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!

Separable Equations

  1. Aug 10, 2008 #1
    Hi I've been working on this problem repeatedly and thought I understood how to solve separable equations problems but I keep getting the wrong answer.

    y' = (1-2x)y^2

    Here's what I got for the problem:

    y'y^2 = (1-2x)

    [tex]\int(1\y^2) dy = \int(1-2x)dx[/tex]

    ln | y^2| = x-x^2 + C

    e^ln|y^2| =e^ x-x^2 +C , e^C = C

    y^2= + e^x-x^2 +C or y^2 = -e^x-x^2 +C

    y=[tex]\sqrt{}+ or - Ce^x-x^2[/tex]

    -1/6=+ or - [tex]\sqrt{}Ce^ 0-0^2[/tex]

    (-1\6)^2 = + or -([tex]\sqrt{}C[/tex])^2

    C = 1\36

    y=+ or - [tex]\sqrt{}1\36 e^x-x^2[/tex]


    Also, I'm not sure what to do when the problem tells you to determine the interval in which the solution is valid.

    Thanks.
     
  2. jcsd
  3. Aug 11, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi Aero6! Welcome to PF! :smile:

    (have a squared: ² and an integral: ∫ :smile:)

    (and fractions in LaTeX are \frac{}{})


    erm … you got too excited at seeing the 1/ in the ∫ … they're not all logs! :wink:

    ∫dy/y² = … ? :smile:
     
  4. Aug 11, 2008 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Shouldn't it be y'/y^2= 1-2x ?

    Yes, it was!

    No, the integral of y-2 is -y-1

     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Separable Equations
  1. Separable Equations (Replies: 2)

  2. Separate the equation (Replies: 6)

  3. Separable equations (Replies: 4)

Loading...