If $J_n(x)$ and $Y_n(x)$ denotes Bessel's functions of order $n$ of first and second kind, then the general solutions of the differential equation $$x\frac{d^2y}{dx^2}-\frac{dy}{dx}+xy=0$$ is given by
$1$. $axJ_1(x)+bxY_1(x)$.
$2$. $aJ_1(x)+bY_1(x)$.
$3$. $aJ_0(x)+bY_0(x)$.
$4$. $axJ_0(x)+bxY_0(x)$.
If we multiply given differential equation by $x$ even then it is not Bessel’s equation. How to solve this question in less time as its objective type question? Indicial equation is $r(r-2)$. I tried to find series solutions corresponding to $r=0$ by letting $y=\sum_0^\infty a_nx^n$ but it’s not like $J_0(x)$ . Thank you .
For a Bessel equation you want that close to $x=0$ you get basis solutions $x^{\pm n}u_\pm(x)$, $u_\pm$ a power series. At infinity the equation should and does reduce to $y''+y=0$. For the first property one needs that the indicial equation is $r^2-n^2=0$.
Currently there are basis solutions for $y$ of the form $y(x)=u(x)$ and $y(x)=x^2v(x)$, $u,v$ power series, $u(0),v(0)\ne 0$. If one splits off a factor $x$, one gets Frobenius series with symmetric leading degrees.
Or in other words, since $r(r-2)=(r-1)^2-1$, you need to consider $y(x)=xz(x)$ to shift the leading power by one down in the Frobenius series for $z(x)$, and to get symmetric roots for the indicial equation. Then the differential equation for $z$ can be computed as $$ 0=x[xz''+2z']-[xz'+z]+x[xz]=x^2z''+xz'+(x^2-1)z. $$ This should now be sufficiently normalized.