Bessel's equation

logistic_guy said:
Solve.

\(\displaystyle x^2y'' + xy' + \left(x^2 - \frac{1}{4}\right)y = 0\)
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The art of solving differential equations lies mostly in intuition. The title of this thread has already given us the most important part to solve a DE. The recognition of the type of the differential equation is the most difficult part, especially when the equation is too complicated. But when you are told that it is Bessel's equation, it is like you were given half the answer!

The general form of Bessel's equation is:

\(\displaystyle x^2y'' + xy' + \left(x^2 - n^2\right)y = 0\)

where \(\displaystyle n\) is the order of Bessel's equation.

The most two famous general solutions to this equation are:

\(\displaystyle y(x) = c_1J_n(x) + c_2J_{-n}(x)\)

\(\displaystyle y(x) = c_1J_n(x) + c_2Y_n(x)\)

Which one to choose depends on \(\displaystyle n\)! If \(\displaystyle n\) is integer, we write the second solution. If \(\displaystyle n\) is not integer, it is the first solution.

We will try to write the original differential equation in the form of the general equation.

\(\displaystyle x^2y'' + xy' + \left(x^2 - \left[\frac{1}{2}\right]^2\right)y = 0\)

If we compare this with what we know, we find that \(\displaystyle n = \frac{1}{2}\). Not integer. Then the solution is:

\(\displaystyle y(x) = c_1J_{1/2}(x) + c_2J_{-1/2}(x)\)

But what the heck does \(\displaystyle J\) mean? Many don't know, so we will try to write the solution with something everyone understand. It happens that:

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

And

\(\displaystyle J_{-1/2}(x) = \sqrt{\frac{2}{\pi x}}\cos x\)

But why? The answer to this question will be given in another Episode. For now just mimic the solutions.

The solution to the original differential equation is then:

\(\displaystyle y(x) = c_1\sqrt{\frac{2}{\pi x}}\sin x + c_2\sqrt{\frac{2}{\pi x}}\cos x\)

Or

\(\displaystyle y(x) = C_1\frac{\sin x}{\sqrt{x}} + C_2\frac{\cos x}{\sqrt{x}}\)
 
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