Решите дифференциальное уравнение: y''+2y(y')^3 =0

[latex]y''+2y(y')^3=0\\\\y'=p(y)\; \to \; \; y''=p\cdot \frac{dp}{dy}\\\\p\cdot \frac{dp}{dy}=-2yp^3\\\\\frac{p\, dp}{p^3}=-2y\, dy\\\\\int \frac{dp}{p^2}=-2\int y\, dy\\\\-\frac{1}{p}=-y^2-C_1\\\\\frac{1}{y'}=y^2+C_1\\\\y'=\frac{1}{y^2+C_1}\\\\\frac{dy}{dx}=\frac{1}{y^2+C_1}\\\\\int (y^2+C_1)dy=\int dx\\\\\frac{y^3}{3}+C_1y=x+C_2[/latex]