Find the general solution of 2y y'' - 3(y')^2 = 4y^2

Find the general solution of $2y\,y'' - 3(y')^2 = 4y^2$.

Step 1: Substitution

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

The equation becomes: \[2y\,p\,\frac{dp}{dy} - 3p^2 = 4y^2. \tag{1}\]

Step 2: Reduce to a linear ODE

Dividing (1) by $y$ gives $2p\dfrac{dp}{dy} = \dfrac{3p^2}{y}+4y$. Try $w = p^2 y^{-3}$: \[\frac{dw}{dy} = 2p\frac{dp}{dy}y^{-3} - 3p^2y^{-4} = \left(\frac{3p^2}{y}+4y\right)y^{-3} - 3p^2y^{-4} = \frac{4}{y^2}.\] So with $w = p^2 y^{-3}$: \[\frac{dw}{dy} = \frac{4}{y^2}. \tag{2}\]

Step 3: Integrate equation (2)

\[ w = \int \frac{4}{y^2}\,dy = -\frac{4}{y} + C_1. \]

Substitute back $w = \dfrac{(y')^2}{y^3}$: \[(y')^2 = y^2(C_1 y - 4). \tag{3}\]

Step 4: Second integration

Take the square root and separate variables: $\frac{dy}{y\sqrt{C_1 y - 4}} = \pm\, dx.$

Use $u = \sqrt{C_1 y - 4}$, so $y = \dfrac{u^2+4}{C_1}$, $dy = \dfrac{2u}{C_1}\,du$: $\frac{dy}{y\sqrt{C_1 y-4}} = \frac{2\,du}{u^2+4}.$

Integrate: \[\int\frac{2\,du}{u^2+4} = \arctan\frac{u}{2} = \arctan\frac{\sqrt{C_1 y-4}}{2}.\]

Thus, absorbing signs into $C_2$: $\sqrt{C_1 y - 4} = 2\tan(x + C_2),\qquad C_1 y - 4 = 4\tan^2(x+C_2).$

Step 5: General solution

where $C_1 \neq 0$ and $C_2$ are arbitrary constants.

Step 6: Verification

Let $\phi = x+C_2$. Then $y' = \dfrac{8\sin\phi}{C_1\cos^3\phi}$ and $y'' = \dfrac{8(\cos^2\phi+3\sin^2\phi)}{C_1\cos^4\phi}$. Check $2yy''-3(y')^2-4y^2$: \[\frac{64(\cos^2\phi+3\sin^2\phi)}{C_1^2\cos^6\phi} - \frac{192\sin^2\phi}{C_1^2\cos^6\phi} - \frac{64\cos^2\phi}{C_1^2\cos^6\phi} = 0.\ \checkmark\]