Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    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


    User Avatar
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook