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)

Title	derivative as parameter for solving differential equations
Canonical name	DerivativeAsParameterForSolvingDifferentialEquations
Date of creation	2013-03-22 18:28:39
Last modified on	2013-03-22 18:28:39
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	10
Author	pahio (2872)
Entry type	Topic
Classification	msc 34A05
Related topic	InverseFunctionTheorem
Related topic	ImplicitFunctionTheorem
Related topic	DAlembertsEquation
Related topic	ClairautsEquation