1. Not finding help here? Sign up for a free 30min 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!

Differential Equation of Order 1

  1. Oct 4, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the particular solution to the initial value problem: (3xy - 4x - 1)dy + y(y - 2)dx = 0; when x=1, y=2


    2. Relevant equations
    dy + (p(x)y - q(x))dx = 0
    e^(-∫p(x)dx) * (c + ∫e^(∫p(x)dx) * q(x)dx)


    3. The attempt at a solution
    Sorry if this is vague, but I just spent 30 min typing out the entire process only to have it deleted. This is where I got stuck:

    p(y) = 3y - 4 / y(y - 2)
    q(y) = 1 / y(y - 2)

    x = [y^(1/2) / (y-2)^(1/2)] * (c + ∫dy/[y^(3/2) * (y - 2)^(1/2)])

    I don't know how to do the remaining integral. I think it's partial fractions but the y^(3/2) is confusing me.
     
  2. jcsd
  3. Oct 4, 2011 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    I can not follow you. If you found the integrating factor which is simply y, (Read: http://www.math.hmc.edu/calculus/tutorials/odes/) you get the exact equation

    y^2(y-2)dx+y(3xy-4x-1)dy.

    You find the potential function U(x,y) by integrating y^2(y-2) with respect to x and adding an "integration constant" in the form g(y), or integrating y(3xy-4x-1) with respect to y, and including the integration constant f(x). Find f(x) and g(y) so that both forms of U(x,y) are identical. The solution is U(x,y)=constant. Substitute the initial condition to get the appropriate value of the constant.

    ehild
     
    Last edited: Oct 4, 2011
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Differential Equation of Order 1
Loading...