2.17.1.2 Example2

\[ \left ( \ln \left ( \frac {y+x}{x+3}\right ) -\frac {y+x}{x+3}\right ) dx+\ln \left ( \frac {y+x}{x+3}\right ) dy=0 \]

Hence \(M=\left ( \ln \left ( \frac {y+x}{x+3}\right ) -\frac {y+x}{x+3}\right ) ,N=\ln \left ( \frac {y+x}{x+3}\right ) \). We see that \(\frac {\partial M}{\partial y}=\frac {3-y}{\left ( y+x\right ) \left ( x+3\right ) }\) and \(\frac {\partial N}{\partial x}=\frac {3-y}{\left ( y+x\right ) \left ( x+3\right ) }\), hence the ode is exact. Eq (5) gives

\begin{align*} \phi & =\int Mdx+f\left ( y\right ) \\ & =\int \left ( \ln \left ( \frac {y+x}{x+3}\right ) -\frac {y+x}{x+3}\right ) dx+f\left ( y\right ) \\ & =\left ( 3-y\right ) \ln \left ( \frac {y-3}{x+3}\right ) +\left ( y+x\right ) \ln \left ( \frac {y+x}{x+3}\right ) +\left ( 3-y\right ) \ln \left ( x+3\right ) -x+f\left ( y\right ) \\ & =\left ( 3-y\right ) \left ( \ln \left ( \frac {y-3}{x+3}\right ) +\ln \left ( x+3\right ) \right ) +\left ( y+x\right ) \ln \left ( \frac {y+x}{x+3}\right ) -x+f\left ( y\right ) \\ & =\left ( 3-y\right ) \ln \left ( y-3\right ) +\left ( y+x\right ) \ln \left ( \frac {y+x}{x+3}\right ) -x+f\left ( y\right ) \end{align*}

Hence (6) gives

\begin{align*} \frac {d}{dy}\left ( \phi \right ) & =N\\ \frac {d}{dy}\left ( \left ( 3-y\right ) \ln \left ( y-3\right ) +\left ( y+x\right ) \ln \left ( \frac {y+x}{x+3}\right ) -x+f\left ( y\right ) \right ) & =\ln \left ( \frac {y+x}{x+3}\right ) \\ \ln \left ( \frac {y+x}{x+3}\right ) -\ln \left ( y-3\right ) +f^{\prime }\left ( y\right ) & =\ln \left ( \frac {y+x}{x+3}\right ) \\ -\ln \left ( y-3\right ) +f^{\prime }\left ( y\right ) & =0\\ f^{\prime }\left ( y\right ) & =\ln \left ( y-3\right ) \end{align*}

Therefore

\begin{align*} f\left ( y\right ) & =\int \ln \left ( y-3\right ) dy\\ & =\ln \left ( y-3\right ) \left ( y-3\right ) +3-y+c_{1}\end{align*}

Hence from above

\begin{align*} \phi & =\left ( 3-y\right ) \ln \left ( y-3\right ) +\left ( y+x\right ) \ln \left ( \frac {y+x}{x+3}\right ) -x+f\left ( y\right ) \\ & =\left ( 3-y\right ) \ln \left ( y-3\right ) +\left ( y+x\right ) \ln \left ( \frac {y+x}{x+3}\right ) -x+\ln \left ( y-3\right ) \left ( y-3\right ) +3-y+c_{1}\\ & =-\left ( y-3\right ) \ln \left ( y-3\right ) +\left ( y+x\right ) \ln \left ( \frac {y+x}{x+3}\right ) -x+\ln \left ( y-3\right ) \left ( y-3\right ) +3-y+c_{1}\\ & =\left ( y+x\right ) \ln \left ( \frac {y+x}{x+3}\right ) -x+3-y+c_{1}\\ & =\left ( y+x\right ) \ln \left ( \frac {y+x}{x+3}\right ) -x-y+c_{2}\end{align*}

But \(\phi =c\), since constant. Hence combining constants the above becomes

\[ \left ( y+x\right ) \ln \left ( \frac {y+x}{x+3}\right ) -x-y=C \]