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Homework Help: Differential equation

  1. Sep 20, 2010 #1
    1. The problem statement, all variables and given/known data

    Solve the differential equation 2y(1 + x2 )y'+ x(1 + y2 ) = 0 where y = 2 when x = 0.
    given answers are
    a. (1 + x2 )(1 + y2 )2= 0
    b. (1 + x2 )2 (1 + y2 ) = 25
    c. (1 + x2 )2 (1 + y2 )2= 0
    d. (1 + x2 )(1 + y2 ) = 0
    e. (1 + x2 )(1 + y2 )2= 25

    3. The attempt at a solution

    2y(1+x2)dy/dx+x(1+y2)=0
    2y(1+x2)dy/dx=-x(1+y2)
    2y/(1+y2)dy = -x/(1+x2)dx

    if i integrate these i end up with ln in the equations. so i dunno how to do it so it relates to the given answers?
     
  2. jcsd
  3. Sep 20, 2010 #2

    phyzguy

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    Science Advisor

    Exponentiate both sides to get rid of the log, and you should get one of the answers.
     
  4. Sep 20, 2010 #3
    if i keep going from where i left off.. after integrating you get:

    ln(1+y2) = - ln(1+x2)/2

    then taking off log of both sides leaves

    (1+y2) = -(1+x2)/2

    take all onto one side gives:

    2(1+y2)+(1+x2)

    where did i go wrong?
     
  5. Sep 20, 2010 #4

    HallsofIvy

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    Science Advisor

    If you are given answers to select from, you don't need to integrate at all. Just put each of the possible answer into the equation and see if it satisfies it!

    "Taking the log off" both sides of [itex]ln(1+y^2)= ln(1+ x^2)/2[/itex] does NOT give "[itex]1+ y^2= (1+ x^2)/2[/itex]: the "2" in the denominator is not inside the logarithm. Instead, write it as [itex]ln(1+ y^2)= ln((1+ x^2)^{1/2})[/itex] so that [itex]1+ y^2= (1+ x^2)^{1/2}[/itex] or [itex](1+ y^2)^2= 1+ x^2[/itex].
     
  6. Sep 20, 2010 #5
    haha i have to show how i got every answer, mathematically. otheriwse yeah id trial and error the questions i couldnt do.

    the denominator there is where i thought i went wrong.

    i think i know the answer but its that one sign i cant get rid of.

    from the start:

    2y(1+x2)dy/dx+x(1+y2)=0
    2y(1+x2)dy/dx=-x(1+y2)
    2y/(1+y2)dy = -x/(1+x2)dx
    then integrate
    ln(1+y2) = - ln(1+x2)/2
    which becomes
    2ln(1+y2) = - ln(1+x2)
    take 2 and into logarithm
    ln((1+y2)2) = - ln(1+x2)

    waittt... ifi take -1 into logarithm then i get an inverse funtion, which will get rid of the plus/minus sign...

    then sub in y=2 and x=0


    Thanks!!!!!!
     
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