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

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

Step 1: Substitution

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

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

Step 2: Reduce to a homogeneous equation

Divide (1) by $p$ ($p \neq 0$): \[\frac{dp}{dy} = \frac{p}{3y-2p}. \tag{2}\]

This is homogeneous in $y,p$. Set $v = \dfrac{p}{y}$, so $p = vy$, $\dfrac{dp}{dy} = v + y\dfrac{dv}{dy}$: \[v + y\frac{dv}{dy} = \frac{v}{3-2v},\qquad y\frac{dv}{dy} = \frac{2v(v-1)}{3-2v}.\]

Step 3: Separate and integrate

\[ \frac{3-2v}{2v(v-1)}\,dv = \frac{dy}{y}. \]

Partial fractions give $\dfrac{3-2v}{2v(v-1)} = -\dfrac{3}{2v}+\dfrac{1}{2(v-1)}$. Integrate: \[-\frac{3}{2}\ln|v| + \frac{1}{2}\ln|v-1| = \ln|y| + \mathrm{const},\qquad \frac{v-1}{v^3} = C_1\,y^2.\]

Step 4: Return to $y$ and $p$

Substitute back $v = p/y$: $\dfrac{v-1}{v^3} = \dfrac{y^2(p-y)}{p^3}$. Thus: \[\frac{p-y}{p^3} = C_1 \implies y = p - C_1\,p^3. \tag{3}\]

Relabelling $-C_1 \to C_1$, this reads $y = p + C_1 p^3$.

Step 5: Find $x$ as a function of $p$

From (3), $\dfrac{dy}{dp} = 1 + 3C_1 p^2$. Since $\dfrac{dy}{dx}=p$, we have $dx = \dfrac{dy}{p} = \dfrac{1}{p}\dfrac{dy}{dp}\,dp$: $dx = \left(\frac{1}{p} + 3C_1 p\right)dp.$

Integrate: \[x = \ln|p| + \frac{3C_1}{2}p^2 + C_2. \tag{4}\]

Step 6: General solution

The general solution is given parametrically by $p$: where $C_1$ and $C_2$ are arbitrary constants.

Step 7: Verification

From $x(p)$ and $y(p)$: $\dfrac{dx}{dp} = \dfrac{1}{p}+3C_1p$, $\dfrac{dy}{dp}=1+3C_1p^2$. Then: \[y' = \frac{dy/dp}{dx/dp} = \frac{1+3C_1p^2}{\frac{1}{p}+3C_1p} = \frac{p(1+3C_1p^2)}{1+3C_1p^2} = p.\ \checkmark\]

Differentiate $y'=p$ with respect to $x$: \[y'' = \frac{dp/dp}{dx/dp} = \frac{1}{\frac{1}{p}+3C_1p} = \frac{p}{1+3C_1p^2}.\]

Check $(y')^2 = (3y-2y')y''$. Since $3y-2p = 3(p+C_1p^3)-2p = p+3C_1p^3 = p(1+3C_1p^2)$: \[(3y-2y')y'' = p(1+3C_1p^2)\cdot\frac{p}{1+3C_1p^2} = p^2 = (y')^2.\ \checkmark\]