# Equation w/ Homogeneous Coefficients - y=ux substitution

1. Nov 12, 2008

### ZachN

Differential Equation w/ Homogeneous Coefficients - y=ux substitution

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} - (x-y){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 + u2x{dx} + ux2{du} - ux{dx} - x2{du} = 0;
xu(2){dx} + x2(u - 1){du} = 0;
-1/x{dx} = (u - 1)/(u2 + 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(x2 + y2) = c

I understand where the Arc tan(y/x) comes from - the (1/(u2 + 1)). I am having trouble with the 1/2 log(x2 + y2) - 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.

Last edited: Nov 12, 2008
2. Nov 12, 2008

### HallsofIvy

Re: Differential Equation w/ Homogeneous Coefficients - y=ux substitution

so -1/x dx= u/(u2+ 1)du - 1/(u2+1 du
The left side give -ln(x), of course. Then anti-derivative of -1/(u2+ 1) is -arctan(u) but to integrate u/(u2+ 1) let v= u2+ 1, dv= 2udu and the integral becomes udu/(u2+ 1)= dv/(2v)= (1/2)ln(v)= (1/2) ln(u2+ 1)= (1/2)ln((y2/x2+ 1)= (1/2)ln((x2+ y2)/x2= (1/2)ln(x2+ y2)- ln(x).

3. Nov 12, 2008

### ZachN

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. Nov 13, 2008

### tiny-tim

Hi ZachN!

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-1y/x and x2 + y2, so the obvious substitution is into polar coordinates, r and θ.

Try it and see.

5. Nov 14, 2008

### ZachN

I'll try polar coordinates.