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Solving a differential equation with intial conditions, please help

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

    Here is the original thing:

    (x[itex]^{2}[/itex]+1)y'+4x(y-1)=0, y(0)=4

    2. Relevant equations



    3. The attempt at a solution

    I thought I knew the procedure.. but I got it wrong. Can someone let me know where I went wrong?

    First I rearrange the equation to get the following in the form y'+p(x)y=g(x)

    y'+([itex]\frac{4x}{x^{2}+1}[/itex])y=[itex]\frac{4x}{x^{2}+1}[/itex]

    So I get the integrating factor to be

    I(x)=e[itex]^{\int\frac{4x}{x^{2}+1}}[/itex]

    That integral comes out to be 2ln(x[itex]^{2}[/itex]+1), which when raised to e, the integrating factor simply becomes

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

    So the new differential equation is:

    d/dx (x[itex]^{2}[/itex]+1)[itex]^{2}[/itex]y=[itex]\frac{4x}{x^{2}+1}[/itex]

    integrating both sides gets me

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

    solving for y:

    y=[itex]\frac{2ln(x^{2}+1)}{(x^{2}+1)^{2}}[/itex]+[itex]\frac{C}{(x^{2}+1)^{2}}[/itex]

    Taking the intial conditions into account I get C=4. When I input this equation into my assignment online, it says I'm wrong.

    Please help, I thought I knew this process pretty well..
     
  2. jcsd
  3. Feb 14, 2012 #2

    vela

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    You forgot to multiply the righthand side by the integrating factor.
     
  4. Feb 14, 2012 #3

    HallsofIvy

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    So you are going to multiply both sides of the equation by that?

    Did you not multiply the right side by [itex](x^2+ 1)^2[/itex]?

     
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