# finding an integrating factor for the nonlinear non-autonomous ODE $(xy)y'+y\ln y - 2xy = 0$

I though i might refresh my ODE knowledge a bit and decided to try the following exercise from "Advanced mathematical methods for scientists and engineers" by C. Bender and S. Orszag

Solve the following ODE using an integrating factor of the form $I=I(y)$:

$$(xy)y'+y\ln y - 2xy = 0 \tag{ODE}$$

Clearly (ODE) is not exact. The integrating factor $I$ is a function such, that $$I(y)\left((xy)y'+y\ln y - 2xy\right) = \frac{d}{dx}F(x,y)$$ for some $F$. The solutions of (ODE) are thus given by $F(x,y)=c$.$^\dagger$ Some initial tests suggested, that $I$ is non-trivial, consider therefore the following necessary and sufficient condition

$$M(x,y) + N(x,y)y' = 0$$ is exact iff $$\frac{\partial}{\partial y}M(x,y) = \frac{\partial}{\partial x}N(x,y)$$

In our case this gives

$$M=I(y) \left(y \ln y - 2xy\right) \quad\quad N=I(y)xy$$ $$\Rightarrow I'(y)+I(y)\cdot \frac{\ln y - y + 1-2x}{y\ln y-2xy} = 0$$

which can formaly be integrated, but does not have a closed algebraic expression.

Did I do anything wrong so far? Am I missing a more obvious integrating factor or is this really the solution Bender and Orszag had in mind?

$^\dagger$ $F$ in general depends on $x$, $y$ and the derivatives of $y$. As the highest derivative must be lower than the order of the ODE though, only $y$ itself remains in our case.

• @Maesumi $\frac{\partial}{\partial y} M = I'(y\ln y-2xy)+I(\ln y + 1 -2x)$. So I divided the whole equation by the factor next to the $I'$ to isolate the $I'$. What did you do with it? – example Feb 24 '13 at 22:39
• my mistake! Oddly enough you can factor a $y$ from the equation and get a simpler equation for I. But still no closed form. – Maesumi Feb 25 '13 at 2:23
• It does not look like there is an analytical solution to this one. – Nasser Oct 15 '13 at 22:38

I don't think you can find an obvious integrating factor for this ODE.

$xyy'+y\ln y-2xy=0$

$x\dfrac{dy}{dx}+\ln y-2x=0$

$x\dfrac{dy}{dx}=2x-\ln y$

$(2x-\ln y)\dfrac{dx}{dy}=x$

Let $u=x-\dfrac{\ln y}{2}$ ,

Then $x=u+\dfrac{\ln y}{2}$

$\dfrac{dx}{dy}=\dfrac{du}{dy}+\dfrac{1}{2y}$

$\therefore2u\left(\dfrac{du}{dy}+\dfrac{1}{2y}\right)=u+\dfrac{\ln y}{2}$

$2u\dfrac{du}{dy}+\dfrac{u}{y}=u+\dfrac{\ln y}{2}$

$2u\dfrac{du}{dy}=\dfrac{(y-1)u}{y}+\dfrac{\ln y}{2}$

$u\dfrac{du}{dy}=\dfrac{(y-1)u}{2y}+\dfrac{\ln y}{4}$

This belongs to an Abel equation of the second kind.

In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.

Let $u=\dfrac{1}{v}$ ,

Then $\dfrac{du}{dy}=-\dfrac{1}{v^2}\dfrac{dv}{dy}$

$\therefore-\dfrac{1}{v^3}\dfrac{dv}{dy}=\dfrac{y-1}{2yv}+\dfrac{\ln y}{4}$

$\dfrac{dv}{dy}=-\dfrac{v^3\ln y}{4}-\dfrac{(y-1)v^2}{2y}$