# Exact Differential Equation Integrating Factor

Finding an integrating factor can be a genuine mathematical art. However, certain differential forms can remind us of differentiation techniques that may aid in the solution of the equation at hand. For example, seeing $$x\ dy + y\ dx$$ reminds us of the product rule, as in $$d(xy) = x\ dx + y\ dy$$ [sic], and $$x\ dy - y\ dx$$ might remind us of the quotient rule, $$d(x/y) = \frac{y\ dx - x\ dy}{y^2}$$. In the equation $$x\ dy + y\ dx + 3xy^2\ dy = 0$$, we are again reminded of the product rule. In fact, if you multiply the equation by $$\frac{1}{xy}$$, then $$\frac{x\ dy\ +\ y\ dx}{xy} + 3y\ dy = 0 \implies d(ln(xy)) + 3y\ dy = 0 \implies ln(xy) + \frac{3}{2}y^2 = C$$. Use these ideas to find a general solution for the differential equation $$(x^2 - y^2)(x\ dy + y\ dx) = 2xy(x\ dy - y\ dx)$$

I'm not really sure where to start with this. I can rewrite the equation as $$(x^2 - y^2)\ d(xy) = 2x^3y\ d(y/x)$$ and solve for one of the differentials, but not in a way where I can integrate the RHS. I noticed that $$x\ dy + y\ dx$$ is cleanly expressible as a dot product, and since $$d\vec r$$ = $$\begin{bmatrix}dx \\ dy\end{bmatrix}$$, integrating $$d(xy)$$ could be viewed as calculating work, where $$\vec f(x,\ y) = \begin{bmatrix}y \\ x\end{bmatrix}$$. However, there doesn't appear to be a curve, so I can't set integration bounds or use Green's Theorem. Multiplying everything out didn't spark any ideas either.

Starting from your original differential equation of

$$(x^2 - y^2)(x\ dy + y\ dx) = 2xy(x\ dy - y\ dx) \tag{1}\label{eq1}$$

divide both sides by $$x^2$$ to get

$$\left(1 - \left(\frac{y}{x}\right)^2\right)(x\ dy + y\ dx) = 2xy\left(\frac{x\ dy - y\ dx}{x^2}\right) \tag{2}\label{eq2}$$

Using the suggestions for the differentials, we get

$$\left(1 - \left(\frac{y}{x}\right)^2\right)d\left(xy\right) = 2xy\left(d\left(\frac{y}{x}\right)\right) \tag{3}\label{eq3}$$

We now have a separable equation in $$xy$$ and $$\frac{y}{x}$$. Making the appropriate adjustments, i.e., dividing both sides by $$xy\left(1 - \left(\frac{y}{x}\right)^2\right)$$, gives

$$\frac{d\left(xy\right)}{xy} = \frac{2d\left(\frac{y}{x}\right)}{1 - \left(\frac{y}{x}\right)^2} \tag{4}\label{eq4}$$

You can now integrate both sides appropriately and simplify the results. I assume you can finish the remaining calculations yourself.

Dividing both sides by $$xy(x^2-y^2)$$, we get $$\frac{d(xy)}{xy}=\frac{d(x+y)}{x+y}-\frac{d(x-y)}{x-y}$$ $$\ln(xy)=\ln(x+y)-\ln(x-y)+C_1$$ Answer: $$\frac{xy(x-y)}{x+y}=C$$