Exact ODE and Finding Integrating Factors

[greentea]

Homework Statement

In my ODE class, we learned how to solve first order ordinary differential equations which are not exact yet but exact after multiplying by the right integrating factor. The integrating factor we learned about take one of the five forms: f(x), f(y), f(xy), f(x/y), and f(y/x). This is to say, the integrating factors can be a function of x only, y only, x*y, x/y, and y/x.

Homework Equations

My teacher is expecting us to solve one of these ODE among many others about 7 minutes flat. Going through all the five possible integrating factors can be very time consuming. I want to know if there are tricks to knowing which one of the five integrating factors are more likely by perhaps examining the structure of the ODE or certain characteristics that they exhibit .

If there aren't any systematically efficient way to solve ODEs that are exact by integrating factors, are there tips or tricks in general are useful in finding the right integrating factor?

Thanks!

 

[Mugged]

Solving math problems is an art. You making a certain step in solving a problem may seem obvious to you but completely surprising to someone else. There is no substitute for experience, so my tip is to look at all practice problems you have and try to guess which is the correct integrating factor before attempting to solve. You should be able to see patterns.

 

[bigfooted]

There are some systematic ways to get integrating factors for certain types of ODE's. The ODE's that you'll encounter are probably chosen in such a way that you can almost immediately guess the integrating factor

The idea is as follows: the first order ode Q(x,y)dx - P(x,y)dy=0
admits a one-parameter group with generator \xi\frac{\partial}{\partial x}+\eta\frac{\partial}{\partial y}
if the function \mu=\frac{1}{\xi Q-\eta P}
is an integrating factor of the ODE.

So you'll need to rewrite your ODE into the Pfaffian form above, and either remember or guess the coefficients in the generator.
We know the coefficients of some ODE's, e.g.:
y'=f(x) \xi=0,\eta=1
y'=f(y) \xi=1,\eta=0
y'=f(y/x) \xi=x,\eta=y
y'=y/x +xf(y/x) \xi=1,\eta=\frac{y}{x}
y'=y/(x+f(y)) \xi=y,\eta=0
xy'=y+f(x) \xi=0,\eta=x

you can check if your generator is correct for your ODE y'(x,y)=\omega(x,y) by substituting it into the equation for the linearized symmetry generator:
\eta_x+(\eta_y-\xi_x)\omega - \xi_y\omega^2=\xi\omega_x+\eta\omega_y