# Solve Bessel's equation through certain substitutions

by ILikeMath
Tags: bessel, function, substitution
 P: 5 1. The problem statement, all variables and given/known data, relevant equation $$x^{2}y'' + xy' + (4x^{4}-\frac{1}{4})y = 0$$ 2. The attempt at a solution I tried substituting z = x2 From this I have $$\frac{dy}{dx} = 2x \frac{dy}{dz}$$ and $$\frac{d}{dx}(2x\frac{dy}{dz}) = 2xy'' + 2y'$$ Then the original equation becomes: $$2z^{3/2}y'' + 4zy' + (4z^{2}-\frac{1}{4})y = 0$$ where derivatives of y are now with respect to the new variable z. This does not look like a Bessel equation and I'm not sure how to make it look like one. Did I use the wrong substitution? I know how to solve once it's in the correct form, but could someone help me get it there please?
 P: 110 try $$z^{2}=4x^{4}$$
 P: 5 If I use z2 = 4x4 I get: z = 2x2 and dz/dx = z' = 4x $$(\frac{dy}{dx}) = \frac{dy}{dz} \frac{dz}{dx} = 4x \frac{dy}{dz}$$ Then $$\frac{d}{dx}(4x\frac{dy}{dz})=4\frac{dy}{dz}+\frac{d}{dz}(\frac{dy}{dz} )\frac{dz}{dx}=4\frac{dy}{dz}+4x\frac{d^{2}y}{dz^{2}}$$ Plugging in: x2(4y' + 4xy'') + x(4xy') + (4x4 - 1/4)y = 0 if z2 = 4x4 (z/2)3/2 y'' + zy' + (z2/4 - 1/16)y = 0 so it still doesn't work. The x3 term messes the whole thing up....