How can you solve this Bessel's equation? From what I've learned, the Bessel's equation in its standard form is
$$x^2y''+xy'+(x^2+v^2)y=0\tag{1}$$
But, for example I have this equation
$$x^2y''+xy'+\frac{1}{4}xy=0\tag{2}$$
How can you solve the 2nd equation by using the Bessel's standard form?
 A: Since WolframAlpha trivializes solving this specific ODE, I'll work in the opposite direction to show how such Bessel-like ODEs arise. Suppose we start with the modified Bessel equation $$z^2y''(z)+zy'(z)+(z^2+v^2)y(z)=0$$
with linearly independent solutions $y(z)=I_0(z),K_0(z)$. We now consider the effect of the substitution $z=bx^a$. Writing $D_z:=d/dz$ and $D_x:=d/dx$, the effect of the chain rule is that
$$zD_z = z\frac{dx}{dz} D_x =z\cdot a z^{a-1} D_x=a x D_x.$$
If we apply these operators a second time, the product rule yields
\begin{align}
(z D_z)^2&=zD_z+z^2 D_z^2,\\( ax D_x)^2&=a^2 x D_x+a^2 x^2 D_x^2.
\end{align}
This is convenient, because the first two terms of the modified Bessel equation can therefore be rewritten as
\begin{align}
z^2 y''(z)+z y'(z)
&=[z D_z+z^2 D_z^2]y(z)\\
&=(z D_z)^2 y(z)\\
&=(a x D_x)^2 y(x)\\
&=a^2 x^2 y''(x)+a^2 x y'(x).
\end{align}
Carrying out the substitution in the third term as well, the ODE now takes the form
$$a^2 x^2 y''(x)+a^2 x y'(x)+(b^2 x^{2/\alpha}+v^2)+v^2)y(z)=0.$$
If we divide through by $a^2$ and introduce $A:=b^2/a^2,B:=v^2/a^2,r:=2/a$ we obtain
$$x^2y''(x)+x y'(x)+(Ax^r+B)y(x)=0.$$
Thus we obtain something that looks very similar to the modified Bessel equation, but with $x^2+v^2$ replaced by a linear function of $x^r$. In the present case, this amounts to choosing $B=0,A=1/4,r=1$ and thus $v=0,a=1/2,b=1$. This justifies the solution set $y(x)=I_0(\sqrt{x}),K_0(\sqrt{x})$ given by WolframAlpha.
