3.5.3.9 Example 9

\(y=\frac {x}{a}y^{\prime }+\frac {b}{ay^{\prime }}\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) which gives

\begin{align} y & =\frac {x}{a}p+\frac {b}{a}p^{-1}\tag {1}\\ & =xf+g\nonumber \end{align}

Where \(f=\frac {p}{a},g=\frac {b}{a}p^{-1}\).  Since \(f\left ( p\right ) \neq p\) then this is d’Almbert ode. Taking derivative w.r.t. \(x\) gives

\begin{align*} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \\ p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\\ p-f & =\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\end{align*}

Using values for \(f,g\) the above simplifies to

\begin{equation} p-\frac {p}{a}=\left ( \frac {x}{a}-\frac {b}{a}p^{-2}\right ) \frac {dp}{dx} \tag {2A}\end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(p=0\). Substituting this in (1) does not generate any solutions due to division by zero. Hence no singular solution exist.

The general solution is found by finding \(p\) from (2A). Since (2A) is not linear in \(p\), then inversion is needed. Writing (2A) as

\[ \frac {p\left ( 1-\frac {1}{a}\right ) }{\frac {x}{a}-\frac {b}{a}p^{-2}}=\frac {dp}{dx}\]

Since this is nonlinear, then inversion is needed

\begin{align*} \frac {dx}{dp} & =\frac {\frac {x}{a}-\frac {b}{a}p^{-2}}{p\left ( 1-\frac {1}{a}\right ) }\\ \frac {dx}{dp}-x\frac {1}{p\left ( a-1\right ) } & =-\frac {b}{a}\frac {1}{p^{3}\left ( 1-\frac {1}{a}\right ) }\end{align*}

This is now linear ode in \(x\left ( p\right ) \). The solution is

\begin{equation} x=\frac {b}{(2a-1)p^{2}}+C_{1}p^{\frac {1}{a-1}} \tag {3}\end{equation}

There are now two choices to take. The first is by solving for \(p\) from the above in terms of \(x\) and then substituting the result in (1) to obtain explicit solution for \(y\left ( x\right ) \), and the second choice is by solving for \(p\) algebraically from (1) and substituting the result in (3). The second choice is easier in this case but gives an implicit solution. Solving for \(p\) from (1) gives

\begin{align*} p_{1} & =\frac {ay+\sqrt {a^{2}y^{2}-4xb}}{2x}\\ p_{1} & =\frac {ay-\sqrt {a^{2}y^{2}-4xb}}{2x}\end{align*}

Substituting each one of these solutions back in (3) gives two implicit solutions

\begin{align*} x & =\frac {b}{(2a-1)\left ( \frac {ay+\sqrt {a^{2}y^{2}-4xb}}{2x}\right ) ^{2}}+C_{1}\left ( \frac {ay+\sqrt {a^{2}y^{2}-4xb}}{2x}\right ) ^{\frac {1}{a-1}}\\ x & =\frac {b}{(2a-1)\left ( \frac {ay-\sqrt {a^{2}y^{2}-4xb}}{2x}\right ) ^{2}}+C_{1}\left ( \frac {ay-\sqrt {a^{2}y^{2}-4xb}}{2x}\right ) ^{\frac {1}{a-1}}\end{align*}