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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) $\displaystyle xy''+y'+K^{2}xy=0$
2) $\displaystyle y''+3\sqrt{x}y=0$
What do you mean "trial solutions"?
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.
May we ask how you came upon the term "trial solution"?
I'll try to find a solution of 1)... but I don't know if it is or not the 'trial solution'(Thinking)...Quote:
Originally Posted by cheme
Using t as independent variable the DE is...
$\displaystyle t\ y^{''} + y^{'} + K^{2}\ t\ y=0$ (1)
If $\displaystyle F(s)= \mathcal{L}\{f(t)\}$ is the Laplace Tranform of f(t), then a basic property is...
$\displaystyle \mathcal{L}\{t\ f(t)\}= -\frac{d}{d s} F(s)$ (2)
Now if we use the property (2) in (1) we obtain...
$\displaystyle -\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...
$\displaystyle (s^{2}+K^{2})\ Y^{'}(s) +s\ Y(s)=0$ (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...
$\displaystyle Y(s)= c\ e^{- \int \frac{s}{s^{2}+K^{2}}\ d s}= c\ e^{-\frac{1}{2}\ \ln (s^{2}+K^{2})}}= \frac{c}{\sqrt{s^{2}+K^{2}}}$ (5)
At this point if You have a look at the manual, You discover that is...
$\displaystyle y(t)= \mathcal{L}^{-1} \{\frac{c}{\sqrt{s^{2}+K^{2}}}\}= c\ J_{0}(K\ t)$ (6)
... where $\displaystyle J_{0}(*)$ 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
$\displaystyle \chi$ $\displaystyle \sigma$
Now we pass to...
$\displaystyle y^{''}+ \sqrt{x}\ y=0$ (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 $\displaystyle \sqrt{x}$, let's suppose that is $\displaystyle y(x)= \sqrt{x}\ g(x)$. In thast case is...
$\displaystyle y^{'}= g^{'}\ \sqrt{x}+ \frac{g}{2\ \sqrt{x}}$ (2)
$\displaystyle y^{''}= g^{''}\ \sqrt{x} + \frac{g^{'}}{\sqrt{x}} - \frac{g}{4 x \sqrt{x}}$ (3)
... and, inserting (3) in (1), we have...
$\displaystyle x^{2}\ g^{''} + x\ g^{'} + (x^{2}-\frac{1}{4})\ g=0$ (4)
The (4) is identified as a Bessel differential equation of order $\displaystyle \nu= \frac{1}{2}$ and its solution is...
$\displaystyle g(x)= c_{1}\ J_{\frac{1}{2}} (x) + c_{2}\ J_{-\frac{1}{2}}(x)$ (5)
... so that the solution of (1) is...
$\displaystyle y(x)=\sqrt{x}\ \{ c_{1}\ J_{\frac{1}{2}} (x) + c_{2}\ J_{-\frac{1}{2}}(x)\}$ (6)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
