How to find the general solution of #x dy/dx = -y^2# ?

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Answer 1 · 2018-05-20T12:55:45

The general solution is #y=-1/(C-lnx)#

The ODE is

#xdy/dx=-y^2#

Rearranging the terms

#dy/y^2=-dx/x#

#intdy/y^2=-intdx/x#

#-1/y=-lnx+C#

#y=-1/(-lnx+C)=-1/(C-lnx)#