2.2.14.1.1 Example1 \[ \left ( 3x^{2}+2xy^{2}\right ) +\left ( 2x^{2}y+4y^{3}\right ) y^{\prime }=0 \] Hence \(M=\left ( 3x^{2}+2xy^{2}\right ) ,N=\left ( 2x^{2}y+4y^{3}\right ) \). We see that \(\frac {\partial M}{\partial y}=4xy\) and \(\frac {\partial N}{\partial x}=4xy\), hence exact. Then (5) gives\begin {align*} \phi & =\int Mdx+f\left ( y\right ) \\ & =\int 3x^{2}+2xy^{2}dx+f\left ( y\right ) \\ & =x^{3}+x^{2}y^{2}+f\left ( y\right ) \end {align*}
Hence (6) gives\begin {align*} \frac {d}{dy}\left ( x^{3}+x^{2}y^{2}+f\left ( y\right ) \right ) & =N\\ 2yx^{2}+f^{\prime }\left ( y\right ) & =2x^{2}y+4y^{3}\\ f^{\prime }\left ( y\right ) & =4y^{3} \end {align*}
Therefore \(f\left ( y\right ) =y^{4}+c_{1}\). Therefore
\begin {align*} \phi & =\int Mdx+f\left ( y\right ) \\ & =x^{3}+x^{2}y^{2}+f\left ( y\right ) \\ & =x^{3}+x^{2}y^{2}+y^{4}+c_{1} \end {align*}
But \(\phi =c\), since constant. Hence combining constants the above becomes\[ x^{3}+x^{2}y^{2}+y^{4}=C \] Which is implicit solution for \(y\left ( x\right ) \).
2.2.14.1.2 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\[ \left ( y+x\right ) \ln \left ( \frac {y+x}{x+3}\right ) -x-y=C \]