Page 2 of 2 FirstFirst 12
Results 16 to 18 of 18

Thread: First-order linear ode

  1. #16
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    7
    Awards
    2
    If you change the K in the denominator to +K, then the answer is y = -2...
    No, because the k in the numerator and the k in the denominator are the same k. They would both have to change.

    I agree with you up to here:

    $\displaystyle \displaystyle -2\,\frac{e^{-\frac{1}{2}x^2} + K}{e^{-\frac{1}{2}x^2} - K}.$

    Replacing $\displaystyle K$ with $\displaystyle -K$ yields

    $\displaystyle \displaystyle -2\,\frac{e^{-\frac{1}{2}x^2} - K}{e^{-\frac{1}{2}x^2} + K}.$

    Multiplying the negative sign in yields

    $\displaystyle \displaystyle 2\,\frac{K-e^{-\frac{1}{2}x^2}}{K+e^{-\frac{1}{2}x^2}}.$

    Now, finally, one more trick: replacing $\displaystyle K$ with $\displaystyle 1/K,$ will, I think, finish off the problem.
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Member
    Joined
    Apr 2010
    Posts
    156
    Thanks. Where does the other solution y = -2 come from?
    Follow Math Help Forum on Facebook and Google+

  3. #18
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    7
    Awards
    2
    Well, you could take the previous step, right before replacing $\displaystyle K$ with $\displaystyle -K$, and then let $\displaystyle K=0$ there. Equivalently, let $\displaystyle K\to\infty$ in the final solution.
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. First-Order Linear PDE
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: Sep 19th 2011, 01:48 AM
  2. Linear program with higher order non-linear constraints.
    Posted in the Advanced Math Topics Forum
    Replies: 2
    Last Post: Sep 12th 2010, 02:36 AM
  3. 2nd Order Linear O.D.E. Help
    Posted in the Differential Equations Forum
    Replies: 15
    Last Post: Oct 20th 2008, 12:22 AM
  4. Replies: 4
    Last Post: Aug 12th 2008, 04:46 AM
  5. Replies: 1
    Last Post: May 11th 2007, 03:01 AM

Search Tags


/mathhelpforum @mathhelpforum