1.1 Introduction

This method is meant to be used for solving first order ode’s that are non-linear.  Let the ode be

\[ f\left ( x,y,y^{\prime }\right ) =0 \]

In the parametric method we let \(y^{\prime }=p\) and let \(x,y\) be functions of parameter \(p\). Hence \(x\equiv x\left ( p\right ) ,y\equiv y\left ( p\right ) \). The above becomes

\begin{equation} f\left ( x,y,p\right ) =0 \tag {1}\end{equation}

To find the solution of the original ode, the idea is to generate two equations in \(x\left ( p\right ) ,y\left ( p\right ) \) and use these two equations to eliminate \(p\).

The first equation is easy to find. It is found by either isolating \(x\left ( p\right ) \) or \(y\left ( p\right ) \) the original ode itself. The second equation is differential equation, either in \(\frac {dy}{dp}\) or \(\frac {dx}{dp}\). Which one to find depends if we have isolated \(x\) or \(y\) at the start. If we have isolated \(x\) then we need to generate \(\frac {dy}{dp}\) equation in order to solve it for \(y\left ( p\right ) \). If we have isolated \(y\) instead, then we need to find \(\frac {dx}{dp}\) equation to solve it for \(x\left ( p\right ) \). But how to obtain \(\frac {dy}{dp}\)or \(\frac {dx}{dp}\)?  This is described below.

Let us assume we wanted to generate \(\frac {dx}{dp}\) ode. Then taking derivative of the original ode w.r.t. \(p\) gives

\begin{align} 0 & =\frac {\partial f\left ( x,y,p\right ) }{\partial p}\nonumber \\ & =\frac {df}{dp}+\frac {\partial f}{\partial x}\frac {dx}{dp}+\frac {\partial f}{\partial y}\frac {dy}{dp}\nonumber \\ & =\frac {df}{dp}+\frac {\partial f}{\partial x}\frac {dx}{dp}+\frac {\partial f}{\partial y}\frac {dy}{dx}\frac {dx}{dp}\nonumber \\ & =\frac {df}{dp}+\frac {\partial f}{\partial x}\frac {dx}{dp}+p\frac {\partial f}{\partial y}\frac {dx}{dp}\nonumber \\ -\frac {\partial f}{\partial x}\frac {dx}{dp}-p\frac {\partial f}{\partial y}\frac {dx}{dp} & =\frac {df}{dp}\nonumber \\ \frac {dx}{dp}\left ( -\frac {\partial f}{\partial x}-p\frac {\partial f}{\partial y}\right ) & =\frac {df}{dp}\nonumber \\ \frac {dx}{dp} & =\frac {\frac {df}{dp}}{-\frac {\partial f}{\partial x}-p\frac {\partial f}{\partial y}}\nonumber \\ & =\frac {-\frac {df}{dp}}{\frac {\partial f}{\partial x}+p\frac {\partial f}{\partial y}} \tag {2A}\end{align}

If instead we wanted \(\frac {dy}{dp}\), then instead we do the following

\begin{align} 0 & =\frac {\partial f\left ( x,y,p\right ) }{\partial p}\nonumber \\ & =\frac {df}{dp}+\frac {\partial f}{\partial x}\frac {dx}{dp}+\frac {\partial f}{\partial y}\frac {dy}{dp}\nonumber \\ & =\frac {df}{dp}+\frac {\partial f}{\partial x}\frac {dx}{dy}\frac {dy}{dp}+\frac {\partial f}{\partial y}\frac {dy}{dp}\nonumber \\ & =\frac {df}{dp}+\frac {\partial f}{\partial x}\frac {1}{p}\frac {dy}{dp}+\frac {\partial f}{\partial y}\frac {dy}{dp}\nonumber \\ -\frac {\partial f}{\partial x}\frac {1}{p}\frac {dy}{dp}-\frac {\partial f}{\partial y}\frac {dy}{dp} & =\frac {df}{dp}\nonumber \\ \frac {dy}{dp}\left ( -\frac {\partial f}{\partial x}\frac {1}{p}-\frac {\partial f}{\partial y}\right ) & =\frac {df}{dp}\nonumber \\ \frac {dy}{dp} & =\frac {\frac {df}{dp}}{-\frac {\partial f}{\partial x}\frac {1}{p}-\frac {\partial f}{\partial y}}\nonumber \\ & =\frac {p\frac {df}{dp}}{-\frac {\partial f}{\partial x}-p\frac {\partial f}{\partial y}}\nonumber \\ & =\frac {-p\frac {df}{dp}}{p\frac {\partial f}{\partial y}+\frac {\partial f}{\partial x}} \tag {3A}\end{align}

Eqs. (2A) or Eq. (3A) can be used as the second equation we talked above above.

It is important to note the following. If given the original ode \(f\left ( x,y,p\right ) =0\) and we decided to isolate \(x\) to obtain \(x=f\left ( y,p\right ) \) then for the second equation we must use (3A) and now it becomes

\begin{equation} \frac {dy}{dp}=\frac {p\frac {df}{dp}}{1-p\frac {\partial f}{\partial y}} \tag {3A}\end{equation}

Then we use the solution \(y\left ( p\right ) \) of the above, with \(x=f\left ( y,p\right ) \,\) to eliminate \(p\) and  find \(y\). The ode generated using (3A) should be simple to solve (quadrature or separable).

If instead, we have an ode \(f\left ( x,y,p\right ) =0\) and we isolated \(y\) instead to obtain \(y=f\left ( x,p\right ) \) then we must now use (2A) to obtain the second equation. (2A) now becomes

\begin{equation} \frac {dx}{dp}=\frac {\frac {df}{dp}}{p-\frac {\partial f}{\partial x}} \tag {2A}\end{equation}

Then we use the solution \(x\left ( p\right ) \) in the above. This gives us the second equation, and with the equation \(y=f\left ( x,p\right ) \,\) to eliminate \(p\) and  find \(y\).

In summary: we use one of these two equations

\begin{align} x & =f\left ( y,p\right ) \nonumber \\ \frac {dy}{dp} & =\frac {p\frac {df}{dp}}{1-p\frac {\partial f}{\partial y}} \tag {3A}\end{align}

Or use the following two equations

\begin{align} y & =f\left ( x,p\right ) \nonumber \\ \frac {dx}{dp} & =\frac {\frac {df}{dp}}{p-\frac {\partial f}{\partial x}} \tag {2A}\end{align}

Once we solve the ode in one of the above two cases, next we have to elminate \(p\) from these. Once \(p\) is elminated, then the solution \(y\) is found. Examples below help illustrate this method.

Sometimes it is easier to isolate \(x\) and use (3A) and sometimes it is easier to isolate \(y\) and use (2A). In theory, both should give same answer, but eliminating \(p\) can be easier using one method compared to the other. Only way is to try and find out. If it is not possible to isolate \(x\) nor \(y\) from the original ode, then this method will not work.

The main difficulties in this method are not in solving the ode (3A) nor the ode (2A) (typically these come out to be basic types) but in eliminating \(p\) from the two equations we obtained. If we ar unable to isolate \(p\) then we just leave the solution in terms of the two equations as is in a parametric form. Normally \(p\) is labeled as \(t\) at the end in this case.

This method only works if in (2A) \(p-\frac {\partial f}{\partial x}\neq 0\) and in (3A) \(1-p\frac {\partial f}{\partial y}\neq 0\).

As always in math, the best way to learn a method is to solve some examples.