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Math Help - differential equation

  1. #1
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    differential equation

    dy/dx - y/(x+1) = x y(1)= 0, x>0
    integral of [Ae^-ln(x+1)(x)]/[Ae^-ln(x+1)]

    y= x/(x+1) - ln(x+1)/(x+1) + c/(x+1)

    0= 1/2 -(ln2)/2 + C/3

    y= x/(x+1) -ln(x+1)/(x+1) + (-3+ln2)/2(x+1)

    is the answer right?
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  2. #2
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    Quote Originally Posted by twilightstr View Post
    dy/dx - y/(x+1) = x y(1)= 0, x>0
    integral of [Ae^-ln(x+1)(x)]/[Ae^-ln(x+1)]

    y= x/(x+1) - ln(x+1)/(x+1) + c/(x+1)

    0= 1/2 -(ln2)/2 + C/3

    y= x/(x+1) -ln(x+1)/(x+1) + (-3+ln2)/2(x+1)

    is the answer right?
    You can always check your answer by substituting it into the differential equation and seeing if it works.
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  3. #3
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    Quote Originally Posted by twilightstr View Post
    dy/dx - y/(x+1) = x y(1)= 0, x>0
    integral of [Ae^-ln(x+1)(x)]/[Ae^-ln(x+1)]

    y= x/(x+1) - ln(x+1)/(x+1) + c/(x+1)

    0= 1/2 -(ln2)/2 + C/3

    y= x/(x+1) -ln(x+1)/(x+1) + (-3+ln2)/2(x+1)

    is the answer right?
    No.

    \frac{dy}{dx} - \frac{1}{x + 1}y = x

    The integrating factor is e^{\int{-\frac{1}{x+1}\,dx}} = e^{-\ln{(x + 1)}} = e^{\ln{(x + 1)^{-1}}} = (x + 1)^{-1}.

    So multiply both sides of the equation by (x + 1)^{-1} and you get

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

    \frac{d}{dx}[(x + 1)^{-1}y] = 1 - \frac{1}{x + 1}.


    Can you go from here?
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  4. #4
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    It is even easier to see that that y does not satisfy y(1)= 0!
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  5. #5
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    thanks but do i have the answer now:

    y= x^2-xln(x+1)+(-1 + ln2)x +x-ln(x+1)-1+ln2
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  6. #6
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    Quote Originally Posted by twilightstr View Post
    thanks but do i have the answer now:

    y= x^2-xln(x+1)+(-1 + ln2)x +x-ln(x+1)-1+ln2
    Let's see...

    \frac{d}{dx}[(x + 1)^{-1}y = 1 - \frac{1}{x + 1}

    (x + 1)^{-1}y = x - \ln{|x + 1|} + C

    y = (x + 1)(x - \ln{|x + 1|} + C).

    We're told y(1) = 0 so

    0 = (1 + 1)(1 - \ln{|1 + 1|} + C)

    0 = 2(1 - \ln{2} + C)

    0 = 1 - \ln{2} + C

    C = \ln{2} - 1.


    So  y = (x + 1)(x - \ln{|x + 1|} + \ln{2} - 1).
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