Find the general solution of y'' + 2y (y')^3 = 0
Find the general solution of $y'' + 2y\,(y')^3 = 0$.
Explanation of Method
Reduce the order via $p = y'$ and the chain-rule identity $y'' = p\,dp/dy$.
Step 1: Substitution
Let $p = y'$. Express $y''$ in terms of $p$ as a function of $y$ using the chain rule: \[y'' = \frac{dp}{dx} = \frac{dp}{dy}\cdot\frac{dy}{dx} = p\,\frac{dp}{dy}.\]
The equation becomes: \[p\,\frac{dp}{dy} + 2y\,p^3 = 0. \tag{1}\]
Step 2: Separate variables
Case 1 · p = 0.
Then $y' = 0$, giving the trivial solution $y = \mathrm{const}$.
Case 2 · p ≠ 0.
Divide (1) by $p$: $\frac{dp}{dy} + 2y\,p^2 = 0.$
Separate variables: $\frac{dp}{p^2} = -2y\,dy.$
Step 3: Integrate both sides
\[ \int p^{-2}\,dp = -\int 2y\,dy,\qquad -\frac{1}{p} = -y^2 + C_1,\qquad \frac{1}{p} = y^2 - C_1. \]
Thus: \[p = y' = \frac{1}{y^2 - C_1}. \tag{2}\]
Step 4: Second integration
Separate variables in (2): $(y^2 - C_1)\,dy = dx.$
Integrate both sides: $\frac{y^3}{3} - C_1 y = x + C_2.$
Step 5: General solution
where $C_1$ and $C_2$ are arbitrary constants.
Step 6: Verification
Differentiate the general solution implicitly with respect to $x$: \[y^2\,y' - C_1\,y' = 1,\qquad y'(y^2 - C_1) = 1,\qquad y' = \frac{1}{y^2 - C_1}.\]
Differentiate again: $y'' = -\frac{2y\,y'}{(y^2-C_1)^2} = -\frac{2y}{(y^2-C_1)^3}.$
Check $y'' + 2y(y')^3$: $-\frac{2y}{(y^2-C_1)^3} + 2y\cdot\frac{1}{(y^2-C_1)^3} = 0.$ $\checkmark$
Result
Final answer:
\[\boxed{\,\frac{y^3}{3} - C_1\,y = x + C_2\,}\]