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Math Help - Can't find my mistake in dif equation

  1. #1
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    Can't find my mistake in dif equation

    I am trying to solve this equation for 2 days and I can't do it.... I solved it but wolfram alpha gives different answers. Please help me to find my mistake
    Can't find my mistake in dif equation-mtk.png - this is my solution. y = c_1x^3-\dfrac{1}{2x}+c_2 - Wolfram alpha answer
    Last edited by darjiaus7; May 17th 2014 at 10:06 AM.
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  2. #2
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    Re: Can't find my mistake in dif equation

    in wolfram alpha input u have given rhs as -2/x^3 while in ur soln u have computed for -2/x^2
    Thanks from topsquark
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  3. #3
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    Re: Can't find my mistake in dif equation

    xy''-y'=-\dfrac{2}{x^2}

    The integrating factor is \dfrac{1}{x^3}, so multiply both sides by that:

    \dfrac{1}{x^2}y'' - \dfrac{1}{x^3}y' = -\dfrac{2}{x^5}

    \left(\dfrac{y'}{x^2}\right)^\prime = -\dfrac{2}{x^5}

    Integrating both sides:

    \dfrac{y'}{x^2} = -\int \dfrac{2}{x^5}dx = \dfrac{1}{2x^4}+C_1

    Multiply both sides by x^2:

    y' = \dfrac{1}{2x^2}+C_1x^2

    Integrate both sides:

    y = -\dfrac{1}{2x}+\dfrac{C_1x^3}{3}+C_2

    Fold the \dfrac{C_1}{3} into a single constant:

    y = c_1x^3-\dfrac{1}{2x}+c_2

    When you entered the problem into WolframAlpha, you entered xy''-2y' = -2/x^3. Notice the RHS of that differential equation has x^3 in the denominator. Your original problem has x^2 in the denominator.
    Thanks from darjiaus7 and topsquark
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  4. #4
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    Re: Can't find my mistake in dif equation

    You are correct that the general solution to the associated homogeneous equation is y= C_1x^3+ C_2 but you are doing the "variation of parameters" incorrectly.

    Yes, we look for a solution to the entire equation of the form y= C_1(x)x^3+ C_2(x), allowing the "parameters", C_1 and C_2 to "vary".

    Then we have y'= C_1'x^3+ 3C_1x^2+ C_2'. We simplify the equation (there are many different possible functions for C_1 and C_2) by requiring that C_1'x^3+ C_2'= 0 leaving y'= 3C_1x^2. Differentiating again, y''= 3C_1'x^2+ 6C_1x and now the differential equation becomes xy''- 2y'= 3C_1'x^3+ 6C_1x^2- 6C_1x^2= 3C_1'x^3= -\frac{2}{x^2}.

    There is your error- you have, although you don't say how you arrived at it, x^2 on the left rather than x^3.
    Thanks from darjiaus7 and topsquark
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  5. #5
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    Re: Can't find my mistake in dif equation

    Can you explaint this part? And why you wrote y' instead of 2y' and got right answer?

    \dfrac{1}{x^2}y'' - \dfrac{1}{x^3}y' = -\dfrac{2}{x^5}

    \left(\dfrac{y'}{x^2}\right)^\prime = -\dfrac{2}{x^5}
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  6. #6
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    Re: Can't find my mistake in dif equation

    Sorry I can't understand integrating factor method, can you say where is error in my solution? ;/
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  7. #7
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    Re: Can't find my mistake in dif equation

    Quote Originally Posted by darjiaus7 View Post
    Can you explaint this part? And why you wrote y' instead of 2y' and got right answer?

    \dfrac{1}{x^2}y'' - \dfrac{1}{x^3}y' = -\dfrac{2}{x^5}

    \left(\dfrac{y'}{x^2}\right)^\prime = -\dfrac{2}{x^5}
    HallsofIvy explained the error with your solution. But, you are correct that I wrote the problem incorrectly. Check post #4 above for HallsofIvy's explanation of where you went wrong.
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  8. #8
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    Re: Can't find my mistake in dif equation

    Ok forget about my solution Now can you explain how \dfrac{1}{x^2}y'' - \dfrac{1}{x^3}y' became \left(\dfrac{y'}{x^2}\right)^\prime
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  9. #9
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    Re: Can't find my mistake in dif equation

    this is easy use product rule of differentiation d(fg)=fd(g)+gd(f)
    here f=1/(x^2) and g=y'
    Thanks from darjiaus7
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