Find the general solution of 3y y'' - 5(y')^2 = 0

Find the general solution of $3y\,y'' - 5(y')^2 = 0$.

Step 1: Substitution

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

The equation becomes: \[3y\,p\,\frac{dp}{dy} - 5p^2 = 0. \tag{1}\]

Step 2: Separate variables

Step 3: Integrate both sides

\[ \int \frac{dp}{p} = \frac{5}{3}\int \frac{dy}{y},\qquad \ln|p| = \frac{5}{3}\ln|y| + \ln|A|,\qquad p = A\,y^{5/3}, \] where $A$ is an arbitrary constant. Thus: \[y' = A\,y^{5/3}. \tag{2}\]

Step 4: Second integration

Separate variables in (2): $y^{-5/3}\,dy = A\,dx.$

Integrate both sides: $\int y^{-5/3}\,dy = \int A\,dx,\qquad -\frac{3}{2}\,y^{-2/3} = Ax + C_2.$

Multiply by $-\dfrac{2}{3}$ and rename the constants: $y^{-2/3} = C_1 x + C_2.$

Step 5: General solution

Raise both sides to the power $-\dfrac{3}{2}$: or equivalently $y^{-2/3} = C_1 x + C_2$, where $C_1$ and $C_2$ are arbitrary constants, valid for $C_1 x + C_2 > 0$.

Step 6: Verification

From $y^{-2/3} = C_1 x + C_2$, differentiate implicitly: $-\frac{2}{3}y^{-5/3}\,y' = C_1,\qquad y' = -\frac{3C_1}{2}\,y^{5/3}.$

Differentiate again: \[y'' = -\frac{3C_1}{2}\cdot\frac{5}{3}\,y^{2/3}\,y' = \frac{15C_1^2}{4}\,y^{7/3}.\]

Check $3y\,y'' - 5(y')^2$: \[3y\cdot\frac{15C_1^2}{4}\,y^{7/3} - 5\cdot\frac{9C_1^2}{4}\,y^{10/3} = \frac{45C_1^2}{4}\,y^{10/3} - \frac{45C_1^2}{4}\,y^{10/3} = 0.\ \checkmark\]