This question is related to the question I asked in this post.
I'm trying to find an integrating factor or some way to turn $-Y^{\prime\prime}(y) - \mu Y^{\prime}(y) = 0$ into Sturm-Liouville form.
On that post, someone suggested the substitution $Y(y) = e^{-\mu y}Z(y)$.
I differentiated $e^{-\mu y} Z(y)$ twice, substituted the expressions I obtained for both $Y^{\prime}$ and $Y^{\prime \prime}$ into the equation and ended up with $-e^{-\mu y} Z^{\prime \prime} (y) - \mu e^{-\mu y} Z^{\prime}(y) + 2 = 0$.
Now, recall that a Sturm-Liouville problem is of the form $-[p(x)u^{\prime}(x)]^{\prime} + q(x)u(x) = \mu r(x) u(x)$ or $\mathbf{-p^{\prime}(x)u^{\prime}(x) - p(x) u^{\prime \prime}(x) + q(x) u(x) = \mu r(x) u(x)}$), and we need for both $p(x)$ and $r(x)$ to be strictly positve on the interval we're dealing with (here it's [0,1]).
So, now I no longer have $r(x)$ strictly positive.
What am I doing wrong?? I'm feeling very frustrated.
Could someone please help me?? And if it's not possible to write this problem in Sturm-Liouville form, could you let me know that? Thank you. I'm really struggling with this.
Here is an alternate approach: $$ -Y''-\mu Y'=0\\ \Downarrow\\ \frac{Y''}{Y'}=-\mu\\ \Downarrow\\ \log(Y')=\eta-\mu y\\ \Downarrow\\ Y'=e^{\eta-\mu y}\\ \Downarrow\\ Y=a+b e^{-\mu y} $$