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Simple 1st order ODE. Need insight

  1. Nov 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Solve the 1st order linear initial value problem.

    [tex](1+x^2)\frac{dy}{dx}\right) + 4xy = x [/tex] , [tex]y(1)=\frac{1}{4}\right)[/tex]

    3. The attempt at a solution

    Why can I get the answer solving the equation one way (below), but not they other?

    Method 1:
    [tex](1+x^2)\frac{dy}{dx}\right) = -x(4y-1)[/tex]

    [tex]\frac{dy}{dx}\right) * \frac{1}{4y-1}\right) = -\frac{x}{1+x^2}\right)[/tex]

    [tex]\frac{1}{4}\right) \int^{}_{} \frac{1}{4y-1}\right) d(4y-1) = -\frac{1}{2}\right) \int^{}_{} \frac{1}{1+x^2}\right d(1+x^2)[/tex]

    (multiply by 4)

    [tex]ln|4y-1| = -2ln|1+x^2| + C[/tex]

    [tex]ln|(4y-1)(1+x^2)^2| = 4C[/tex]

    [tex] 4y-1 = \frac{e^4^C}{(1+x^2)^2}\right)[/tex]

    Satisfying y(0)=1/4, C=0

    Therefore: y = 1/4


    Method 2:

    [tex](1+x^2)\frac{dy}{dx}\right) = -x(4y-1)[/tex]

    [tex]\frac{dy}{dx}\right) * \frac{1}{4y-1}\right) = -\frac{x}{1+x^2}\right)[/tex]

    [tex]\frac{1}{4}\right) \int^{y(x)}_{1/4} \frac{1}{4y-1}\right) d(4y-1) = -\frac{1}{2}\right) \int^{x}_{1} \frac{1}{1+x^2}\right d(1+x^2)[/tex]

    [tex]( ln|4y-1| - ln|0| ) = -2(ln|1+x^2| - ln|2|)[/tex]

    Here's where i'm confused. should I ignore ln0? or should I solve with infinity in my equation.
     
  2. jcsd
  3. Nov 19, 2009 #2
    Remember, don't divide by zero :)
     
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