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Problem regarding '+ or -' with squares and square roots in a differential equation

  1. Feb 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Solve:

    y'=x+y, y(0)=2

    2. Relevant equations



    3. The attempt at a solution

    I THINK my method is correct.. but I messed up somewhere.

    I rearrange for y'-y=x

    integrate both sides gets me:

    y-[itex]\frac{y^{2}}{2}[/itex]=[itex]\frac{x^{2}}{2}[/itex]

    after completing the square I get

    (y-1)[itex]^{2}[/itex]=-x[itex]^{2}[/itex]+1

    But this is where I mess up. To solve for y, I square root each side. But then I get '+ or -' on the right side:

    y=[itex]\pm[/itex][itex]\sqrt{-x^{2}+1}[/itex]+1+c

    so solving for both cases gets me c=0 OR c=2. But I can only have once answer. Where did I go wrong?
     
    Last edited: Feb 14, 2012
  2. jcsd
  3. Feb 14, 2012 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    Re: Problem regarding '+ or -' with squares and square roots in a differential equati

    Going from y'-y=x to [itex]y-\frac{y^2}{2} =\frac{x^2}{2}[/itex] is wrong. The DE reads as [itex] dy - y dx = x dx,[/itex] so when you integrate on the left you don't get y - y^2/2. You need to use an "integrating factor"; see http://www.ucl.ac.uk/Mathematics/geomath/level2/deqn/de8.html , or read your textbook.

    RGV
     
  4. Feb 14, 2012 #3
    Re: Problem regarding '+ or -' with squares and square roots in a differential equati

    Thanks! Your hint helped me get the right answer!
     
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