Wronskian of two linearly independent solutions $y_1$ and $y_2$ Let $P$ be a continuous function on $\mathbb{R}$ and $W$ the Wronskian of two linearly independent solutions $y_1$ and $ y_2$ of the ODE: $\frac{d^2y}{dx^2}+(1+x^2)\frac{dy}{dx}+P(x)y=0$, $x\in \mathbb{R}$. Let $W(1)=a$, $W(2)=b$ and $W(3)=c$ then
$\\1. \;a\lt0$ and $b\gt0$
$\\2.\; a<b<c $ or $a>b>c$
$\\3.\;\frac{a}{|a|}= \frac{b}{|b|}=\frac{c}{|c|}$
$\\4.\; 0<b<c $ or $a>b>0$
My attempt: by Abel's theorem W=$e^{\int pdx}$=$e^{\int (1+x^2)dx}$=$e^{x+\frac{x^3}{3}}$
$W(1)=e^\frac{4}{3}=$a, $W(2)=e^\frac{14}{3}=b$, $W(3)=e^{12}=c$. Therefore options $2$ and $3$ are correct. Am i right?


Abel's formula: Let functions $p_1(x)$ and $p_2(x)$ in $y''(x)+p_1(x)y'(x)+p_2(x)y(x)=0$, $x\in I$ be defined and continuous on an interval $I$. Let $\phi_1$ and $\phi_2$ be two linearly independent solutions of the ODE existing on $I$ containing a point $x_0$. Then $W(\phi_1,\phi_2)(x)=W(\phi_1,\phi_2)(x_0)\exp(-\int_{x_0}^{x}  p_1(x)dx)$.

Now we have $a=W(1)=W(x_0)\exp(-\int_{x_0}^{1}  (1+x^2)dx)=ke^{-\frac4 3}$, where $k=W(x_0)e^{x_0+\frac{x_{0}^3}{3}}$. Here note that $k$ may be positive or negative according to $W(x_0)$ is positive or negative.
Similarly $b=W(2)=ke^{-\frac{14}{3}}$ and $c=W(3)=ke^{-12}$.
Thus  $a>b>c$ if $k>0$ and $a<b<c$ if $k<0$.
Also $\frac{a}{|a|}=\frac{k}{|k|}=\frac{b}{|b|}=\frac{c}{|c|}$. Thus option $2$ and $3$ are correct. 
Now $a<0$ implies $k<0$ and $b>0$ implies $k>0$. So both are contradict and hence $1$ is not true.
Also for $k<0$, $a<b$ but  $0>a$. Thus $4$ is also not true.
