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Homederivative as parameter for solving differential equations

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# 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

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

I. Consider first the equation

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

for which we suppose that $p\mapsto f(p)$ and its derivative $p\mapsto f^{{\prime}}(p)$ are continuous and $f^{{\prime}}(p)\neq 0$ 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

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

defines from $[p_{1},\,p_{2}]$ onto $[x_{1},\,x_{2}]$ a bijection

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

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

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

and therefore, the solution of (2) is

$\displaystyle y=\int g(x)\,dx+C.$ | (6) |

If we cannot express $g(x)$ in a closed form, 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,

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

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

$\displaystyle\begin{cases}\displaystyle x=f(p),\\ \displaystyle y=\int\!p\,f^{{\prime}}(p)\,dp+C.\end{cases}$ | (7) |

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

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

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

$\displaystyle\begin{cases}\displaystyle x=\int\!\frac{f^{{\prime}}(p)}{p}\,dp+% C,\\ \displaystyle y=f(p).\end{cases}$ | (9) |

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

In

$\displaystyle x\;=\;f(y,\,p)$ | (10) |

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

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

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

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

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

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

## Mathematics Subject Classification

34A05*no label found*

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