It can be shown that Jo has infinitely many zeros for x > 0. In particular, the first three zeros are approximately 2.405, 5.520, and 8.653 (see Figure 5.7.1). Let A;, j = 1, 2, 3, ..., denote the zeros of Jo; it follows that x = 0, 1, Jo(A;æ) = x = 1. Verify that y = Jo(A;x) satisfies the differential equation 1 y" + y' + X3y = 0, x > 0.

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