ODE types reductible to the variables separable case

There are certain of non-linear ordinary differential equations of first order (http://planetmath.org/ODE) which may by a suitable substitution be to a form where one can separate (http://planetmath.org/SeparationOfVariables) 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 (http://planetmath.org/HomogeneousFunction).  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

${\displaystyle \frac{dy}{dx}}=f\left({\displaystyle \frac{y}{x}}\right).$(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  $x=Cx(t)$,  $y=Cy(t)$  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

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

The equation takes the form

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

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

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

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

$$x=C{e}^{\int \frac{dt}{f(t)-t}},y=Ct{e}^{\int \frac{dt}{f(t)-t}}.$$

II.  Equation of the form  y${}^{\mathrm{\prime }}$= 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

$ax+by+c:=u$(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

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

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

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