How do you solve the differential #y'xln(x)=y#?

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Answer 1 · 2016-11-18T15:56:35

#y=Ce^( 1/4x^2(2 ln x - 1) #, where #C >0#, and so is y.

To make ln x real, x > 0.

Separating variables and integrating,

#int 1/y dy=int x ln x dx#. So,

#ln y #

#= int ln x d(x^2/2)#

#=1/2x^2 ln x-1/2 int x^2 d(ln x)#

#=1/2x^2 ln x- 1/2 intx^2/x dx#

#=1/2x^2 ln x -x^2/4+A#

So,

#y =e^Ae^( 1/4x^2(2 ln x - 1) #

#=Ce^( 1/4x^2(2 ln x - 1) #, where #C > 0#. and so is y.