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Решите задачу:

$3y'y''=y+(y')^3+1\\y'=z\\y''=z'z\\3z^2z'=y+z^3+1\\3z^2z'-z^3=y+1\\t=z^3;t'=3z^2z'=\ \textgreater \ z'=\frac{t'}{3z^2}\\t'-t=y+1\\t=uv;t'=u'v+v'u\\u'v+v'u-uv=y+1\\u'v+u(v'-v)=y+1\\\begin{cases}v'-v=0\\u'v=y+1\end{cases}\\\frac{dv}{dy}-v=0\\\frac{dv}{dy}=v|*\frac{dy}{v}\\\frac{dv}{v}=dy\\\int\frac{dv}{v}=\int dy\\ln|v|=y\\v=e^y\\u'e^y=y+1\\du=e^{-y}(y+1)dy\\\int du=\int e^{-y}(y+1)dy\\u=-e^{-y}(y+1)-e^{-y}+C_1\\t=-y-2+C_1e^y$

$z^3=-y-2+C_1e^y\\z=\sqrt[3]{-y-2+C_1e^y}\\y'=\sqrt[3]{-y-2+C_1e^y}\\0=\sqrt[3]{2-2+C_1e^{-2}}\\0=\sqrt[3]{C_1e^{-2}}\\C_1e^{-2}=0\\C_1=0\\y'=-\sqrt[3]{y+2}\\\frac{dy}{dx}=-\sqrt[3]{y+2}\\-\frac{dy}{\sqrt[3]{y+2}}=dx\\-\int \frac{dy}{\sqrt[3]{y+2}}=\int dx\\-\frac{3(y+2)^\frac{2}{3}}{2}=x+C_2\\-\frac{3(-2+2)^\frac{2}{3}}{2}=0+C_2\\C_2=0\\-\frac{3(y+2)^\frac{2}{3}}{2}=x\\(y+2)^\frac{2}{3}=-\frac{2x}{3}\\y+2=\sqrt{(-\frac{2x}{3})^3}\\y=\sqrt{(-\frac{2x}{3})^3}-2$

$y=\sqrt{(-\frac{2x}{3})^3}-2\\y'=-\sqrt{(-\frac{2x}{3})}\\y''=\frac{1}{3\sqrt{(-\frac{2x}{3})}}\\3*-\sqrt{(-\frac{2x}{3})}*\frac{1}{3\sqrt{(-\frac{2x}{3})}}=\sqrt{(-\frac{2x}{3})^3}-2-\sqrt{(-\frac{2x}{3})^3}+1\\-1=-1$