# 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

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

I.  Consider first the equation

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

for which we suppose that  $p\mapsto f(p)$  and its derivative  $p\mapsto f^{\prime}(p)$  are continuous and  $f^{\prime}(p)\neq 0$  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

 $\displaystyle x=f(p)$ (3)

defines from  $[p_{1},\,p_{2}]$  onto  $[x_{1},\,x_{2}]$  a bijection

 $\displaystyle p=g(x)$ (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

 $\displaystyle y=\int g(x)\,dx+C.$ (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,

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

Accordingly, the solution of the given differential equation may be presented on  $[p_{1},\,p_{2}]$  as

 $\displaystyle\begin{cases}\displaystyle x=f(p),\\ \displaystyle y=\int\!p\,f^{\prime}(p)\,dp+C.\end{cases}$ (7)

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

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

where $p$ changes on some interval  $[p_{1},\,p_{2}]$  where $f(p)$ and $f^{\prime}(p)$ are continuous and  $p\cdot f^{\prime}(p)\neq 0$,  in the parametric presentation

 $\displaystyle\begin{cases}\displaystyle x=\int\!\frac{f^{\prime}(p)}{p}\,dp+C,% \\ \displaystyle y=f(p).\end{cases}$ (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

 $\displaystyle x\;=\;f(y,\,p)$ (10)

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

 $\frac{dx}{dy}\;=\;\frac{1}{p}\;=\;f^{\prime}_{y}(y,\,p)\!+\!f^{\prime}_{p}(y,% \,p)\frac{dp}{dy}.$

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

 $\displaystyle\frac{dp}{dy}\;=\;\frac{\frac{1}{p}-f^{\prime}_{y}(y,\,p)}{f^{% \prime}_{p}(y,\,p)}\;:=\;g(y,\,p).$ (11)

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

 $\displaystyle x\;=\;f(y,\,p(y,C))$ (12)
Title derivative as parameter for solving differential equations DerivativeAsParameterForSolvingDifferentialEquations 2013-03-22 18:28:39 2013-03-22 18:28:39 pahio (2872) pahio (2872) 10 pahio (2872) Topic msc 34A05 InverseFunctionTheorem ImplicitFunctionTheorem DAlembertsEquation ClairautsEquation