Author Topic: TUT0301  (Read 3953 times)

Xuefen luo

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« on: October 11, 2019, 02:14:55 PM »
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 $W=\frac{c}{x}$