?