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: Separation of variables to solve DE

  1. Aug 2, 2010 #1
    1. The problem statement, all variables and given/known data

    y' + (2/x)y = 3/x^2

    2. Relevant equations

    separation of variables

    3. The attempt at a solution

    First I turned it into

    dy/dx + (2/x)y = 3/x^2 dx


    then multiplied both sides by dx

    dy + (2/x)y = 3/x^2 dx

    I then tried to divide both sides by 2/x and got

    dy + y = 3/2x

    Do I just integrate both sides now?
     
    Last edited: Aug 2, 2010
  2. jcsd
  3. Aug 2, 2010 #2
    Separation of variables won't help (you also made an error when multiplying through the equation with dx). It looks like you'll have to solve this using an integrating factor.
     
  4. Aug 2, 2010 #3
    You also made an error when dividing by (2/x). Make sure that what you do to one term goes for all terms.
     
  5. Aug 3, 2010 #4
    The integrating factor is [itex] e^{\int{\frac{2}{x}dx}[/itex] = [itex]x^2[/itex]

    We than use this and multiply both sides by [itex]x^2[/itex]

    which gives


    [itex]x^2y'+2xy = 3[/itex]

    or

    [itex]yx^2= 3x+C[/itex]

    divide both sides by [itex]x^2[/itex]

    [itex]y = \frac{3}{x}+\frac{C}{x^2}[/itex]
     
  6. Aug 3, 2010 #5
    Is this the general solution?
     
  7. Aug 3, 2010 #6
    yes, it's the general solution.
     
  8. Aug 3, 2010 #7
    Also can be solved as a non-homogeneous Euler equation.
     
  9. Aug 3, 2010 #8

    I see you integrated on the right side wrt x to get 3x+C

    but how did [itex]x^2y'+2xy [/itex] become [itex]yx^2[/itex]? did we integrate the LHS wrt x or y?
     
  10. Aug 3, 2010 #9
    Given a function [tex] f(x,y) [/tex], its total differential with respect to x is [tex] \frac{df}{dx} = \frac{\partial f}{\partial x}\frac{dx}{dx} + \frac{\partial f}{\partial y}\frac{dy}{dx}. [/tex]
     
  11. Aug 3, 2010 #10

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You have to treat both sides of the equation the same way. If you integrate the RHS wrt x, you have to integrate the LHS wrt x as well.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook