Let we have $$y(x) = x^n\int_0^{\pi}\cos(x \cos\varphi)\sin^{2n}\varphi d\varphi$$
How to check that $y(x)$ satisfies following differential equation : $$x^2y'' + xy' + (x^2 -n^2)y = 0$$
Prove that $y(x) = x^n\int_0^{\pi}\cos(x \cos\varphi)\sin^{2n}\varphi d\varphi$ solves $x^2y'' + xy' + (x^2 -n^2)y = 0$
0
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integration
ordinary-differential-equations
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1diffentiate $y(x)$ and plug-in the differential equation – 2017-02-06
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0What attempts have you made to solve the problem? Can you calculate $y'$? – 2017-02-06
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0you might need to integrate by parts – 2017-02-06
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1but because you are an experienced user who puts no effort in his question i vote to close – 2017-02-06
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0these are like Chebyshev polynomials. – 2017-02-06
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0@MyGlasses It is Bessel functions. See my answer. – 2017-02-06
1 Answers
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Context: It is the differential equation of a multiple of Bessel function $J_n$ (http://mathworld.wolfram.com/BesselDifferentialEquation.html) where this function is under a certain integral representation that can be found formula (10.9.4) here.
It suffices to check that you can differentiate under the integral sign, i.e., (not taking into account the first $x^n$) that you can write:
$$\dfrac{d}{dx}\int_0^{\pi}\cos(x \cos\varphi)\sin^{2n}(\varphi) d\varphi=\int_0^{\pi}\dfrac{d}{dx}\cos(x \cos\varphi)\sin^{2n}(\varphi) d\varphi$$
$$=-\int_0^{\pi}\sin(x \cos\varphi)\cos\varphi\sin^{2n}(\varphi) d\varphi$$
See theorem 11.3 in this document.
Can you proceed from that?
An advice: start with $n=1$.