Integrating factor of two variables

Question

I am trying to solve the following: $$ (x + x^2 + y^2) dy - ydx = 0. $$ with an integrating factor involving both x and y. Indeed, it seems that an integrating factor of only one variable would not be possible. As $f(x,y) = \frac{y}{x+x^2 + y^2}$ is not homogeneous, it seems that route is closed as well.

My tactic thus far has been to try functions of several forms, such as $x^\alpha y^\beta$ and a few polynomials, but have always come up sort, ending up with higher and higher powers of both $x$ and $y$, meaning that I have not been able to find constants such that any variation brings about an exact DEQ.

I have two questions:

1. Is there a general ansatz when dealing with $M$ and $N$ as polynomials of differing degree and with differing constants?
2. Can anyone point me in a direction where I might be able to find something like an integrating factor for this?

I'd be most appreciative of any ideas!

Answer 1

Integrating factor is $\mu=\frac{1}{x^2+y^2}$

How i find it ?

Let $\mu=f(t)$, $t=x^2+y^2$, $P=-y$, $Q=x+x^2+y^2$.

Then $\mu'_x=f'(t)2x$, $\mu'_y=f'(t)2y$.

From $(P\mu)'_y=(Q\mu)'_x$ we get differential equation $$tf'(t)+f(t)=0$$ with solution $$f=\frac{C}{t}$$

Answer 2

Multiply both sides with a function of $x$ and derive both coefficients of the differential forms with respect to the other variables

$$ d \ \left((a(x) (x^2 + x + y^2) \ dy - a(x) \ y \ dx)\right) = \left(\ \partial_x \left(a(x)\ (x^2 + x + y^2)\ \right) \ + \ \partial_y\left(\ a(x)\ y \right) \ \right) \ dx \wedge dy$$

If the 2-form is zero, the 1-form with the factor a(x) is the derivative of a function f

$$d \ f = \partial_x f dx + \partial_y f dy = - a(x) y dx +a(x) (x^2 + x + y^2) dy)$$

Answer 3

All the credit for finding a simple integrating factor goes to Aleksas Domarkas.

I only contribute this answer for two reasons:

• It seems like Aleksas used an intermediate step that is left to solve by more pedestrian readers (like me).

• The ODE really interesting and related again to the mother of all differential forms in $\mathbb R^2\setminus\{0\}$ that are closed but not exact (see below).

Notation: $$ Q\,dy+P\,dx=0\,,\quad Q(x,y)=x+x^2+y^2\,,\quad P(x,y)=-y\,. $$ Ansatz for the integrating factor: $$ \mu(x,y)=g(x^2+y^2)\,. $$ This gives $$ \mu_x=2\,x\,g'\,,\quad \mu_y=2\,y\,g'\,. $$ The condition $$ \partial_y(\mu P)=\partial_x(\mu Q) $$ gives $$ \mu_yP+\mu\,P_y=\mu_xQ+\mu\,Q_x\,, $$ that is: $$ -2\,y^2g'-g=2\,x\,g'\,(x+x^2+y^2)+g\,(1+2x)\,. $$ This simplifies to $$ 0=2\,(y^2+x^2)\,g'+2\,x\,g'(x^2+y^2)+2\,g+2\,x\,g\,. $$ Setting $t=x^2+y^2$ this simplifies further to $$ 0=t\,g'+g+x\,(t\,g'+g) $$ from which we conclude $0=t\,g'+g$ and thus $g(t)=1/t\,.$ One integrating factor is therefore $$ \boxed{\quad\phantom{\Big|}\mu(x,y)=\frac1{x^2+y^2}\,\quad} $$ (cf. Aleksas).

Now solving the ODE:

The traditional method of finding a function $f(x,y)$ whose gradient is $(\mu\,Q,\mu\,P)$ won't work in this case as I will show in the following.

The ODE $\mu\,Q\,dy+\mu\,P\,dx=0$ now takes the form $$ \underbrace{\frac{x\,dy-y\,dx}{x^2+y^2}}_{\textstyle\boldsymbol{\omega}}+dy=0\,. $$ The first term $\boldsymbol{\omega}$ is the mother of all differential forms on $\mathbb R^2\setminus\{0\}$ which is closed but not exact: The function $$ \phi(x,y)=\arccos\Big(\frac{x}{\sqrt{x^2+y^2}}\Big) $$ (or in polar coordinates $\phi(r,\theta)=-\theta$) solves $d\phi=\boldsymbol{\omega}$ but has a branch cut where it is discontinuous. In other words, it is defined only on, say, the domain $\mathbb R^2\setminus(-\infty,0)\,.$ Obviously $$ f(x,y)=\phi(x,y)+y $$ is a function on that same domain that satisfies $df=\boldsymbol{\omega}+dy\,.$

It follows that the solutions of the original ODEs $Q\,dy+P\,dx=0$ and $\mu\,Q\,dy+\mu\,P\,dx=0$ are implicitly given as the level curves of $f(x,y)=0\,.$