Finding trial solutions of 2nd order homogeneous DE.

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September 25th 2011, 10:30 AM
cheme

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) $y''+3\sqrt{x}y=0$
September 25th 2011, 02:50 PM
Jester

Re: Finding trial solutions of 2nd order homogeneous DE.

What do you mean "trial solutions"?
September 25th 2011, 06:37 PM
cheme

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.
September 25th 2011, 07:52 PM
Jester

Re: Finding trial solutions of 2nd order homogeneous DE.

May we ask how you came upon the term "trial solution"?
September 26th 2011, 01:38 AM
chisigma

Re: Finding trial solutions of 2nd order homogeneous DE.

Quote:

Originally Posted by cheme

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'(Thinking)...

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

$\chi$ $\sigma$
September 26th 2011, 08:15 PM
chisigma

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

$\chi$ $\sigma$