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
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