I'm doing some practice problems, and with the help of my solutions manual and wolfram alpha I've worked out a solution to(adsbygoogle = window.adsbygoogle || []).push({});

[tex] (x+1)\frac{dy}{dx} +xy = e^{-x} [/tex]

However, I don't understand why we can drop the absolute value bars when we calculate the integrating factor:

[tex] e^{\int \frac{x}{x+1} dx} = e^{x- \ln|x+1|} = \frac{e^x}{x+1} [/tex]

I understand that if we don't drop the absolute value bars in the last step, certain terms won't cancel when we multiply our equation by this integrating factor. But why are we justified in doing so? Usually we say something about how x > 0, but this isn't an initial value problem and x isn't specified to be in any particular interval. Can you please explain why we are justified in dropping the absolute value bars? How do we know that we aren't losing solutions when we do this?

(For a full solution to this see

http://www.wolframalpha.com/input/?i=(x+1)dy/dx++xy+=e^(-x) )

and click "show steps" under "Differential Equation Solutions".

Thanks so much!

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# Homework Help: Linear Differential Equation: Dropping Absolute Value Bars

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