# Ordinary Differential Equations/Without x or y

## Equations without y

Consider a differential equation of the form

${\displaystyle F(x,y')=0}$ .

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

${\displaystyle x=f(y')}$

Then differentiating by y,

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

Which makes it become

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

The two equations

${\displaystyle x=f(y')}$

and

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

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

${\displaystyle F(x,y')=0}$

parametrically as ${\displaystyle x=f(t),y'=g(t)}$ ,

then one can differentiate the first equation:

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

So that

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

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

## Equations without x

Similarly, if the equation

${\displaystyle F(y,y')=0}$ .

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:

${\displaystyle y=f(y')}$

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

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

${\displaystyle y=f(t),y'=g(t),}$

then the parametric solution is

${\displaystyle y=f(t),}$

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

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