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Thread: y''= (2/x)y' - (2/x^2)y - 1/x^2

  1. #1
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    Question y''= (2/x)y' - (2/x^2)y - 1/x^2

    Hey everyone, I need help with this past exam paper as I haven't been given any solutions to it so I am unsure how to find the solutions to these problems, so if someone could please show me a stepy by step answer

    Consider the following initial value problem:

    y''= (2/x)y' - (2/x^2)y - 1/x^2, y(1)=0, y'(1)=1

    By observing that the first two terms on the right-hand side of the equation form a total derivative of a function, find the analytical solution of the problem.
    Last edited by mr fantastic; Apr 3rd 2011 at 12:00 PM. Reason: Re-titled.
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  2. #2
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    Well do what the question suggests...

    $\displaystyle \displaystyle \frac{d^2y}{dx^2} = \frac{2}{x}\,\frac{dy}{dx} - \frac{2}{x^2}\,y - \frac{1}{x^2}$

    $\displaystyle \displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{2}{x}\,y\right) - \frac{1}{x^2}$.


    Now integrate both sides...
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  3. #3
    MHF Contributor chisigma's Avatar
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    The DE can be written as...

    $\displaystyle x^{2} y^{''} - 2 x y^{'} + 2 y = 1 $ (1)

    ... so that it is 'Euler's type'. The solution of the 'incomplete' DE is...

    $\displaystyle y= c_{1} t^{\alpha_{1}} + c_{2} t^{\alpha_{2}}$ (2)

    ... where $\displaystyle \alpha_{1}$ and $\displaystyle \alpha_{2}$ are the solutions of the 'characteristic equation'...

    $\displaystyle \alpha\ (\alpha-1) -2 \alpha + 2=0$ (3)

    ... i.e. $\displaystyle \alpha_{1}= -1$ , $\displaystyle \alpha_{2}= -2$. A solution of the 'complete DE' is $\displaystyle y= \frac{x^{2}}{4} + \frac{1}{2}$ so that the general solution of the 'complete DE' is...

    $\displaystyle \displaystyle y= \frac{c_{1}} {x} + \frac{c_{2}}{x^{2}} + \frac{x^{2}}{4} + \frac{1}{2}$ (4)

    The value of $\displaystyle c_{1}$ and $\displaystyle c_{2}$ can be derived from the 'initial conditions'...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  4. #4
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    Quote Originally Posted by chisigma View Post
    The DE can be written as...

    $\displaystyle x^{2} y^{''} - 2 x y^{'} + 2 y = 1 $ (1)

    ... so that it is 'Euler's type'. The solution of the 'incomplete' DE is...

    $\displaystyle y= c_{1} t^{\alpha_{1}} + c_{2} t^{\alpha_{2}}$ (2)

    ... where $\displaystyle \alpha_{1}$ and $\displaystyle \alpha_{2}$ are the solutions of the 'characteristic equation'...

    $\displaystyle \alpha\ (\alpha-1) -2 \alpha + 2=0$ (3)

    ... i.e. $\displaystyle \alpha_{1}= -1$ , $\displaystyle \alpha_{2}= -2$. A solution of the 'complete DE' is $\displaystyle y= \frac{x^{2}}{4} + \frac{1}{2}$ so that the general solution of the 'complete DE' is...

    $\displaystyle \displaystyle y= \frac{c_{1}} {x} + \frac{c_{2}}{x^{2}} + \frac{x^{2}}{4} + \frac{1}{2}$ (4)

    The value of $\displaystyle c_{1}$ and $\displaystyle c_{2}$ can be derived from the 'initial conditions'...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    As useful as this post is, it doesn't follow the method the OP was asked to follow...
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  5. #5
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    Thanks a lot!!
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