Equation w/ Homogeneous Coefficients - y=ux substitution

[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.

Homework Statement

(x+y){dx} - (x-y){dy} = 0

Homework Equations

y=ux, {dy} = u{dx} + x{du}

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.

 

[HallsofIvy]

I am teaching myself, this problem is from ODEs by Tenenbaum and Pollard. This is not homework for a class.

Homework Statement

(x+y){dx} - (x-y){dy} = 0

Homework Equations

y=ux, {dy} = u{dx} + x{du}

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}

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).

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.

 

[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.

 

[tiny-tim]

(x+y){dx} - (x-y){dy} = 0
…
the answer given by the text is:

Arc tan(y/x) - 1/2log(x² + y²) = c

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.

 

[ZachN]

I'll try polar coordinates.