derivative as parameter for solving differential equations

The solution of some differential equations of the forms  $x=f(\frac{dy}{dx})$  and  $y=f(\frac{dy}{dx})$  may be expressed in a parametric form by taking for the parameter the derivative

$p:={\displaystyle \frac{dy}{dx}}.$(1)

I.  Consider first the equation

$x=f({\displaystyle \frac{dy}{dx}}),$(2)

for which we suppose that  $p\mapsto f(p)$  and its derivative  $p\mapsto f^{\prime }(p)$  are continuous and $f^{\prime }(p)\ne 0$  on an interval  $[p_1,p_2]$.  It follows that on the interval, the function  $p\mapsto f(p)$  changes monotonically from  $f(p_1):=x_1$  to  $f(p_2):=x_2$, whence conversely the equation

$x=f(p)$(3)

defines from  $[p_1,p_2]$  onto  $[x_1,x_2]$  a bijection

$p=g(x)$(4)

which is continuously differentiable.  Thus on the interval  $[x_1,x_2]$,  the differential equation (2) can be replaced by the equation

${\displaystyle \frac{dy}{dx}}=g(x),$(5)

and therefore, the solution of (2) is

$y={\displaystyle \int g(x)𝑑x}+C.$(6)

If we cannot express $g(x)$ in a , we take $p$ as an independent variable through the substitution (3), which maps  $[x_1,x_2]$  bijectively onto  $[p_1,p_2]$.  Then (6) becomes a function of $p$, and by the chain rule,

$$\frac{dy}{dp}=g(f(p))f^{\prime }(p)=p{f}^{\prime }(p).$$

Accordingly, the solution of the given differential equation may be presented on  $[p_1,p_2]$  as

$\{\begin{array}{cc}x=f(p),\hfill & \\ y={\displaystyle \int p{f}^{\prime }(p)𝑑p}+C.\hfill & \end{array}$(7)

II.  With corresponding considerations, one can write the solution of the differential equation

$y=f({\displaystyle \frac{dy}{dx}}):=f(p),$(8)

where $p$ changes on some interval  $[p_1,p_2]$  where $f(p)$ and $f^{\prime }(p)$ are continuous and $p\cdot f^{\prime }(p)\ne 0$,  in the parametric presentation

$\{\begin{array}{cc}x={\displaystyle \int \frac{f^{\prime }(p)}{p}𝑑p}+C,\hfill & \\ y=f(p).\hfill & \end{array}$(9)

III.  The procedures of I and II may be generalised for the differential equations of  $x=f(y,p)$  and  $y=f(x,p)$;  let’s consider the former one.

In

$x=f(y,p)$(10)

we regard $y$ as the independent variable and differentiate with respect to it:

$$\frac{dx}{dy}=\frac{1}{p}=f_y^{\prime }(y,p)+f_p^{\prime }(y,p)\frac{dp}{dy}.$$

Supposing that the partial derivative  $f_p^{\prime }(y,p)$ does not vanish identically, we get

${\displaystyle \frac{dp}{dy}}={\displaystyle \frac{\frac{1}{p}-f_y^{\prime }(y,p)}{f_p^{\prime }(y,p)}}:=g(y,p).$(11)

If  $p=p(y,C)$  is the general solution of (11), we obtain the general solution of (10):

$x=f(y,p(y,C))$(12)