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$x^2y''+xy'+(x^2-v^2)y=0$

Dividing both sides by $x^2$, then we have:

$y''+ \frac{1}{x}y'+\frac{(x^2-v^2)}{x^2}y=0$

Since $W=ce^{\int -p(x)dx}$, and $p(x)=\frac{1}{x}$ in this case, we have:

$W=ce^{-\int \frac{1}{x}dx}=ce^{-ln(x)+C}=ce^{ln(x^{-1})+C}=cx^{-1}e^C$

We know $e^C$ is just a constant, so we can just subsume it into $c$. Then the Wronskian is $W=\frac{c}{x}$

Dividing both sides by $x^2$, then we have:

$y''+ \frac{1}{x}y'+\frac{(x^2-v^2)}{x^2}y=0$

Since $W=ce^{\int -p(x)dx}$, and $p(x)=\frac{1}{x}$ in this case, we have:

$W=ce^{-\int \frac{1}{x}dx}=ce^{-ln(x)+C}=ce^{ln(x^{-1})+C}=cx^{-1}e^C$

We know $e^C$ is just a constant, so we can just subsume it into $c$. Then the Wronskian is $W=\frac{c}{x}$