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Substituting y = vx in Differential Equations


Date: 02/16/2001 at 06:20:44
From: Andrew Thomson
Subject: Differential equations

This equation is supposed to be solvable by separating the variables, 
but I can't figure it out.

     (y^2 - x^2)dy + 2xy*dx = 0


Date: 02/16/2001 at 15:19:08
From: Doctor Anthony
Subject: Re: Differential equations

Try putting your equation in this form:

     dy/dx = 2xy/(x^2 - y^2)

The right-hand side can be written as a function of y/x:

     dy/dx = 2(y/x)/(1 - (y/x)^2) = f(y/x)

For an equation of the form dy/dx = f(y/x), making the substitution 
y = vx is guaranteed to render it separable. (An easier test for this 
condition is to see whether f(x,y) = f(vx,vy).) The product rule gives

     dy/dx = v + x(dv/dx)

Plug this into your equation:

     (v^2*x^2 - x^2)(v + x(dv/dx)) + 2x^2*v = 0

Divide through by x^2:

     (v^2 - 1)(v + x(dv/dx)) + 2v = 0

     v + x(dv/dx) = -2v/(v^2-1)
         x(dv/dx) = -2v/(v^2-1) - v
         x(dv/dx) = [-2v - v(v^2-1)]/(v^2-1)
         x(dv/dx) = -v(2 + v^2 - 1)/(v^2-1)
         x(dv/dx) = -v(v^2+1)/(v^2-1)

     (v^2-1)dv      dx
     ---------  = - --
      v(v^2+1)       x

      -1     2v         dx
     [-- + -----]dv = - --
       v   v^2+1         x

Integrate:

     -ln(v) + ln(v^2+1) = - ln(x) + C

          ln[(v^2+1)/v] = ln(1/x) + ln(A)  where A is a constant

              (v^2+1)/v =  A/x

                v + 1/v = A/x

Substitute v = y/x:

     y/x + x/y = A/x

     y^2 + x^2   A 
     --------- = -
        xy       x

     y^2 + x^2 = Ay

     x^2 + y^2 - Ay  = 0

which is the equation of a circle.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Calculus

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