3.5.3.19 Example 19

\[ y=xf\left ( p\right ) +g\left ( p\right ) \]

This problem is meant to show what to do when we are unable to solve explicitly for \(x\left ( p\right ) \) when doing inversion. Taking derivative the above becomes

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

Inversion is needed. Hence gives

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

This is now linear in \(x\).

\[ \frac {dx}{dp}-\frac {xf^{\prime }}{p-f}=\frac {g^{\prime }}{p-f}\]

Integrating factor is \(\mu =e^{\int \frac {f^{\prime }\left ( p\right ) }{p-f}dp}\). Hence the above becomes

\begin{align} \frac {d}{dp}\left ( x\mu \right ) & =\mu \frac {g^{\prime }}{p-f}\nonumber \\ x\mu & =\int \mu \frac {g^{\prime }}{p-f}dp+c_{1}\nonumber \\ x & =\frac {1}{\mu }\int \mu \frac {g^{\prime }}{p-f}dp+c_{1}\mu \tag {1}\end{align}

Now we solve for \(p\) from \(y=xf\left ( p\right ) +g\left ( p\right ) \) and plug-in the result into the above. To show how this work, lets apply the earlier problem to the above which was to solve \(x-yy^{\prime }=a\left ( y^{\prime }\right ) ^{2}\). From that problem we found that

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

And we had \(f=\frac {1}{p},g=-ap\). Using these value we now find

\begin{align*} \mu & =e^{\int \frac {f^{\prime }\left ( p\right ) }{p-f}dp}\\ & =e^{\int \frac {-\frac {1}{p^{2}}}{p-\frac {1}{p}}dp}\\ & =\frac {p}{\sqrt {p^{2}-1}}\end{align*}

Hence

\begin{align*} x & =\frac {\sqrt {p^{2}-1}}{p}\int \frac {p}{\sqrt {p^{2}-1}}\frac {-a}{p-\frac {1}{p}}dp+c_{1}\frac {p}{\sqrt {p^{2}-1}}\\ & =-\frac {a\sqrt {p^{2}-1}}{p}\int \frac {p^{2}}{\left ( p^{2}-1\right ) ^{\frac {3}{2}}}dp+c_{1}\frac {p}{\sqrt {p^{2}-1}}\\ & =-\frac {a\sqrt {p^{2}-1}}{p}\left ( -\frac {p}{\sqrt {p^{2}-1}}+\ln \left ( p+\sqrt {p^{2}-1}\right ) \right ) +c_{1}\frac {p}{\sqrt {p^{2}-1}}\\ & =a-\frac {a\sqrt {p^{2}-1}}{p}\ln \left ( p+\sqrt {p^{2}-1}\right ) +c_{1}\frac {p}{\sqrt {p^{2}-1}}\end{align*}

Substituting each one of the above value for \(p\) in (2) gives the two solutions. For example, using \(p_{1}=\frac {1}{2}\frac {-y+\sqrt {4ax+y^{2}}}{a}\) gives

\[ x=a-\frac {a\sqrt {\left ( \frac {1}{2}\frac {-y+\sqrt {4ax+y^{2}}}{a}\right ) ^{2}-1}}{\frac {1}{2}\frac {-y+\sqrt {4ax+y^{2}}}{a}}\ln \left ( \frac {1}{2}\frac {-y+\sqrt {4ax+y^{2}}}{a}+\sqrt {\left ( \frac {1}{2}\frac {-y+\sqrt {4ax+y^{2}}}{a}\right ) ^{2}-1}\right ) +c_{1}\frac {\frac {1}{2}\frac {-y+\sqrt {4ax+y^{2}}}{a}}{\sqrt {\left ( \frac {1}{2}\frac {-y+\sqrt {4ax+y^{2}}}{a}\right ) ^{2}-1}}\]

And same for the other \(p_{2}\).

In the above example it was possible to evaluate the integrals in \(p\), then replace \(p\) by its solution from the original ode. What if this was not possible? Let say we have integral

\[ \int ap^{2}dp \]

And for some reason we are not able to the integration. In this case we first replace the above with

\[ \int ^{p}a\tau ^{2}d\tau \]

And only now replace \(p\) with its solution as the upper limit.