Solve the Bessel differential equation

Question

Show that $J_{n}(x) / x^{n}$ is a solution of $$\frac{d^{2} y}{d x^{2}}+\left(\frac{1+2 n}{x}\right) \frac{d y}{d x}+y=0$$ and that $\sqrt{(x)} J_{n}(k x)$ is a solution of $$\frac{d^{2} y}{d x^{2}}+\left(k^{2}-\frac{4 n^{2}-1}{4 x^{2}}\right) y=0$$ where, in both cases, $n$ is a positive integer.

Answer 1

$$y''+\left(\frac{1+2 n}{x}\right)y'+y=0\tag 1$$ Change of function :

$y(x)=\frac{u(x)}{x^n} \quad;\quad y'=\frac{u'}{x^n}-n\frac{u}{x^{n+1}} \quad;\quad y''=\frac{u''}{x^n}-2n\frac{u'}{x^{n+1}}+n(n+1)\frac{u}{x^{n+2}} $

Putting them into the ODE : $$\frac{u''}{x^n}-2n\frac{u'}{x^{n+1}}+n(n+1)\frac{u}{x^{n+2}}+\left(\frac{1+2 n}{x}\right)\left(\frac{u'}{x^n}-n\frac{u}{x^{n+1}}\right) +\frac{u}{x^n}=0$$ $$u''-2n\frac{u'}{x}+n(n+1)\frac{u}{x^2}+\left(\frac{1+2 n}{x}\right)\left(u'-n\frac{u}{x}\right) +u=0$$ After simplification : $$u''+\frac{u'}{x}+(1-\frac{1}{x^2})u=0$$ This is the Bessel equation on standard form, which solution is : $$u(x)=c_1J_1(x)+c_2Y_1(x)$$ with the Bessel functions of first and second kind. $$y(x)=c_1\frac{J_1(x)}{x^n}+c_2\frac{Y_1(x)}{x^n}$$ Thus $\frac{J_1(x)}{x^n}$ is a solution of Eq.$(1)$.

Proceed on the same manner for Eq.$(2)$ , with the change of function $y(x)=\sqrt{x}\:u(kx)$ and the change of variable $X=kx$.