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Math Help - Problem in defferential equation

  1. #1
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    Problem in defferential equation

    1. Find the general solution of linear differential equation

    dy/dx y/(x+1) = x


    2. Given that (x) dy/dx 2y = x^3 ln x, find y in terms of x such that y=2 at x=1.

    Please help me to solve these.
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  2. #2
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    Quote Originally Posted by geton View Post
    1. Find the general solution of linear differential equation

    dy/dx y/(x+1) = x


    2. Given that (x) dy/dx 2y = x^3 ln x, find y in terms of x such that y=2 at x=1.

    Please help me to solve these.
    1. Use the integrating factor method. I.F. = \displaystyle e^{-\int \frac{1}{x+1} \, dx} = e^{-\ln|x + 1|} = e^{\ln \frac{1}{|x + 1|}} = \frac{1}{x + 1}.

    ------------------------------------------------------------------------------------

    2. Divide through by x. Then use the integrating factor method. I.F. = \displaystyle e^{-2 \int \frac{1}{x} \, dx} = e^{-2 \ln|x|} = e^{\ln \frac{1}{x^2}} = \frac{1}{x^2}.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    1. Use the integrating factor method. I.F. = \displaystyle e^{-\int \frac{1}{x+1} \, dx} = e^{-\ln|x + 1|} = e^{\ln \frac{1}{|x + 1|}} = \frac{1}{x + 1}.

    ------------------------------------------------------------------------------------

    2. Divide through by x. Then use the integrating factor method. I.F. = \displaystyle e^{-2 \int \frac{1}{x} \, dx} = e^{-2 \ln|x|} = e^{\ln \frac{1}{x^2}} = \frac{1}{x^2}.

    1. Multiplying by (x+1)^-1 gives:

    (x+1)^-1 dy/dx - y (x+1)^-2 = x(x+1)^-1

    y(x+1)^1 d/dx = ∫ x(x+1)^-1

    How can I integrated this?



    2. After Integrating:

    yx^-2 = x(ln x - 1) + C

    Since y=2 at x=1 this gives:

    C=3

    That is: yx^-2 = x(ln x - 1) + 3

    So y =x^3 (ln x - 1) + 3 x^2


    Is it right?
    Last edited by geton; January 25th 2008 at 09:56 PM.
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  4. #4
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    Quote Originally Posted by geton View Post
    [snip]
    ∫ x(x+1)^-1

    How can I integrated this?
    [snip]
    \frac{x}{x + 1} = \frac{(x + 1) - 1}{x + 1} = 1 - \frac{1}{x + 1}
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  5. #5
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    Thank you mr fantastic. But is my 2nd problem's solution right?
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  6. #6
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    Yes I found my mistake on 2nd problem

    Problem resolved.
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  7. #7
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    Quote Originally Posted by geton View Post
    Yes I found my mistake on 2nd problem

    Problem resolved.
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