Solution to Bessel's Differential Equation

[Parmenides]

Hey everyone. Need some more pairs of eyes for this one:

"For each positive integer $n$, the Bessel Function $J_n(x)$ may be defined by:
$$J_n(x) = \frac{x^n}{1\cdot3\cdot5\cdots(2n-1)\pi}\int^1_{-1}(1 - t^2)^{n-\frac{1}{2}}\cos(xt)dt$$

Prove that $J_n(x)$ satisfies Bessel's differential equation:
$${J_n}^{\prime\prime} + \frac{1}{x}{J_n}^{\prime} + \Big(1 - \frac{n^2}{x^2}\Big)J_n = 0$$"

This problem is following a question where I proved differentiating under the integral sign. Thus, the way to do this is to simply calculate the derivatives of $J_n(x)$ and plug back into the equation. However, I can't seem to get all of the terms to cancel. I believe that the only thing that needs "calculus" methods is to calculate the derivative, with respect to x, of all the $x$ terms:
$$\frac{\partial}{\partial{x}}(x^ncos(xt)) = (nx^{n-1}cos(xt) - tx^nsin(xt))$$
and
$$\frac{\partial^2}{\partial{x^2}}(x^ncost(xt)) = \frac{\partial}{\partial{x}}(nx^{n-1}cos(xt) - tx^nsin(xt) = (n(n-1)x^{n-2}cos(xt) - tnx^{n-1}sin(xt)) - (ntx^{n-1}sin(xt) + t^2x^ncos(xt))$$
The rest would appear to be algebra. But again, not everything seems to cancel. That $t^2$ term seems to be messing things up for the $x^n$ terms and the $x^{n-1}$ terms seem to just combine. Unless there's an identity I'm overlooking, or my differentiation is wrong, I'm quite lost. Ideas? Thank you.

 

[lippyka]

Hint: You need to use integration by parts to get the derivatives correct. Letting u = (1 - t^2)^(n-1/2) and dv = cos(xt)dt, we can use integration by parts to calculate the derivatives of J_n(x):du = -2t(1 - t^2)^(n-3/2)dt and v = \frac{sin(xt)}{x}, thus \frac{\partial}{\partial{x}}(x^ncos(xt)) = (nx^{n-1}cos(xt) - tx^nsin(xt))and\frac{\partial^2}{\partial{x^2}}(x^ncost(xt)) = \Big(n(n-1)x^{n-2}cos(xt) - tnx^{n-1}sin(xt)\Big) - \Big(ntx^{n-1}sin(xt) + t^2x^ncos(xt)\Big)Substituting these derivatives back into the equation and using the identity sin^2(x) + cos^2(x) = 1, we can simplify to J_n''(x) + \frac{1}{x}J_n'(x) + \Big(1 - \frac{n^2}{x^2}\Big)J_n(x) = \frac{2ntx^{n-1}sin(xt)}{1\cdot3\cdot5\cdots(2n-1)\pi} \int^1_{-1}(1 - t^2)^{n-\frac{1}{2}}dtwhich is equal to 0 since the integral is evaluated from -1 to 1.