1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: I need help solving this differential equation

  1. Feb 18, 2010 #1
    Given that,
    f(x,y)= x^3 + 3x^2y + 4xy^2 +2y^3

    Prove that x df/dx + y df/dy = 3f

    I need help at working this out. I have been trying to figure it out all day.
  2. jcsd
  3. Feb 18, 2010 #2
    First, are you able to calculate ∂f/∂x, ∂f/∂y and 3f?
  4. Feb 18, 2010 #3
    I am not too sure . I have tried to differentiate it but got mixed up variables.
  5. Feb 18, 2010 #4
    I differentiated df/dx and got 3x^2 + 6xy +4y^2 and for
    df/dy i got 3x^2 + 8xy + 6y^2.
    Is this right?
  6. Feb 18, 2010 #5
    Since f is a function of two variables, x and y, the derivatives of f with respect to x or y are called "partial derivatives", and denoted ∂f/∂x rather than df/dx, for example.

    To calculate a partial derivative, then all you have to do is differentiate with respect to the desired variable, viewing all others as constant.

    For example: ∂f/∂x = 3x2 + 6xy + 4y2. The "y" is treated as constant in the partial derivative of f with respect to x.

    Now, all there is to do is find the partial of f with respect to y and plug them into the equation and see if the equality holds. :smile:
  7. Feb 18, 2010 #6
    Looks like we're posting at the same time. Yes, those are what I got as well.
  8. Feb 18, 2010 #7
    I found the partial of f with respect to y. so now i got 2 equations, one with respect to x and one with respect to y. what do you plug into the equations exactly and what is 3f?
  9. Feb 18, 2010 #8
    Your equation is x (∂f/∂x) + y (∂f/∂y) = 3f. You want to plug your derivatives in for the respective functions in the equation (and distribute the x and y), which will give you the left side of the equation. One the right, plug in the x^3 + 3x^2y + 4xy^2 +2y^3 for f and distribute the 3 across it. Both sides should be equal.
  10. Feb 18, 2010 #9
    Can you show me what you mean in numbers instead of words because i am having trouble understanding
  11. Feb 18, 2010 #10
    Sure, you have x ∂f/∂x + y ∂f/∂y = 3f.

    Also, ∂f/∂x = 3x2 + 6xy + 4y2 and ∂f/∂y = 3x2 + 8xy + 6y2.

    On the left side of the equation:
    x(∂f/∂x) + y(∂f/∂y) = x(3x2 + 6xy + 4y2) + y(3x2 + 8xy + 6y2) = 3x3 + 9x2y + 12xy2 + 6y3

    On the right side:
    3f = 3(x3 + 3x2y + 4xy2 +2y3) = 3x3 + 9x2y + 12xy2 + 6y3.

    Both are equal.
  12. Feb 18, 2010 #11
    Thank you for your help!
  13. Feb 19, 2010 #12


    User Avatar
    Homework Helper
    Gold Member

    Isn't this Euler's theorem? (one of them).

    The polynomial is homogeneous. That means all the terms are of the form aixiyn-i.

    Not difficult to prove that for any homogeneous polynomial of degree n, f = [tex]\Sigma[/tex] aixiyn-i

    x (∂f/∂x) + y (∂f/∂y) = nf
  14. Feb 20, 2010 #13
    hey, this is simple eulers theorem application. that is, for a 2 variable fxn,if f(x,y) is a fxn homogeneous of degree n, xdf/dx + ydf/dy = nf(x.y) (note, its all partial derivatives here. so it should be a del operator notation rather than d).

    all u need to do is, substitute x=xt and y=yt in the fxn. u will get l.h.s = t^3f(x,y). the degree of t will give the degree of the eqn, if at all homogeneous. and by euler's theorem, the result follows.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook