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Numerade Educator



Problem 35 Hard Difficulty

(a) Show that $ J_0 $ (the Bessel function of order 0 given in Example 4) satisfies the differential equation
$ x^2 J''_0 (x) + x J'_0 (x) + x^2 J_0 (x) = 0 $
(b) Evaluate $ \int^1_0 J_0 (x) dx $ correct to three decimal places.


a. $0=0$
b. 0.920

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Video Transcript

So what we want to do is we're looking at the vessel function of order zero. Um and we're gonna have a lot of notation that we're dealing with eso where we're gonna have the equation. Zero equals X squared J not double prime of X plus x j not prime of X plus X squared Jane, not of X. So about this, we want to plug in the different values. We know that J not double Prime of X is equal to the summation of negative one to the n two n times to end minus one times X to the to end minus two over to to the to and times and factorial squirt. And that's from an equal zero to infinity. Then we have J nut. Ah, prime of X. We know that's gonna be equal to the, uh, summation from unequal zero to infinity of negative one to the end times to an X to the two and minus one. And that's all over to to the two n times and factorial squared. Then Lastly, we have j not vex, which is going to be equal to the summation from an equal zero to infinity of again negative one to the end Times X to the to end over to to the to end in factorial squared. So when we plug in these terms, we can put them all using the same denominator. They all share the same denominator so we can combine the fraction. Then we do that. We can cancel out a lot of terms on doing that whole process. What we end up getting is much more simplified form, which is that zero equals de summation, from an equal zero to infinity of negative one to the N four end squared X to the two n over to to the to and in factorial squared plus summation for my an equal 02 Infinity again of negative one end times X, the two and plus two over and no surprise. It's the same denominator. Then, looking at those two Siri's, we see some important terms. The one on the left right here is going to look a little something like this. Zero minus X squared plus X to the fourth over four minus X to the sixth over 64. So we see that happening right there, and we see that it's very similar to just J not of X. The only difference is a one instead of zero and the signs. So this would be, um minus right here. This would be plus right here and we don't have any X squared shown. So there's the major differences. But that's because we multiplied the X squared times, the one So it's clearly just a multiplication of X squared times, times that. So because we notice that this right here is just a negative x squared times, Jane, not Vex, we can rewrite it as such. And then on the right side, we see that the right portion right here is just equal toe X squared times J not X if we look at it. So what we end up finding is that if we add the negative form and the positive form, obviously we're gonna get zero equals zero. And that is a true statement, which means that we verified this. Now what we have to do is just evaluate, uninterested, incorporating these battle functions. So what we're gonna have is the integral from 0 to 1 of J not vax DX. We already know Jeannot from before we already to find it right here If we wanna look back to that. So when we do that, we're gonna end up having the integral of 01 of this summation right here. DX. So now that we have that, we see that this is gonna be equal. Thio. Just the summation from an equal zero to infinity of negative one to the end of X the two n plus one over to to the two n n factorial squared times 21 plus one. And that's gonna be all evaluated at zero and one. When we do that to three decimal places we're gonna end up getting is 0.920 So this will be our final answer for part B that ISS.