Solve.
\(\displaystyle x^2y'' + xy' + \left(x^2 - \frac{25}{4}\right)y = 0\)
We all now can solve this differential equation quickly. With full of confidence we can say that the general solution is:
\(\displaystyle y(x) = c_1J_{5/2}(x) + c_2J_{-5/2}(x)\)
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.
\(\displaystyle 2nJ_n(x) = xJ_{n + 1}(x) + xJ_{n - 1}(x)\)
This time let us try \(\displaystyle n = \frac{3}{2}\) and see what we get.
\(\displaystyle 2\frac{3}{2}J_{3/2}(x) = xJ_{3/2 + 1}(x) + xJ_{3/2 - 1}(x)\)
\(\displaystyle 3J_{3/2}(x) = xJ_{5/2}(x) + xJ_{1/2}(x)\)
This gives:
\(\displaystyle J_{5/2}(x) = \frac{3}{x}J_{3/2}(x) - J_{1/2}(x)\)
From previous exercises, we know that:
\(\displaystyle J_{3/2}(x) = \frac{1}{x}\sqrt{\frac{2}{\pi x}}\sin x - \sqrt{\frac{2}{\pi x}}\cos x\)
And
\(\displaystyle J_{1/2}(x) = \sqrt{\frac{2}{\pi x}}\sin x\)
Then
\(\displaystyle J_{5/2}(x) = \frac{3}{x}\left(\frac{1}{x}\sqrt{\frac{2}{\pi x}}\sin x - \sqrt{\frac{2}{\pi x}}\cos x\right) - \sqrt{\frac{2}{\pi x}}\sin x\)
Again let us try \(\displaystyle n = -\frac{3}{2}\) and see what that leads us.
\(\displaystyle -2\frac{3}{2}J_{-3/2}(x) = xJ_{-3/2 + 1}(x) + xJ_{-3/2 - 1}(x)\)
\(\displaystyle -3J_{-3/2}(x) = xJ_{-1/2}(x) + xJ_{-5/2}(x)\)
This gives:
\(\displaystyle J_{-5/2}(x) = -\frac{3}{x}J_{-3/2}(x) - J_{-1/2}(x)\)
From previous exercises, we know that:
\(\displaystyle J_{-3/2}(x) = -\sqrt{\frac{2}{\pi x}}\sin x - \frac{1}{x}\sqrt{\frac{2}{\pi x}}\cos x\)
And
\(\displaystyle J_{-1/2}(x) = \sqrt{\frac{2}{\pi x}}\cos x\)
Then
\(\displaystyle J_{-5/2}(x) = -\frac{3}{x}\left(-\sqrt{\frac{2}{\pi x}}\sin x - \frac{1}{x}\sqrt{\frac{2}{\pi x}}\cos x\right) - \sqrt{\frac{2}{\pi x}}\cos x\)
Now we can write the general solution in terms of trigonometric functions.
\(\displaystyle y(x) = c_1\left[\frac{3}{x}\left(\frac{1}{x}\sqrt{\frac{2}{\pi x}}\sin x - \sqrt{\frac{2}{\pi x}}\cos x\right) - \sqrt{\frac{2}{\pi x}}\sin x\right] + c_2\left[-\frac{3}{x}\left(-\sqrt{\frac{2}{\pi x}}\sin x - \frac{1}{x}\sqrt{\frac{2}{\pi x}}\cos x\right) - \sqrt{\frac{2}{\pi x}}\cos x\right]\)
We can simplify it further if we want in terms of the arbitrary constants \(\displaystyle C_1, C_2, C_3, C_4, C_5, \text{and} \ C_6\). I think that you got the idea!


