Homework Help: How to find an integrating factor for this problem?

AdrianZ · Oct 24, 2011

1. The problem statement, all variables and given/known data
Find an integrating factor for:
xdy - (y + x^2 + 9y^2)dx = 0

2. Relevant equations
P(x,y)dx + Q(x,y)dy = 0
Δμ = μ_yP - μ_xQ.
where μ is the integrating factor.
3. The attempt at a solution
well, I don't know what I should do. I can use the formula I wrote but that would lead to solving a PDE. any ideas?

AdrianZ · Oct 25, 2011

No ideas?

obafgkmrns · Oct 25, 2011

It would take a pretty lucky guess to come up with an integrating factor directly. However, you can use a Riccati transform and then back-track to find an integrating factor if that's your pleasure.

If you substitute y(x) = - x (dw/dx) / (9 w) where w is a function of x, you'll wind up with a linear 2nd order ODE in w.

Wikipedia has a decent description: http://en.wikipedia.org/wiki/Riccati_equation

AdrianZ · Oct 25, 2011

I don't know how to solve a linear 2nd order ODE yet. I mean I know but our professor hasn't started the second chapter which is about 2nd order linear equations. so I can't transform it into a 2nd order linear ODE. Is there a way to solve it without transforming it into a 2nd order linear ODE by finding an integrating factor?

obafgkmrns · Oct 25, 2011

I sure don't see an obvious way. Is it possible that the equation should be

y dy - (y + x^2 + 9 y^2) dy = 0 ?

An integrating factor would be simple to find in that case.

AdrianZ · Oct 25, 2011

obafgkmrns said: ↑

I sure don't see an obvious way. Is it possible that the equation should be

y dy - (y + x^2 + 9 y^2) dy = 0 ?

An integrating factor would be simple to find in that case.

No :(

lurflurf · Oct 25, 2011

group in terms of 3y/x to make the integrating factor easy to see
x dy - (y + x^2 + 9y^2)dx = 0
(x²/3)d(3y/x)-x²((3y/x)²+1)dx=0

AdrianZ · Oct 25, 2011

lurflurf said: ↑

group in terms of 3y/x to make the integrating factor easy to see
x dy - (y + x^2 + 9y^2)dx = 0
(x²/3)d(3y/x)-x²((3y/x)²+1)dx=0

Now It's clear what I should do next. but I don't understand how you obtained the second equation from the first that directly.

lurflurf · Oct 25, 2011

It helps that I knew that
x dy-y dx=x²d(y/x)
from there it was clear x and y/x would be better variables that x and y.
These take a bit of practice.
There are a few others, but I keep in mind these common examples
d(xy)=x dy+y dx
y² d(x/y)=y dx-x dy
x² d(y/x)=x dy-y dx
(x²+y²) d(arctan(y/x))=x dy-y dx
[tex] x^{1-p}y^{1-q}{d}(x^py^q)=pydx+qxdy[/tex]