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# ODE types reductible to the variables separable case

There are certain types of non-linear ordinary differential equations of first order which may by a suitable substitution be reduced to a form where one can separate the variables.

I. So-called homogeneous differential equation

This means the equation of the form

$X(x,\,y)dx+Y(x,\,y)dy=0,$ |

where $X$ and $Y$ are two homogeneous functions of the same degree. Therefore, if the equation is written as

$\frac{dy}{dx}=-\frac{X(x,\,y)}{Y(x,\,y)},$ |

its right hand side is a homogeneous function of degree 0, i.e. it depends only on the ratio $y\!:\!x$, and has thus the form

$\displaystyle\frac{dy}{dx}=f\left(\frac{y}{x}\right).$ | (1) |

Accordingly, if this ratio is constant, then also $\frac{dy}{dx}$ is constant; thus all lines $\frac{y}{x}=$ constant are isoclines of the family of the integral curves which intersect any such line isogonally.

We can infer as well, that if one integral curve is represented by $x=x(t)$, $y=y(t)$, then also $x=Cx(t)$, $y=Cy(t)$ represents an integral curve for any constant $C$. Hence the integral curves are homothetic with respect to the origin; therefore some people call the equation (1) a similarity equation.

For generally solving the equation (1), make the substitution

$\frac{y}{x}:=t;\quad y=tx;\quad\frac{dy}{dx}=t+x\frac{dt}{dx}.$ |

The equation takes the form

$\displaystyle t+x\frac{dt}{dx}=f(t)$ | (2) |

which shows that any root $t_{\nu}$ of the equality $f(t)=t$ gives a singular solution $y=t_{\nu}x$. The variables in (2) may be separated:

$\frac{dx}{x}=\frac{dt}{f(t)\!-\!t}$ |

Thus one obtains $\ln{|x|}=\int\!\frac{dt}{f(t)\!-\!t}+\ln{C}$, whence the general solution of the homogeneous differential equation (1) is in a parametric form

$x=Ce^{{\int\!\frac{dt}{f(t)\!-\!t}}},\quad y=Cte^{{\int\!\frac{dt}{f(t)\!-\!t}% }}.$ |

II. Equation of the form y$\,{}^{{\prime}}$= f(ax+by+c)

It’s a question of the equation

$\displaystyle\frac{dy}{dx}=f(ax+by+c),$ | (3) |

where $a$, $b$ and $c$ are given constants. If $ax+by$ is constant, then $\frac{dy}{dx}$ is constant, and we see that the lines $ax+by=$ constant are isoclines of the intgral curves of (3).

Let

$\displaystyle ax+by+c:=u$ | (4) |

be a new variable. It changes the equation (3) to

$\displaystyle\frac{du}{dx}=a+bf(u).$ | (5) |

Here, one can see that the real zeros $u$ of the right hand side yield lines (4) which are integral curves of (3), and thus we have singular solutions. Moreover, one can separate the variables in (5) and integrate, obtaining $x$ as a function of $u$. Using still (4) gives also $y$. The general solution is

$x=\int\frac{du}{a+bf(u)}+C,\quad y=\frac{1}{b}\left(u-c-a\int\frac{du}{a+bf(u)% }-aC\right).\\$ |

Example. In the nonlinear equation

$\frac{dy}{dx}=(x-y)^{2},$ |

which is of the type II, one cannot separate the variables $x$ and $y$. The substitution $x-y:=u$ converts it to

$\frac{du}{dx}=1-u^{2},$ |

where one can separate the variables. Since the right hand side has the zeros $u=\pm 1$, the given equation has the singular solutions $y$ given by $x-y=\pm 1$. Separating the variables $x$ and $u$, one obtains

$dx=\frac{du}{1-u^{2}},$ |

whence

$x=\int\frac{du}{(1+u)(1-u)}=\frac{1}{2}\int\left(\frac{1}{1+u}+\frac{1}{1-u}% \right)du=\frac{1}{2}\ln\left|\frac{1+u}{1-u}\right|+C.$ |

Accordingly, the given differential equation has the parametric solution

$x=\ln\sqrt{\left|\frac{1\!+\!u}{1\!-\!u}\right|}+C,\quad y=\ln\sqrt{\left|% \frac{1\!+\!u}{1\!-\!u}\right|}-u\!+\!C.$ |

# References

- 1 E. Lindelöf: Differentiali- ja integralilasku III 1. Mercatorin Kirjapaino Osakeyhtiö, Helsinki (1935).

## Mathematics Subject Classification

34A09*no label found*34A05

*no label found*

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