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Math Help - O.D.E. Problem

  1. #1
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    O.D.E. Problem

    I need some help on knowing how to tackle this problem. I missed the class where we discussed this so i'm a bit clueless on how to do it .


    One solution of the differential equation

    (4x^2)y'' +(4x)y' + ((4x^2) -1)y = 0

    is y1(x) = x^(-1/2)sinx. Find a second independent solution y2(x)


    Any help is appreciated
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  2. #2
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    you tired using the wronskian?

    Bobak
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  3. #3
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    have no idea what that is but i'm gonna look into it
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  4. #4
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    here is a quick summary of things you need to know about it.

    for a second order equation W(x) = \left|\begin{array}{cc}y_{1} & y_{2} \\y'_{1} & y'_{2} \end{array}\right| = y_{1} y'_{2} - y_{2}y'_{1}

    if you write your differential equation in the form y'' + p(x) y' + q(x) y = 0 then you can show that W'(x) + p(x)W = 0 so W(x) = e^{ - \int p(x) dx}

    so given y_{1} you can contruct a first order differential equation in  y_{2}

    Bobak
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  5. #5
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    Oh wow that is so helpful . I'll give it a try and report back~
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  6. #6
    MHF Contributor chisigma's Avatar
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    Let's write the equation as...

    y^{''}+ \frac{y^{'}}{x} + (1-\frac{1}{4\cdot x^{2}})\cdot y=0 (1)

    The general solution of (1) is...

    y(x)= c_{1}\cdot u(x) + c_{2}\cdot v(x) (2)

    ... where u(*) and v(*) are two linearly independent sulution of (1), c_{1} and c_{2} two arbitrary constants.

    If u(*) and v(*) are both solution of (1) then...

    u^{''}+ \frac{u^{'}}{x} + (1-\frac{1}{4\cdot x^{2}})\cdot u=0

    v^{''}+ \frac{v^{'}}{x} + (1-\frac{1}{4\cdot x^{2}})\cdot v=0 (3)

    If we multiply the first equation by v and the second by u and take the difference we obtain...

    x\cdot (u^{''}\cdot v - v^{''}\cdot u) + u^{'}\cdot v - v^{'}\cdot u = \frac {d}{dx} [x\cdot (u^{'}\cdot v - v^{'}\cdot u)]=0 (4)

    The solution of (4) is 'very easy'...

    x\cdot (u^{'}\cdot v - v^{'}\cdot u)= c_{2} (5)

    ... where c_{2} is an arbitrary constant. Deviding both terms of (5) by x\cdot v^{2} we obtain...

    \frac{u^{'}\cdot v - v^{'}\cdot u}{v^{2}}= \frac{d}{dx} \frac{u}{v}= \frac{c_{2}}{x\cdot v^{2}} (6)

    Also the (6) is relatively easy to solve and we obtain...

    \frac{u}{v}= c_{1} + c_{2}\cdot \int \frac{dx}{x\cdot v^{2}} (7)

    ... and from it...

    u=c_{1}\cdot v + c_{2}\cdot v\cdot \int \frac{dx}{x\cdot v^{2}} (8)

    The (8) allows us to obtain, if you know a solution v(*) of the (1), an other solution u(*) independent from it as follows...

    u= v \cdot \int \frac{dx}{x\cdot v^{2}} (9)

    In our case is...

     v(x)= \frac{\sin x}{\sqrt x} (10)

    ... so that what we have to do now is to compute the integral in (9) and then find u(*)

    Kind regards

    \chi \sigma
    Last edited by chisigma; April 2nd 2009 at 12:19 AM.
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  7. #7
    MHF Contributor chisigma's Avatar
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    Some useful comments… remembering the so called ‘Bessel equation’ …

    x^{2}\cdot y^{''} + x \cdot y^{'} + (x^{2}- n^{2})\cdot y = 0 (1)

    … it is easy to verify that the proposed ODE is the Bessel equation with n=\frac{1}{2}, the solution of which is …

    y(x)= c_{1}\cdot J_{\frac{1}{2}} (x) + c_{2}\cdot J_{-\frac{1}{2}} (x) (2)

    With a little of patience you can find that is…

    J_{\frac{1}{2}} (x)= \frac {\sin x}{\sqrt x}

    J_{-\frac{1}{2}}(x) = \frac{\cos x}{\sqrt x} (3)

    … functions that are represented here…




    Setting J_{- \frac{1}{2}} (x)= u(x) and  J_{\frac{1}{2}} (x)= v(x) you can verify that is [unless the sign] …

    u(x) = v(x)\cdot \int \frac{dx}{x\cdot v^{2}(x)}

     v(x) = u(x)\cdot \int \frac{dx}{x\cdot u^{2}(x)} (4)

    In is interesting to note that, if u(*) and v(*) are linearly independent solution of the equation…

    y^{''} + \frac{y^{'}}{x} + g(x)\cdot y=0 (5)

    … with g(*) an arbitrary function of x, the relations (4) hold in any case…

    Kind regards

    \chi \sigma
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