ответ:
пошаговое объяснение:
(x^2-1)*y'+2xy^2=0, y(0)=1
(x^2-1)*y'=-2xy^2
y'/y^2=-2x/(x^2-1)
∫y'/y^2 dx=-2∫x/(x^2-1) dx
∫1/y^2 dy=-∫1/(x^2-1) d(x^2-1)
-1/y +c2=-ln|x^2-1|+c1
1/y=ln|x^2-1|+c3
y=1/[ln|x^2-1|+c3]
y(0)=1/[ln|0^2-1|+c3]=1/[ln|-1|+c3]=1/[ln|1|+c3]=1/[0+c3]=1/c3=1 => c3=1
y=1/[ln|x^2-1|+1]
