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I need to reduce to sturm-liouville using the 'integrating factor' method:

$$My=x^2y''+3xy+y$$ where y=y(x). I am looking at my notes and I know the Sturm-Liouville form is $$\left(P(x)y'\right)'+Q(x)y$$ However, I don't understand what to do. If someone can help me in DETAIL that will be much appreciated.

Thanks!!

I have a sense at what to do but my professor notes are a little hard to understand.

Arbuja
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geeking4math
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1 Answers1

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Sorry- my question was really about the "3xy" term and I mistyped. Do you mean $M(y)= x^2y''+ 3xy'+ y$? If so then let p(x) be the factor you want: we want $x^2p(x)y''+ 3xp(x)y'= (x^2p(x)y')'$. By the product rule that right side is $x^2p(x)y''+ 2xp(x)y'+ x^2p'(x)y'= x^2p(x)y''+ (2xp(x)+ x^2p'(x))y'$ which must equal $x^2p(x)y''+ 3xp(x)y'$. That is, we must have $2xp(x)+ x^2p'(x)= 3xp(x)$ which gives $p'(x)= p(x)/x$ so $dp/p= dx/x$. Integrating both sides, $ln(p)= ln(x)+ C$ and, taking the exponential, $p(x)= C'x$ where $C'= e^C$. Since we only need a single function, take C'= 1. The integrating factor is x: $xM(y)= x^3y''+ 3x^2y'+ xy= (x^3y')'+ xy$.

user247327
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