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Thread: 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. $\displaystyle y = c_1x^3-\dfrac{1}{2x}+c_2$ - Wolfram alpha answer
    Last edited by darjiaus7; May 17th 2014 at 09:06 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
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  3. #3
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    Re: Can't find my mistake in dif equation

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

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

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

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

    Integrating both sides:

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

    Multiply both sides by $\displaystyle x^2$:

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

    Integrate both sides:

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

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

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

    When you entered the problem into WolframAlpha, you entered $\displaystyle xy''-2y' = -2/x^3$. Notice the RHS of that differential equation has $\displaystyle x^3$ in the denominator. Your original problem has $\displaystyle x^2$ 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 $\displaystyle y= C_1x^3+ C_2$ but you are doing the "variation of parameters" incorrectly.

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

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

    There is your error- you have, although you don't say how you arrived at it, $\displaystyle x^2$ on the left rather than $\displaystyle x^3$.
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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?

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

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

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

    $\displaystyle \left(\dfrac{y'}{x^2}\right)^\prime = -\dfrac{2}{x^5}$
    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 $\displaystyle \dfrac{1}{x^2}y'' - \dfrac{1}{x^3}y' $ became $\displaystyle \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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