# 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.