Answer
$y(x)=(C_1e^x+C_2)^{\frac{1}{3}}$
Work Step by Step
We are given:
$$y''+2y^{-1}(y')^2=y'$$
Suppose that $a= \dfrac{dy}{dx}$ and $ \dfrac{da}{dx}= \dfrac{d^2y}{dx^2}$
Integrating factor; $I.P.=e^{\int \frac{2}{y} dy}=e^{2 \ln y}=y^{2}$
Now, $\dfrac{d}{dx}(ay^{2})=y^2$
Integrate to obtain: $ay^{2}=y^3+C \implies dx=\frac{y^2dy}{y^3+C}$
Continue to integrate:
$x+C_1=\frac{1}{3}\ln(y^3+C_2)$
$y^3=C_1e^x+C_2$
Therefore, the general solution is: $y(x)=(C_1e^x+C_2)^{\frac{1}{3}}$
