Bessel's equation - 4

logistic_guy

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Solve.

x2y+xy+(x2254)y=0\displaystyle x^2y'' + xy' + \left(x^2 - \frac{25}{4}\right)y = 0
 
Solve.

x2y+xy+(x2254)y=0\displaystyle x^2y'' + xy' + \left(x^2 - \frac{25}{4}\right)y = 0
We all now can solve this differential equation quickly. With full of confidence we can say that the general solution is:

y(x)=c1J5/2(x)+c2J5/2(x)\displaystyle y(x) = c_1J_{5/2}(x) + c_2J_{-5/2}(x)

Writing this solution in terms of trigonometric functions is a little bit challenging. But we love Challenges😍

We start with our secret weapon which is the recurrence relation.

2nJn(x)=xJn+1(x)+xJn1(x)\displaystyle 2nJ_n(x) = xJ_{n + 1}(x) + xJ_{n - 1}(x)

This time let us try n=32\displaystyle n = \frac{3}{2} and see what we get.

232J3/2(x)=xJ3/2+1(x)+xJ3/21(x)\displaystyle 2\frac{3}{2}J_{3/2}(x) = xJ_{3/2 + 1}(x) + xJ_{3/2 - 1}(x)

3J3/2(x)=xJ5/2(x)+xJ1/2(x)\displaystyle 3J_{3/2}(x) = xJ_{5/2}(x) + xJ_{1/2}(x)

This gives:

J5/2(x)=3xJ3/2(x)J1/2(x)\displaystyle J_{5/2}(x) = \frac{3}{x}J_{3/2}(x) - J_{1/2}(x)

From previous exercises, we know that:

J3/2(x)=1x2πxsinx2πxcosx\displaystyle J_{3/2}(x) = \frac{1}{x}\sqrt{\frac{2}{\pi x}}\sin x - \sqrt{\frac{2}{\pi x}}\cos x

And

J1/2(x)=2πxsinx\displaystyle J_{1/2}(x) = \sqrt{\frac{2}{\pi x}}\sin x

Then

J5/2(x)=3x(1x2πxsinx2πxcosx)2πxsinx\displaystyle J_{5/2}(x) = \frac{3}{x}\left(\frac{1}{x}\sqrt{\frac{2}{\pi x}}\sin x - \sqrt{\frac{2}{\pi x}}\cos x\right) - \sqrt{\frac{2}{\pi x}}\sin x

Again let us try n=32\displaystyle n = -\frac{3}{2} and see what that leads us.

232J3/2(x)=xJ3/2+1(x)+xJ3/21(x)\displaystyle -2\frac{3}{2}J_{-3/2}(x) = xJ_{-3/2 + 1}(x) + xJ_{-3/2 - 1}(x)

3J3/2(x)=xJ1/2(x)+xJ5/2(x)\displaystyle -3J_{-3/2}(x) = xJ_{-1/2}(x) + xJ_{-5/2}(x)

This gives:

J5/2(x)=3xJ3/2(x)J1/2(x)\displaystyle J_{-5/2}(x) = -\frac{3}{x}J_{-3/2}(x) - J_{-1/2}(x)

From previous exercises, we know that:

J3/2(x)=2πxsinx1x2πxcosx\displaystyle J_{-3/2}(x) = -\sqrt{\frac{2}{\pi x}}\sin x - \frac{1}{x}\sqrt{\frac{2}{\pi x}}\cos x

And

J1/2(x)=2πxcosx\displaystyle J_{-1/2}(x) = \sqrt{\frac{2}{\pi x}}\cos x

Then

J5/2(x)=3x(2πxsinx1x2πxcosx)2πxcosx\displaystyle J_{-5/2}(x) = -\frac{3}{x}\left(-\sqrt{\frac{2}{\pi x}}\sin x - \frac{1}{x}\sqrt{\frac{2}{\pi x}}\cos x\right) - \sqrt{\frac{2}{\pi x}}\cos x

Now we can write the general solution in terms of trigonometric functions.

y(x)=c1[3x(1x2πxsinx2πxcosx)2πxsinx]+c2[3x(2πxsinx1x2πxcosx)2πxcosx]\displaystyle y(x) = c_1\left[\frac{3}{x}\left(\frac{1}{x}\sqrt{\frac{2}{\pi x}}\sin x - \sqrt{\frac{2}{\pi x}}\cos x\right) - \sqrt{\frac{2}{\pi x}}\sin x\right] + c_2\left[-\frac{3}{x}\left(-\sqrt{\frac{2}{\pi x}}\sin x - \frac{1}{x}\sqrt{\frac{2}{\pi x}}\cos x\right) - \sqrt{\frac{2}{\pi x}}\cos x\right]

We can simplify it further if we want in terms of the arbitrary constants C1,C2,C3,C4,C5,and C6\displaystyle C_1, C_2, C_3, C_4, C_5, \text{and} \ C_6. I think that you got the idea!

💪🎃🎃
 
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