помогите решить y''-yy'=0

Так.
y''-yy'=0
p(y)=y’ => y’’=pp’
pp’-yp=0 => pp’=yp
1)y'=0 => y=C
2)p’=y => p=∫ydy=y/2+C
y’=y/2+C => dy/(y+C)=dx/2
(1/√C)•arctg(y/√C)=x/2+C₁
arctg(y/√C)=(x/2+C₁)•√C
y/√C=tg((x/2+C₁)•√C)
y/=tg((x/2+C₁)•√C)•√C