# derivative as parameter for solving differential equations

The solution of some differential equations^{} of the forms $x=f(\frac{dy}{dx})$ and $y=f(\frac{dy}{dx})$ may be expressed in a parametric form by taking for the parameter the derivative

$p:={\displaystyle \frac{dy}{dx}}.$ | (1) |

I. Consider first the equation

$x=f({\displaystyle \frac{dy}{dx}}),$ | (2) |

for which we suppose that $p\mapsto f(p)$ and its derivative $p\mapsto {f}^{\prime}(p)$ are continuous^{} and
${f}^{\prime}(p)\ne 0$ on an interval $[{p}_{1},{p}_{2}]$. It follows that on the interval, the function^{} $p\mapsto f(p)$ changes monotonically from $f({p}_{1}):={x}_{1}$ to $f({p}_{2}):={x}_{2}$, whence conversely the equation

$x=f(p)$ | (3) |

defines from $[{p}_{1},{p}_{2}]$ onto $[{x}_{1},{x}_{2}]$ a bijection

$p=g(x)$ | (4) |

which is continuously differentiable. Thus on the interval $[{x}_{1},{x}_{2}]$, the differential equation (2) can be replaced by the equation

$\frac{dy}{dx}}=g(x),$ | (5) |

and therefore, the solution of (2) is

$y={\displaystyle \int g(x)\mathit{d}x}+C.$ | (6) |

If we cannot express $g(x)$ in a , we take $p$ as an independent variable through the substitution (3), which maps $[{x}_{1},{x}_{2}]$ bijectively onto $[{p}_{1},{p}_{2}]$. Then (6) becomes a function of $p$, and by the chain rule^{},

$$\frac{dy}{dp}=g(f(p)){f}^{\prime}(p)=p{f}^{\prime}(p).$$ |

Accordingly, the solution of the given differential equation may be presented on $[{p}_{1},{p}_{2}]$ as

$\{\begin{array}{cc}x=f(p),\hfill & \\ y={\displaystyle \int p{f}^{\prime}(p)\mathit{d}p}+C.\hfill & \end{array}$ | (7) |

II. With corresponding considerations, one can write the solution of the differential equation

$y=f({\displaystyle \frac{dy}{dx}}):=f(p),$ | (8) |

where $p$ changes on some interval $[{p}_{1},{p}_{2}]$ where $f(p)$ and ${f}^{\prime}(p)$ are continuous and $p\cdot {f}^{\prime}(p)\ne 0$, in the parametric presentation

$\{\begin{array}{cc}x={\displaystyle \int \frac{{f}^{\prime}(p)}{p}\mathit{d}p}+C,\hfill & \\ y=f(p).\hfill & \end{array}$ | (9) |

III. The procedures of I and II may be generalised for the differential equations of $x=f(y,p)$ and $y=f(x,p)$; let’s consider the former one.

In

$x=f(y,p)$ | (10) |

we regard $y$ as the independent variable and differentiate with respect to it:

$$\frac{dx}{dy}=\frac{1}{p}={f}_{y}^{\prime}(y,p)+{f}_{p}^{\prime}(y,p)\frac{dp}{dy}.$$ |

Supposing that the partial derivative^{} ${f}_{p}^{\prime}(y,p)$ does not vanish identically, we get

$\frac{dp}{dy}}={\displaystyle \frac{\frac{1}{p}-{f}_{y}^{\prime}(y,p)}{{f}_{p}^{\prime}(y,p)}}:=g(y,p).$ | (11) |

If $p=p(y,C)$ is the general solution of (11), we obtain the general solution of (10):

$x=f(y,p(y,C))$ | (12) |

Title | derivative as parameter for solving differential equations |
---|---|

Canonical name | DerivativeAsParameterForSolvingDifferentialEquations |

Date of creation | 2013-03-22 18:28:39 |

Last modified on | 2013-03-22 18:28:39 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 10 |

Author | pahio (2872) |

Entry type | Topic |

Classification | msc 34A05 |

Related topic | InverseFunctionTheorem |

Related topic | ImplicitFunctionTheorem |

Related topic | DAlembertsEquation |

Related topic | ClairautsEquation |