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Math Help - Finding trial solutions of 2nd order homogeneous DE.

  1. #1
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    Finding trial solutions of 2nd order homogeneous DE.

    I am trying to find the trial solutions to the following differential equations. I would be able to solve them, but I am not sure how to go about simply determining trial solutions.

    1) xy''+y'+K^{2}xy=0

    2) y''+3\sqrt{x}y=0
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  2. #2
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    Re: Finding trial solutions of 2nd order homogeneous DE.

    What do you mean "trial solutions"?
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  3. #3
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    Re: Finding trial solutions of 2nd order homogeneous DE.

    That is what I don't understand. I guess no one will be able to assist on this. I thought trial solution was something specific for certain types of problems.
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    Re: Finding trial solutions of 2nd order homogeneous DE.

    May we ask how you came upon the term "trial solution"?
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    Re: Finding trial solutions of 2nd order homogeneous DE.

    Quote Originally Posted by cheme View Post
    I am trying to find the trial solutions to the following differential equations. I would be able to solve them, but I am not sure how to go about simply determining trial solutions.

    1) xy''+y'+K^{2}xy=0

    2) y''+3\sqrt{x}y=0
    I'll try to find a solution of 1)... but I don't know if it is or not the 'trial solution'...

    Using t as independent variable the DE is...

    t\ y^{''} + y^{'} + K^{2}\ t\ y=0 (1)

    If F(s)= \mathcal{L}\{f(t)\} is the Laplace Tranform of f(t), then a basic property is...

    \mathcal{L}\{t\ f(t)\}= -\frac{d}{d s} F(s) (2)
    Now if we use the property (2) in (1) we obtain...

    -\frac{d}{ds}\{s^{2}\ Y(s) -s\ y(0)-y^{'}(0)\}+ s\ Y(s) -y(0) -K^{2}\ \frac{d}{ds} Y(s)=0 (3)

    ... that with some steps becomes...

    (s^{2}+K^{2})\ Y^{'}(s) +s\ Y(s)=0 (4)

    ... so that we have now a first order linear DE in s, of course more 'approachable' then (1). The (4) can be solved with 'standard procedure' obtaining...

    Y(s)= c\ e^{- \int \frac{s}{s^{2}+K^{2}}\ d s}= c\ e^{-\frac{1}{2}\ \ln (s^{2}+K^{2})}}= \frac{c}{\sqrt{s^{2}+K^{2}}} (5)

    At this point if You have a look at the manual, You discover that is...

    y(t)= \mathcal{L}^{-1} \{\frac{c}{\sqrt{s^{2}+K^{2}}}\}= c\ J_{0}(K\ t) (6)

    ... where J_{0}(*) is the Bessel function of first kind of order 0. Only one consideation: in (6) we have only one 'arbitrary constant' c even if the (1) is a linear ODE of order two. The [probable] reason is that the procedure finds only the solutions of (1) that are L-transformable, so that the other independent solution, probably with a singularity in t=0, is 'desaparecida'...

    Kind regards

    \chi \sigma
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  6. #6
    MHF Contributor chisigma's Avatar
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    Re: Finding trial solutions of 2nd order homogeneous DE.

    Now we pass to...

    y^{''}+ \sqrt{x}\ y=0 (1)

    ... which is a little modified respect to the ODE 2) in the original post [1 instead of 3...]. Here we only want to show how to proceed in cases like this...

    In order to eliminate the 'unconfortable' term \sqrt{x}, let's suppose that is y(x)= \sqrt{x}\ g(x). In thast case is...

    y^{'}= g^{'}\ \sqrt{x}+ \frac{g}{2\ \sqrt{x}} (2)

    y^{''}= g^{''}\ \sqrt{x} + \frac{g^{'}}{\sqrt{x}} - \frac{g}{4 x \sqrt{x}} (3)

    ... and, inserting (3) in (1), we have...

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

    The (4) is identified as a Bessel differential equation of order \nu= \frac{1}{2} and its solution is...

    g(x)= c_{1}\ J_{\frac{1}{2}} (x) + c_{2}\ J_{-\frac{1}{2}}(x) (5)

    ... so that the solution of (1) is...

    y(x)=\sqrt{x}\ \{ c_{1}\ J_{\frac{1}{2}} (x) + c_{2}\ J_{-\frac{1}{2}}(x)\} (6)

    Kind regards

    \chi \sigma
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