Find general and special solutions to $$y= 2xy'-3(y')^2.$$

Substituting $y'=p$, we get an equation of the form $y=2xp–3{p^2}$

Differentiating both sides, we get

$dy=2xdp+2pdx-6pdp$

$dy=pdx$ $\Rightarrow$ $pdx=2xdp+2pdx-6pdp$ $\Rightarrow$ $–pdx=2xdp–6pdp$

Dividing by $p$, we get $-dx={2x \over p}dp-6dp$ $\Rightarrow$ $ {dx \over dp} + {2 \over p}x-6=0 \tag{1}$

We get a linear differential equation. To solve $(1)$, we use an integrating factor given by $\mu=e^{\int{2 \over p}dp}=e^{2\ln|p|}=p^2$

Rewrite $(1)$ as ${2 \over p}x+{dx \over dp}=6$ and multiply by the integrating factor, $\mu(p)=p^2$ to get,

$$p^2{2 \over p}x+p^2{dx \over dp}=p^2 6$$

$${d \over dp}(p^2 x)=p^2 6$$

$$x={{2p^3+C} \over p^2}$$

Thus, the general solution to (1) is

$$x={{2p + {C \over p^2}}}\tag{2}$$

Substituting $(2)$ into the Lagrange Equation, we get

$$y={2\left({2p+\frac{C}{{{p^2}}}}\right)p–3p^2}=p^2+ \frac{2C}p$$

Thus,

\begin{equation} \left\{\begin{aligned} &x={{2p + {C \over p^2}}}\\ &y=p^2+ \frac{2C}p \end{aligned}\right. \tag{3}\end{equation}

gives us a solution in the parametric form.

The Lagrange equation, $y= 2xy'-3(y')^2$, can also have a special solution (or solutions)

$\varphi (p)-p=0$

$2p–p=0$ $\Rightarrow$ $p(2-p)=0$ $\Rightarrow$ $p_1=0$ and $p_2=2$

Thus, the special solutions are

$$y=x\varphi (0)+\psi (0)=x \cdot 0+0=0.$$