Find the general solution of 2y y'' + (y')^2 + (y')^4 = 0

Find the general solution of $2y\,y'' + (y')^2 + (y')^4 = 0$.

Step 1: Substitution

Let $p = y'$. Express $y''$ using the chain rule: $y'' = p\,\frac{dp}{dy}.$

The equation becomes: \[2y\,p\,\frac{dp}{dy} + p^2 + p^4 = 0. \tag{1}\]

Step 2: Separate variables

Step 3: Integrate both sides

Decompose $\dfrac{1}{p(1+p^2)} = \dfrac{1}{p} - \dfrac{p}{1+p^2}$ and integrate: \[\ln|p| - \frac{1}{2}\ln(1+p^2) = -\frac{1}{2}\ln|y| + \mathrm{const},\qquad \frac{p^2}{1+p^2} = \frac{C_1}{y}.\]

Solve for $p^2$: \[p^2 y = C_1(1+p^2) \implies p^2(y-C_1) = C_1 \implies p^2 = \frac{C_1}{y-C_1}. \tag{2}\]

Step 4: Second integration

From (2), $y' = \pm\sqrt{\dfrac{C_1}{y-C_1}}$. Separate variables: $\sqrt{y-C_1}\,dy = \pm\sqrt{C_1}\,dx.$

Integrate: $\frac{2}{3}(y-C_1)^{3/2} = \pm\sqrt{C_1}\,(x+C_2).$

Step 5: General solution

or equivalently $(y-C_1)^3 = \dfrac{9C_1}{4}(x+C_2)^2$, where $C_1 > 0$ and $C_2$ are arbitrary constants, valid for $y > C_1$.

Step 6: Verification

From $y' = \sqrt{C_1/(y-C_1)}$, differentiate: $y'' = -\frac{C_1}{2(y-C_1)^2}.$

Check $2yy'' + (y')^2 + (y')^4$: \[\frac{-yC_1}{(y-C_1)^2} + \frac{C_1}{y-C_1} + \frac{C_1^2}{(y-C_1)^2} = \frac{C_1\left[-y+(y-C_1)+C_1\right]}{(y-C_1)^2} = 0.\ \checkmark\]