example of derivative as parameter

For solving the (nonlinear) differential equation

$x={\displaystyle \frac{y}{3p}}-2p{y}^2$(1)

with  $p=\frac{dy}{dx}$,  according to III in the parent entry (http://planetmath.org/DerivativeAsParameterForSolvingDifferentialEquations), we differentiate both sides in regard to $y$, getting first

$$\frac{1}{p}=\frac{1}{3p}-\left(\frac{y}{3p^2}+2y^2\right)\frac{dp}{dy}-4py.$$

Removing the denominators, we obtain

$$2p+(y+6p^2{y}^2)\frac{dp}{dy}+12p^{3}y=0.$$

The left hand side can be factored:

$(y{\displaystyle \frac{dp}{dy}}+2p)(1+6p^{2}y)=0$(2)

Now we may use the zero rule of product; the first factor of the product in (2) yields  $y\frac{dp}{dy}=-2p$, i.e.

$$2\int \frac{dy}{y}=-\int \frac{dp}{p}+\ln C,$$

whence  $y^2=\frac{C}{p}$,  i.e.  $p=\frac{C}{y^2}$.  Substituting this into the original equation (1) we get  $x={\displaystyle \frac{y^3}{3C}}-2C$.  Hence the general solution of (1) may be written

$$y^3=3Cx+6C^2.$$

The second factor in (2) yields  $6p^{2}y=-1$,  which is substituted into (1) multiplied by $3p$:

$$3px=y-(-y)$$

Thus we see that  $p=\frac{2y}{3x}$, which is again set into (1), giving

$$x=\frac{y\cdot 3x}{3\cdot 2y}-\frac{4y^3}{3x}.$$

Finally, we can write it

$$3x^2=-8y^3,$$

which (a variant of the so-called semicubical parabola) is the singular solution of (1).