Bessel's differential equation

iolantham

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Dec 12, 2009
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Hello everyone!

I'm trying to solve \(\displaystyle y"+x^{2} y=0\), given \(\displaystyle y=u\sqrt{x}\) and \(\displaystyle \frac{x^{2}}{2}=z\). When I replace y with \(\displaystyle u\sqrt{x}\) I get \(\displaystyle u"+\frac{u'}{x} +u (x^{2}-\frac{1}{4x^{2}})\). The only difference to a Bessel function would be the term multiplying u: \(\displaystyle x^{2}-\frac{1}{(4x^{2})}\). If I replace just that one by z, I get \(\displaystyle (2z-\frac{1}{8z})\). If I factor 2 out I get \(\displaystyle 2(z-\frac{1}{16z})\). This still doesn't look like Bessel's equation to me because it's not in the form \(\displaystyle 1-\frac{n^{2}}{z}\) (thinking that z is a function of \(\displaystyle x^{2}\)). Is it Bessel's equation?
Further, \(\displaystyle u(x)=C_{1} J_{n}(x)+C_{2}J_{n}(x)\) and then \(\displaystyle y(x)=\sqrt{x}[C_{1}J_{n}(z)+C_{2}J_{n}(z)]=\sqrt{x}[C_{1}J_{n}(\frac{x^{2}}{2})+C_{2}J_{n}(\frac{x^{2}}{2})]\). Why is n=1/4?
Thank you in advance for looking and for your help.
 
That still didn't answer my question. The term before the function u is not in the form of 1-n[sup:wkf9ss5s]2[/sup:wkf9ss5s]/x[sup:wkf9ss5s]2[/sup:wkf9ss5s], so it doesn't look like Bessel's function. Then how do you get n=1/4?
Thanks for your reply.
 
check your y" term carefully. Please show us - how did you calculate that. I don't get same terms.
 
iolantham said:
Hello everyone!
I get u"+u'/x +u (x[sup:2f2rptnw]2[/sup:2f2rptnw]-1/(4x[sup:2f2rptnw]2[/sup:2f2rptnw])).... If I replace just that one by z, I get (2z-1/(8z)).

When you replace x^2/2 by z, you should also replace your du/dx by something involving du/dz. This will give you a completely different equation from the one you got.
 
Thank you, but my boyfriend has already solved the exercise for me before you suggested the same method. Have a nice day!
 
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