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The Bessel equation $$\frac{d^2y}{dx^2}+\frac{1}{x}\frac{dy}{dx}+(1-\frac{\nu^2}{x^2})y=0$$ is equivalent to the differential equation $$\frac{d^2z(x)}{dx^2}+z(x)=\frac{\mu}{x^2}z(x), \ \ \ \ where \ \ \ \mu=\nu^2-\frac{1}{4}. $$ by using the Liouville-transformation $$y(x) = x^{-\frac{1}{2}}z(x)$$ Prove, using variation of parameters, that for $0<a\leq x$ holds: $$z(x)=z(a)cos(x-a)+z'(a)sin(x-a)+\int_a^xsin(x-\xi)\frac{\mu}{\xi^2}z(\xi)d\xi$$ Furthermore, that $$|z(x)|\leq r + \int_a^x\frac{\mu}{\xi^2}|z(\xi)|d\xi, \ \ \ \ where \ \ r=\sqrt{z(a)^2+z'(a)^2}$$ And, using Gronwall's lemma, that $$|z(x)| \leq r\ e^{\frac{\mu}{a}-\frac{\mu}{x}} \leq r\ e^{\frac{\mu}{a}}$$ Notably: for every $a>0$, the function $z(x)$ is bounded on $[a,\infty[$

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