
#1
Apr609, 12:27 PM

P: 267

[tex] \frac{N_{x}(x,y)  M_{y}(x,y)}{M(x,y)} [/tex] Where M (x,y) = 4(x3/y2)+(3/y)]dx, and N = 3(x/y2)+4y dy When I try and solve this, I get a very complex integrating factor. Does anyone have any suggestions on solving these types of problems? 



#2
Apr609, 12:43 PM

PF Gold
P: 619

[tex]
\frac{M_{y}(x,y)  N_{x}(x,y)}{N(x,y)} [/tex] is equally valid 



#3
Apr609, 01:48 PM

P: 267





#4
Apr2909, 05:04 PM

P: 267

Help with integrating factor
I still am having trouble with integrating factors....any ideas on the problem above?




#5
Apr2909, 05:18 PM

P: 239

What did you get for [tex] \frac{N_{x}(x,y)  M_{y}(x,y)}{M(x,y)}[/tex]?
If your integrating factor say, [tex] \mu [/tex] is a function of x only, then [tex] \frac{d \mu}{dx} = \frac{N_{x}(x,y)  M_{y}(x,y)}{M(x,y)} \mu [/tex] From here you can solve for the exact equation after multiply the integrating factor by the original equation. 



#6
Apr2909, 05:46 PM

P: 267

N_{x}=3y^{2} I think I see what I've been doing wrong now....I've always canceled out the two 3y^{2}, when I should have added. I can't believe I did that..... The reason I posted again, was partly due to another problem involving integrating factors. xsin(y) dy + (2+x)cos(y) dy  Find the integrating factor to make this exact. Since I thought I didn't understand them well, I figured learning how to do the originally posted problem would help. Now, I'm not sure if this problem can be solved as written, since both derivates are with respect to y. 


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