Ordinary Differential Equations/Without x or y

Equations without y

Consider a differential equation of the form

$F(x,y^{\prime })=0$.

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

$x=f(y^{\prime })$

Then differentiating by y,

$\frac{1}{y^{\prime }}=\frac{df}{d{y}^{\prime }}\frac{d{y}^{\prime }}{dy}$

Which makes it become

$y=C+\int y^{\prime }\frac{df}{d{y}^{\prime }}d{y}^{\prime }$.

The two equations

$x=f(y^{\prime })$

and

$y=C+\int y^{\prime }\frac{df}{d{y}^{\prime }}d{y}^{\prime }$

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

$F(x,y^{\prime })=0$

parametrically as $x=f(t),y^{\prime }=g(t)$,

then one can differentiate the first equation:

$\frac{1}{y^{\prime }}\frac{dy}{dt}=f^{\prime }(t)$

So that

$y=C+\int g(t)f^{\prime }(t)dt$

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

$F(y,y^{\prime })=0$.

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:

$y=f(y^{\prime })$

$x=C+\int \frac{f^{\prime }(y^{\prime })}{y^{\prime }}d{y}^{\prime }$

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

$y=f(t),y^{\prime }=g(t),$

then the parametric solution is

$y=f(t),$

$x=C+\int \frac{f^{\prime }(t)}{g(t)}dt$

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