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Ordinary Differential Equation by substitution.

  1. Feb 17, 2009 #1
    1. The problem statement, all variables and given/known data
    dy/dx = (x+3y)/(3x+y)

    I have to solve the given differential equation by using an appropriate substitution...


    3. The attempt at a solution

    I used algebra to make the equation (3x+y)dy - (x+3y)dx = 0

    then made x = vy and dx = vdy + ydv.

    then plugged in to get (3vy+y)dy - (vy+3y)(vdy+ydv)

    and now I'm looking for a common factor to cancel and I'm stuck, any help? Am I even headed in the right direction...thanks.
     
  2. jcsd
  3. Feb 17, 2009 #2

    Mark44

    Staff: Mentor

    It doesn't seem you're heading in the right direction. Here's what I would try.

    Starting with this equation,
    (3x+y)dy = (x+3y)dx

    integrate the left side with respect to y. Since you're integrating with respect to y, you'll end up with whatever you get plus an arbitrary function of x alone.
    Then, integrate the right side with respect to x, similar to above. This time around you'll need to include an arbitrary function of y alone.

    The left and right sides have to be equal for all pairs of (x, y) values, so you can equate the arbitrary functions of x or y on either side with what's on the opposite side.

    This should give you the equation h(x, y) = 0.

    To check your answer, take the derivative implicitly.
     
  4. Feb 18, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Looking at the symmetry of "3x+ y" and "x+ 3y", which has "average" value 2x+ 2y= 2(x+ y), let u= x+ y and v= x- y. Then x= (1/2)u+ (1/2)v and y= (1/2)u- (1/2)v, dx= (1/2)du+ (1/2)dv, dy= (1/2)du- (1/2) dv. Put those into the equation. you should wind up with udv= vdu.
     
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