Solving a differential equation in terms of Bessel Functions

Question

I need to solve the following differential equation in terms of Bessel functions $$4x^2y''+4xy'+(x-4)y=0$$

I know I need to use the transformation $x=z^2$ but I am unsure on how to even solve the differential equation

Answer 1

First rewrite the equation using differential operators: $$4x^2~\mathrm{D}_x^2(y)+4x~\mathrm{D}_x(y)+(x-4)y=0$$ Using the subsitution $x\equiv z^2$, we can see that $$\mathrm{D}_x=\frac{\mathrm{d}z}{\mathrm{d}x}\mathrm{D}_z$$ And from the definition of $z$, $$1=2z\frac{\mathrm{d}z}{\mathrm{d}x}\implies \frac{\mathrm{d}z}{\mathrm{d}x}=\frac{1}{2z}$$ So $$\mathrm{D}_x=\frac{1}{2z}\mathrm{D}_z$$ Thus, $$4z^4\frac{1}{2z}\mathrm{D}_z\left(\frac{1}{2z}\mathrm{D}_z(y)\right)+4z^2\frac{1}{2z}\mathrm{D}_z(y)+(z^2-4)y=0$$ Using the quotient rule, $$\mathrm{D}_z\left(\frac{1}{2z}\frac{\mathrm{d}y}{\mathrm{d}z}\right)=\frac{1}{2}\frac{\frac{\mathrm{d}^2 y}{\mathrm{d}z^2}\cdot z-\frac{\mathrm{d}y}{\mathrm{d}z}}{z^2}$$ Thus, $$2z^3\left(\frac{1}{2}\frac{\frac{\mathrm{d}^2 y}{\mathrm{d}z^2}\cdot z-\frac{\mathrm{d}y}{\mathrm{d}z}}{z^2}\right)+2z\frac{\mathrm{d}y}{\mathrm{d}z}+(z^2-4)y=0$$ $$z^2\frac{\mathrm{d}^2y}{\mathrm{d}z^2}+z\frac{\mathrm{d}y}{\mathrm{d}z}+(z^2-4)y=0$$ This is obviously a Bessel ODE, with solutions $$y(z)=a\cdot J_2(z)+b\cdot Y_2(z)$$ Or, $$y(x)=a\cdot J_2\left(\sqrt{x}\right)+b\cdot Y_2\left(\sqrt{x}\right)$$