Originally Posted by

**VonNemo19** Oh, right. Sorry. Integrating factor. My bad.

So, what about going this route...

$\displaystyle \ln|y|=-\ln|\sin(x)|+c_1$

$\displaystyle \Rightarrow{y}=\pm{e}^{-\ln|\sin(x)|+c_1}$

$\displaystyle \Rightarrow{y}=c{e}^{\ln|\sin(x)|^{-1}}$

$\displaystyle \Rightarrow{y}=c{e}^{\ln|\csc(x)|}$

But, $\displaystyle y'+\cot(x)y=0$ is of the form $\displaystyle y'+P(x)y=0$ and we are only seeking a solution on an interval $\displaystyle I$ for which the coefficient function $\displaystyle P(x)$ is continuous. So, we take $\displaystyle I=(0,\infty)$ and therefore we are justified in removing the absolute value signs, giving

$\displaystyle {y}=c{e}^{\ln\csc(x)}$

or

$\displaystyle {y}=c{\csc(x)}$

What about that? Page 53. Right under equation (2).

Thanks.