Reducing Order: Is it Valid to Say G(x) = 0?

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The discussion centers on the validity of stating that the differential equation (DE) y" - xy' = 0 is in standard form with G(x) = 0. The standard form of a second-order linear DE is defined as y" + P(x)y' + G(x)y = 0, where P(x) and G(x) are continuous functions on the interval I. The user confirms that it is indeed valid to express G(x) as 0 in this context, despite initially questioning the solution y = e^x, which is not a solution to the DE. The conversation highlights the importance of verifying solutions when applying reduction of order techniques.

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bishy
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I'm making an assumption while trying to solve DEs by reduction of order. I've got a short form equation that I can use to reduce it if and only if I can place the DE into standard form. The standard form with generic notation would be y"+ P(x)y' +G(x)y = 0 where P(x) and G(x) are continuous and on the interval I. I am not sure if I am able to say the following, therefore my question would be is this valid:

Given the DE y" - xy' = 0; y=e^x; I= (0,infinity] is it valid to state that the DE is under standard form where G(x) = 0?
 
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Yes, of course it is! I am wondering what you mean by "y= e^x" immediately after the differential equation. That is clearly not a solution to the equation.
 
ah ok. that's what I thought, I just wasn't sure. As for it not being a solution, I made it up and never bothered checking it. It's not a big a deal, most of the work I have done is confirmed with this.
 

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