What is a general solution to the differential equation #(y^2+xy^2)y'=1#?

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Answer 1 · 2016-07-11T17:55:50

# y^3/3 =ln (1+x) + C#

separate it

#(y^2+xy^2)y'=1#

#y^2(1+x)y'=1#

#y^2\ y'=1/(1+x)#

#int y^2\ y' \ dx =int \ 1/(1+x) \ dx#

#int y^2\ dy =int \ 1/(1+x) \ dx#

# y^3/3 =ln (1+x) + C#

personally i'd leave it there but you can "tidy" ie clutter it more as

# y =root3( 3ln (1+x) + C)#