What is a solution to the differential equation #dy/dx=y^2#?

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Answer 1 · 2016-07-09T15:43:57

#y = 1/ (C-x)#

this is a separable equation which can be re-written as

#1/y^2 dy/dx = 1#

and we can integrate

#int \ 1/y^2 dy/dx \ dx = int dx#

or, if you like

#int \ 1/y^2 \ dy = int dx#

so

#- 1/y = x - C#

#y = 1/ (C-x)#