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Equation w/ Homogeneous Coefficients  y=ux substitution 
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#1
Nov1208, 12:20 PM

P: 24

I am teaching myself, this problem is from ODEs by Tenenbaum and Pollard. This is not homework for a class.
1. The problem statement, all variables and given/known data (x+y){dx}  (xy){dy} = 0 2. Relevant equations y=ux, {dy} = u{dx} + x{du} 3. The attempt at a solution Substitution should lead to a separable equation in x and u: (x+ux){dx} + (ux  x)(u{dx} + x{du}) = 0; x(u+1)dx + u^{2}x{dx} + ux^{2}{du}  ux{dx}  x^{2}{du} = 0; xu(^{2}){dx} + x^{2}(u  1){du} = 0; 1/x{dx} = (u  1)/(u^{2} + 1){du} Okay, I am assuming that I am correct up to this point but the answer given by the text is: Arc tan(y/x)  1/2log(x^{2} + y^{2}) = c I understand where the Arc tan(y/x) comes from  the (1/(u^{2} + 1)). I am having trouble with the 1/2 log(x^{2} + y^{2})  where does the log(x) go? I have a couple of other problems in the same form as this which arise in isogonal trajectories and I don't want to just skip over this because I am obviously having a problem with these integrations. 


#2
Nov1208, 01:32 PM

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PF Gold
P: 39,564

The left side give ln(x), of course. Then antiderivative of 1/(u^{2}+ 1) is arctan(u) but to integrate u/(u^{2}+ 1) let v= u^{2}+ 1, dv= 2udu and the integral becomes udu/(u^{2}+ 1)= dv/(2v)= (1/2)ln(v)= (1/2) ln(u^{2}+ 1)= (1/2)ln((y^{2}/x^{2}+ 1)= (1/2)ln((x^{2}+ y[sup]2[sup])/x^{2}= (1/2)ln(x^{2}+ y^{2}) ln(x). 


#3
Nov1208, 02:00 PM

P: 24

Yes, thank you  I was not splitting up the ration into two equations and then integrating. I will try to be more observant from now on.



#4
Nov1308, 05:01 PM

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P: 26,148

Equation w/ Homogeneous Coefficients  y=ux substitution
As an alternative method … always look at the answer … it may give you a clue as to an easy substitution … in this case, the answer uses tan^{1}y/x and x^{2} + y^{2}, so the obvious substitution is into polar coordinates, r and θ. Try it and see. 


#5
Nov1408, 09:56 AM

P: 24

I'll try polar coordinates.



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