Results 1 to 2 of 2

Math Help - "simple" initial value problem

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    7

    Exclamation "simple" initial value problem

    I need help with this problem:

    Setting y/x=u, solve:

    xy' = (y-x)^3 + y; y(1) = 3/2

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by ntfabolous View Post
    I need help with this problem:

    Setting y/x=u, solve:

    xy' = (y-x)^3 + y; y(1) = 3/2

    Thanks!

    Using the hint given and assuming u is a function of x.

    \frac{y}{x}=u \iff y=xu

    Taking the derivative we get

    y'=u+x\frac{du}{dx}

    Now subbing this into the equation we get

    x\left( u+x\frac{du}{dx}\right)=(ux-x)^3+ux

    Simplifing and factoring we get

    x^2\frac{du}{dx}=x^3(u-1)^3 \iff \frac{du}{dx}=x(u-1)^3

    \frac{du}{(u-1)^3}=xdx

    Integrating both sides we get

    \frac{1}{(u-1)^2}=\frac{1}{2}x^2+c

    From here it is just back subbing good luck.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 6th 2011, 03:00 PM
  2. Replies: 2
    Last Post: April 24th 2011, 07:01 AM
  3. Replies: 1
    Last Post: October 25th 2010, 04:45 AM
  4. Stumped by "simple" polynomial problem
    Posted in the Number Theory Forum
    Replies: 0
    Last Post: November 4th 2009, 08:05 PM
  5. Replies: 5
    Last Post: April 30th 2009, 03:18 AM

Search Tags


/mathhelpforum @mathhelpforum