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# example of derivative as parameter

For solving the (nonlinear) differential equation

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

with $p=\frac{dy}{dx}$, according to III in the parent entry, 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:

$\displaystyle(y\frac{dp}{dy}+2p)(1+6p^{2}y)=0$ | (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 $\displaystyle x=\frac{y^{3}}{3C}-2C$. 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).

## Mathematics Subject Classification

34A05*no label found*

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