Linear Differentiation: 1 = xy = xy', y' - y = 1/x

paulxzt

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Aug 30, 2006
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1 = xy = xy'

y' - y = 1/x

I figured out I(x) to be e^-x and multiplied it into the equation above to get:

e^-x*y' - e^-x*y = e^-x(1/x)
(e^-x*y)' = e^-x (1/x)
Integrate both sides

e^x*y = S e^-x(1/x)

can someone help me integrate the right side? i used integration by parts with u = 1/x and dV = e^-x.

du = -dx/x^2 , V = -e^-x

-(1/x)*e^-x - integral of (e^-x)/x^2 dx

is there an easier way to integrate this from the beginning? if i keep using integration by parts again, i end up with the same function i had in the beginning.. which means i'd be integrating the same thing over and over..

thanks
 
paulxzt said:
1 = xy = xy'

y' - y = 1/x
What are you supposed to do with these two equations? What were the instructions?

Thank you! :D

Eliz.
 
Correction:

paulxzt said:
1 + xy = xy'

y' - y = 1/x
Assuming you do not have any other mistakes:

S(1/x)(e^(-x)) dx = ln x - x/(1 * 1!) + x^2/(2 * 2!) - x^3/(3 * 3!) .......
 
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