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Math Help - initial value DE problem

  1. #1
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    initial value DE problem

    dy/dx= (x+1)y/2x^2 where y(1)=1, for all x>=1

    i ended up with

    y^2/2= 1/2 ln (2x^2)+1/4 tan-1 (x/2) + K


    anythoughts?
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  2. #2
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    I think this DE is separable

     \frac{dy}{dx}= \frac{(x+1)y}{2x^2}

     \frac{dy}{y}= \frac{(x+1)}{2x^2}~dx

     \int\frac{1}{y}~dy= \int \frac{(x+1)}{2x^2}~dx

     \int\frac{1}{y}~dy= \int \frac{x}{2x^2}+\frac{1}{2x^2}~dx

     \int\frac{1}{y}~dy= \int \frac{1}{2x}+\frac{1}{2x^2}~dx

    Did you use this method?
    Last edited by pickslides; December 15th 2009 at 03:20 PM. Reason: bad latex
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  3. #3
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    i did it that way before and came up with:

    ln [y] = ln [2x] + 1/4 tan-1 (x/2) + K

    still doesnt seem right
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  4. #4
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    I think that solution is better, from here raise each side of the equation to the power of e . This will eliminate the \ln(y) on the left hand side.

    Before this, may I ask how you arrived at the \tan^{-1}(x) term? I can't see this being correct.
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  5. #5
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    table of integrals du / a^2 + u^2 = 1/a tan-1 u/a + C
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  6. #6
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    I see, I'm not sure if you have applied that the correct way here.

    consider

     \int\frac{1}{y}~dy= \int \frac{1}{2x}+\frac{1}{2x^2}~dx

     \ln(y)= \int \frac{1}{2x}~dx+\int \frac{x^{-2}}{2}~dx

     \ln(y)= \frac{\ln(x)}{2}- \frac{x^{-1}}{2}+C
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  7. #7
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    Smile

    i definitely overcomplicated that one obviously

    cheers
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