# 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

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

Removing the denominators, we obtain

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

The left hand side can be factored:

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

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

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

$${y}^{3}=3Cx+6{C}^{2}.$$ |

The second factor in (2) yields $6{p}^{2}y=-1$, 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

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

Finally, we can write it

$$3{x}^{2}=-8{y}^{3},$$ |

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

Title | example of derivative as parameter |
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Canonical name | ExampleOfDerivativeAsParameter |

Date of creation | 2013-03-22 18:29:03 |

Last modified on | 2013-03-22 18:29:03 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 6 |

Author | pahio (2872) |

Entry type | Example |

Classification | msc 34A05 |

Synonym | example of solving an ODE |