logistic_guy
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Solve.
\(\displaystyle x^2y'' + xy' + \left(x^2 - \frac{9}{4}\right)y = 0\)
\(\displaystyle x^2y'' + xy' + \left(x^2 - \frac{9}{4}\right)y = 0\)
Bessel's equation - 3
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logistic_guy
Senior Member
- Joined
- Apr 17, 2024
- Messages
- 2,214
If you followed the first thread of Bessel's equation, you would say instantly that the general solution to this differential equation is:
\(\displaystyle y(x) = c_1J_{3/2}(x) + c_2J_{-3/2}(x)\)
You are right, but most of us are simple people and we still don't understand what the heck \(\displaystyle J\) means! Therefore, you need to write the solution in terms of trigonometric functions or anything else that we can understand.
You are very lucky buddy as soon you will understand everything about \(\displaystyle J\) or \(\displaystyle J_n\). For now it is enough that you know that \(\displaystyle J\) is called Bessel's function of the first kind and when we write it like this \(\displaystyle J_n\) it is called Bessel's function of the first kind of order \(\displaystyle n\).
Now I will give you a magical recurrence relation for Bessel functions that will help you a lot to solve tons of Bessel's equations. And I promise you that the proof of this relation will be given in another Episode. For now just mimic how to use it!
The recurrence relation is:
\(\displaystyle 2nJ_n(x) = xJ_{n+1}(x) + xJ_{n-1}(x)\)
Let us try to plug \(\displaystyle n = \frac{1}{2}\) and see what we get.
\(\displaystyle 2\frac{1}{2}J_{1/2}(x) = xJ_{1/2+1}(x) + xJ_{1/2-1}(x)\)
\(\displaystyle J_{1/2}(x) = xJ_{3/2}(x) + xJ_{-1/2}(x)\)
This gives:
\(\displaystyle J_{3/2}(x) = \frac{1}{x}J_{1/2}(x) - J_{-1/2}(x)\)
From previous lessons, we know that:
\(\displaystyle J_{1/2}(x) = \sqrt{\frac{2}{\pi x}}\sin x\)
And
\(\displaystyle J_{-1/2}(x) = \sqrt{\frac{2}{\pi x}}\cos x\)
Then
\(\displaystyle J_{3/2}(x) = \frac{1}{x}\sqrt{\frac{2}{\pi x}}\sin x - \sqrt{\frac{2}{\pi x}}\cos x\)
If we do the same steps but with \(\displaystyle n = -\frac{1}{2}\), we get:
\(\displaystyle J_{-3/2}(x) = -\sqrt{\frac{2}{\pi x}}\sin x - \frac{1}{x}\sqrt{\frac{2}{\pi x}}\cos x\)
Then, the general solution in terms of trigonometric functions is:
\(\displaystyle y(x) = c_1\left(\frac{1}{x}\sqrt{\frac{2}{\pi x}}\sin x - \sqrt{\frac{2}{\pi x}}\cos x\right) + c_2\left(-\sqrt{\frac{2}{\pi x}}\sin x - \frac{1}{x}\sqrt{\frac{2}{\pi x}}\cos x\right)\)
Or
\(\displaystyle y(x) = \left(C_1\frac{\sin x}{x^{3/2}} + C_3\frac{\cos x}{\sqrt{x}}\right) + \left(C_2\frac{\sin x}{\sqrt{x}} + C_4\frac{\cos x}{x^{3/2}}\right)\)
Or
\(\displaystyle y(x) = C_1\frac{\sin x}{x^{3/2}} + C_2\frac{\sin x}{\sqrt{x}} + C_3\frac{\cos x}{\sqrt{x}} + C_4\frac{\cos x}{x^{3/2}}\)
Welcome to the world of differential equations
\(\displaystyle y(x) = c_1J_{3/2}(x) + c_2J_{-3/2}(x)\)
You are right, but most of us are simple people and we still don't understand what the heck \(\displaystyle J\) means! Therefore, you need to write the solution in terms of trigonometric functions or anything else that we can understand.
You are very lucky buddy as soon you will understand everything about \(\displaystyle J\) or \(\displaystyle J_n\). For now it is enough that you know that \(\displaystyle J\) is called Bessel's function of the first kind and when we write it like this \(\displaystyle J_n\) it is called Bessel's function of the first kind of order \(\displaystyle n\).
Now I will give you a magical recurrence relation for Bessel functions that will help you a lot to solve tons of Bessel's equations. And I promise you that the proof of this relation will be given in another Episode. For now just mimic how to use it!
The recurrence relation is:
\(\displaystyle 2nJ_n(x) = xJ_{n+1}(x) + xJ_{n-1}(x)\)
Let us try to plug \(\displaystyle n = \frac{1}{2}\) and see what we get.
\(\displaystyle 2\frac{1}{2}J_{1/2}(x) = xJ_{1/2+1}(x) + xJ_{1/2-1}(x)\)
\(\displaystyle J_{1/2}(x) = xJ_{3/2}(x) + xJ_{-1/2}(x)\)
This gives:
\(\displaystyle J_{3/2}(x) = \frac{1}{x}J_{1/2}(x) - J_{-1/2}(x)\)
From previous lessons, we know that:
\(\displaystyle J_{1/2}(x) = \sqrt{\frac{2}{\pi x}}\sin x\)
And
\(\displaystyle J_{-1/2}(x) = \sqrt{\frac{2}{\pi x}}\cos x\)
Then
\(\displaystyle J_{3/2}(x) = \frac{1}{x}\sqrt{\frac{2}{\pi x}}\sin x - \sqrt{\frac{2}{\pi x}}\cos x\)
If we do the same steps but with \(\displaystyle n = -\frac{1}{2}\), we get:
\(\displaystyle J_{-3/2}(x) = -\sqrt{\frac{2}{\pi x}}\sin x - \frac{1}{x}\sqrt{\frac{2}{\pi x}}\cos x\)
Then, the general solution in terms of trigonometric functions is:
\(\displaystyle y(x) = c_1\left(\frac{1}{x}\sqrt{\frac{2}{\pi x}}\sin x - \sqrt{\frac{2}{\pi x}}\cos x\right) + c_2\left(-\sqrt{\frac{2}{\pi x}}\sin x - \frac{1}{x}\sqrt{\frac{2}{\pi x}}\cos x\right)\)
Or
\(\displaystyle y(x) = \left(C_1\frac{\sin x}{x^{3/2}} + C_3\frac{\cos x}{\sqrt{x}}\right) + \left(C_2\frac{\sin x}{\sqrt{x}} + C_4\frac{\cos x}{x^{3/2}}\right)\)
Or
\(\displaystyle y(x) = C_1\frac{\sin x}{x^{3/2}} + C_2\frac{\sin x}{\sqrt{x}} + C_3\frac{\cos x}{\sqrt{x}} + C_4\frac{\cos x}{x^{3/2}}\)
Welcome to the world of differential equations
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