Equation w/ Homogeneous Coefficients - y=ux substitution

In summary, we are given a differential equation with homogeneous coefficients and are asked to use the y=ux substitution to solve it. After substitution, the equation becomes separable in x and u. We integrate both sides and simplify to get the solution -1/x dx= u/(u2+ 1)du - 1/(u2+1 du = -ln(x) = (1/2)ln(x2+ y2) - ln(x). The given answer uses the substitution into polar coordinates, r and θ, leading to the solution Arc tan(y/x) - 1/2log(x2 + y2) = c.
  • #1
ZachN
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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.
 
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  • #2


ZachN said:
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.
 
  • #3
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
ZachN said:
(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! :smile:

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. :smile:
 
  • #5
I'll try polar coordinates.
 

1. What is an equation with homogeneous coefficients?

An equation with homogeneous coefficients is a type of mathematical equation where all the coefficients of the variables are of the same degree. This means that each term in the equation has the same number of variables raised to the same power.

2. What is the purpose of using u=x substitution in solving equations with homogeneous coefficients?

The purpose of using u=x substitution is to transform the equation into a simpler form, making it easier to solve. By substituting u for x, the equation becomes a polynomial in one variable, which can be solved using traditional methods.

3. How do you perform the u=x substitution in an equation with homogeneous coefficients?

To perform the u=x substitution, you need to identify the terms in the equation that have the same degree. Then, you can choose one of these terms to substitute for u. Once you have substituted u for x, you can solve the equation using traditional algebraic methods.

4. Can any equation with homogeneous coefficients be solved using the u=x substitution method?

No, not all equations with homogeneous coefficients can be solved using the u=x substitution method. This method is only applicable to equations where all the coefficients are of the same degree. If an equation has variables raised to different powers, this method cannot be used.

5. Are there any other methods for solving equations with homogeneous coefficients?

Yes, there are other methods for solving equations with homogeneous coefficients, such as the method of undetermined coefficients and the method of variation of parameters. These methods are more advanced and are often used for solving higher-order homogeneous differential equations.

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