Ordinary Differential Equations/Without x or y

Equations without y
Consider a differential equation of the form

$$F(x, y')=0$$.

If we can solve for y', then we can simply integrate the equation to get the a solution in the form y=f(x). However, sometimes it may be easier to solve for x. In that case, we get

$$x=f(y')$$

Then differentiating by y,

$${1 \over y'}={df \over dy'}{dy' \over dy}$$

Which makes it become

$$y=C+\int y' {df \over dy'} dy'$$.

The two equations

$$x=f(y')$$

and

$$y=C+\int y' {df \over dy'} dy'$$

is a parametric solution in terms of y'. To obtain an explicit solution, we eliminate y' between the two equations.

If it is possible to express

$$F(x, y')=0$$

parametrically as $$x=f(t), y'=g(t)$$,

then one can differentiate the first equation:

$$\frac{1}{y'}\frac{dy}{dt}=f'(t)$$

So that

$$y=C+\int g(t)f'(t)dt$$

to obtain a parametric solution in terms of $$t$$. If it is possible to eliminate $$t$$, then one can obtain an integral solution.

Equations without x
Similarly, if the equation

$$F(y, y')=0$$.

can be solved for y, write y=f(y'). Then the following solution, which can be obtained by the same process as above is the parametric solution:

$$y=f(y')$$

$$x=C+\int \frac{f'(y')}{y'} dy'$$

In addition, if one can express y and y' parametrically

$$y=f(t), y'=g(t),$$

then the parametric solution is

$$y=f(t),$$

$$x=C+\int \frac {f'(t)}{g(t)} dt$$

so that if the parameter $$t$$ can be eliminated, then one can obtain an integral solution.