2.18.3.2 Example 2

This is same example as above but with initial conditions \(y\left ( x_{0}\right ) =y_{0}\) to show how to handle IC when unable to do the integration.

\begin{align*} -\frac {\sqrt {\left ( x^{2}-1\right ) \left ( y^{2}-1\right ) }}{\left ( x^{2}-1\right ) }dx+dy & =0\\ y\left ( x_{0}\right ) & =y_{0}\end{align*}

The solution found in above example is

\[ \frac {\sqrt {y^{2}-1}\ln \left ( y+\sqrt {y^{2}-1}\right ) }{\sqrt {y-1}\sqrt {y+1}}+\int ^{x}\frac {\sqrt {\left ( \tau ^{2}-1\right ) \left ( y^{2}-1\right ) }}{\sqrt {y-1}\sqrt {y+1}\left ( \tau ^{2}-1\right ) }d\tau +c=0 \]

At \(y\left ( x_{0}\right ) =y_{0}\) the above becomes

\[ \frac {\sqrt {y_{0}^{2}-1}\ln \left ( y_{0}+\sqrt {y_{0}^{2}-1}\right ) }{\sqrt {y_{0}-1}\sqrt {y_{0}+1}}+\int _{x_{0}}^{x}\frac {\sqrt {\left ( \tau ^{2}-1\right ) \left ( y^{2}-1\right ) }}{\sqrt {y_{0}-1}\sqrt {y_{0}+1}\left ( \tau ^{2}-1\right ) }d\tau +c=0 \]

Substituting this value of \(c\) in the solution gives

\[ \frac {\sqrt {y^{2}-1}\ln \left ( y+\sqrt {y^{2}-1}\right ) }{\sqrt {y-1}\sqrt {y+1}}+\int _{x0}^{x}\frac {\sqrt {\left ( \tau ^{2}-1\right ) \left ( y^{2}-1\right ) }}{\sqrt {y-1}\sqrt {y+1}\left ( \tau ^{2}-1\right ) }d\tau =\frac {\sqrt {y_{0}^{2}-1}\ln \left ( y_{0}+\sqrt {y_{0}^{2}-1}\right ) }{\sqrt {y_{0}-1}\sqrt {y_{0}+1}}+\int _{x_{0}}^{x}\frac {\sqrt {\left ( \tau ^{2}-1\right ) \left ( y^{2}-1\right ) }}{\sqrt {y_{0}-1}\sqrt {y_{0}+1}\left ( \tau ^{2}-1\right ) }d\tau \]

Or

\[ \left ( \frac {\sqrt {y^{2}-1}\ln \left ( y+\sqrt {y^{2}-1}\right ) }{\sqrt {y-1}\sqrt {y+1}}-\frac {\sqrt {y_{0}^{2}-1}\ln \left ( y_{0}+\sqrt {y_{0}^{2}-1}\right ) }{\sqrt {y_{0}-1}\sqrt {y_{0}+1}}\right ) +\int _{x0}^{x}\frac {\sqrt {\left ( \tau ^{2}-1\right ) \left ( y^{2}-1\right ) }}{\sqrt {y-1}\sqrt {y+1}\left ( \tau ^{2}-1\right ) }-\frac {\sqrt {\left ( \tau ^{2}-1\right ) \left ( y^{2}-1\right ) }}{\sqrt {y_{0}-1}\sqrt {y_{0}+1}\left ( \tau ^{2}-1\right ) }d\tau =0 \]