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I want to find the zeros of the solution to this ODE

DSolve[{y''[x] - (I x - 2) y[x] == 0, y[0]==0,y'[0]==1}, y[x],x]

I use

ResourceFunction["Intercepts"]

and get a product of several hypergeometric functions. Are there alternative ways of getting the zeros?`For instance, such as the Bessel zeros, or the sine cosine zeros at $n\pi$ and $n\pi+\frac{\pi}{2}$?

Thanks

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  • 1
    $\begingroup$ You could use something like the the technique in this answer to get zeroes within a range. $\endgroup$ May 28 at 16:48
  • $\begingroup$ There is AiryAiZero just like BesselJZero, but Mma cannot solve Solve[BesselJ[1, x] == BesselJ[2, x], x, Method -> Reduce] or even Solve[BesselJ[1, x] == 1/10, x, Method -> Reduce], even though it can solve Solve[BesselJ[1, x] == 0 x, Method -> Reduce]. On a bounded interval, it can solve Solve[BesselJ[1, x] == BesselJ[2, x] && 0 < x < 10, x, Method -> Reduce]. $\endgroup$
    – Michael E2
    May 28 at 17:16

1 Answer 1

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Maybe something like this, with a bounded domain:

func = y[x] /. 
   First@
    DSolve[{y''[x] - (I x - 2) y[x] == 0, y[0] == 0, y'[0] == 1}, 
     y[x], x];

Solve[func == 0 && -1/10 < Re[x] < 4 && -1 < Im[x] < 2, x]
(*
{{x -> 0},
 {x -> 
   Root[{AiryAi[-(-1)^(2/3) (-2 + I #1)] AiryBi[2 (-1)^(2/3)] - 
       AiryAi[2 (-1)^(2/3)] AiryBi[-(-1)^(2/3) (-2 + I #1)] &, 
     1.9742593 + 0.4276746 I}]},
 {x -> 
   Root[{AiryAi[-(-1)^(2/3) (-2 + I #1)] AiryBi[2 (-1)^(2/3)] - 
       AiryAi[2 (-1)^(2/3)] AiryBi[-(-1)^(2/3) (-2 + I #1)] &, 
     3.4529450 + 1.0382391 I}]}}
*)

Root objects are an exact representation of algebraic or, in this case, transcendental roots. They are numeric and may be calculated to arbitrary precision.

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