Find the general solution of 2x y' y'' = (y')^2 - 1

Find the general solution of $2x\,y'\,y'' = (y')^2 - 1$.

Step 1: Substitution

Let $p = y'$, so $p' = y''$. The equation becomes: \[2x\,p\,p' = p^2 - 1. \tag{1}\]

This is a first-order separable ODE in $p(x)$.

Step 2: Separate variables

\[ \frac{2p\,dp}{p^2-1} = \frac{dx}{x}. \]

Step 3: Integrate both sides

Left side: use $u = p^2 - 1$, $du = 2p\,dp$: $\int \frac{2p\,dp}{p^2-1} = \int \frac{du}{u} = \ln|p^2-1|.$

Right side: $\int \frac{dx}{x} = \ln|x| + \ln|C_1|.$

Thus: \[\ln|p^2-1| = \ln|C_1 x|,\qquad p^2 - 1 = C_1 x,\qquad p^2 = 1 + C_1 x.\]

\[ y' = \pm\sqrt{1 + C_1 x}. \tag{2} \]

Step 4: Second integration

Integrate (2) using $u = 1 + C_1 x$, $du = C_1\,dx$: $\int \sqrt{1+C_1 x}\,dx = \frac{1}{C_1}\int \sqrt{u}\,du = \frac{2}{3C_1}(1+C_1 x)^{3/2}.$

Thus: $y = \pm\frac{2}{3C_1}(1+C_1 x)^{3/2} + C_2.$

Step 5: General solution

Absorbing the sign into $C_1$:

Step 6: Verification

Differentiate: $y' = (1+C_1 x)^{1/2},\qquad y'' = \frac{C_1}{2}(1+C_1 x)^{-1/2}.$

Check $2x\,y'\,y'' - (y')^2 + 1$: \[2x\cdot(1+C_1 x)^{1/2}\cdot\frac{C_1}{2}(1+C_1 x)^{-1/2} - (1+C_1 x) + 1 = C_1 x - 1 - C_1 x + 1 = 0.\ \checkmark\]