The solution of the differential equation $\left(1+e^{\frac{x}{y}}\right)dx+e^{\frac{x}{y}}\left(1+\frac{x}{y}\right)dy=0$ is

The given differential equation is

$\begin{array}{r}\left(1+{\mathrm{e}}^{\frac{x}{y}}\right)\mathrm{dx}+{\mathrm{e}}^{\frac{x}{y}}\left(1-\frac{x}{y}\right)\mathrm{dy}=0\\ \begin{array}{ll}\frac{dx}{dy}& =\frac{e^{\frac{x}{y}}\left(\frac{x}{y}-1\right)}{\left(e^{x/y}+1\right)}\dots \left(i\right)\\ & =g\left(\frac{x}{y}\right)\\ \because & \frac{dx}{dy}=g\left(\frac{x}{y}\right)\end{array}\end{array}$

$\therefore$ eq. (i) is the homogeneous differential equation 

so put, $\frac{x}{y}=v$

i. e. $x=vy\Rightarrow \frac{dx}{dy}=v+y\frac{dv}{dy}$

Then, eq. (i) becomes

$\begin{array}{l}\mathrm{v}+\mathrm{y}\frac{\mathrm{dv}}{\mathrm{dy}}=\frac{{\mathrm{e}}^{\mathrm{v}}(\mathrm{v}-1)}{{\mathrm{e}}^{\mathrm{v}}+1}\\ \Rightarrow \mathrm{y}\frac{\mathrm{dv}}{\mathrm{dy}}=\frac{{\mathrm{e}}^{\mathrm{v}}(\mathrm{v}-1)}{{\mathrm{e}}^{\mathrm{v}}+1}-\mathrm{v}\\ \Rightarrow \mathrm{y}\frac{\mathrm{dv}}{\mathrm{dy}}=\frac{{\mathrm{ve}}^{\mathrm{v}}-{\mathrm{e}}^{\mathrm{v}}-{\mathrm{ve}}^{\mathrm{v}}-\mathrm{v}}{{\mathrm{e}}^{\mathrm{v}}+1}-\mathrm{v}\\ \Rightarrow \frac{{\mathrm{e}}^{\mathrm{v}}+1}{{\mathrm{e}}^{\mathrm{v}}+\mathrm{v}}\mathrm{dv}=-\frac{1}{\mathrm{y}}\mathrm{dy}\end{array}$

On integrating both sides, we get 

$\begin{array}{l}\int \frac{{\mathrm{e}}^{\mathrm{v}}+1}{{\mathrm{e}}^{\mathrm{v}}+\mathrm{v}}\mathrm{dv}=-\int \frac{1}{\mathrm{y}}\mathrm{dy}\\ \text{ Put }{\mathrm{e}}^{\mathrm{v}}+\mathrm{v}=\mathrm{t}\\ \Rightarrow {\mathrm{e}}^{\mathrm{v}}+1=\frac{\mathrm{dt}}{\mathrm{dv}}\\ \Rightarrow \mathrm{dv}=\frac{\mathrm{dt}}{{\mathrm{e}}^{\mathrm{v}}+1}\\ \therefore \int \frac{{\mathrm{e}}^{\mathrm{v}}+1}{\mathrm{t}}\frac{\mathrm{dt}}{{\mathrm{e}}^{\mathrm{v}}+1}-\log \left|\mathrm{y}\right|+\log \mathrm{C}\\ \Rightarrow \log \left|\mathrm{t}\right|+\log \left|\mathrm{y}\right|=\log \mathrm{C}\end{array}$

$\begin{array}{l}\Rightarrow \log \left|{\mathrm{e}}^{\mathrm{v}}+\mathrm{v}\right|+\log \left|\mathrm{y}\right|=\log \mathrm{C}\left(\because \mathrm{t}={\mathrm{e}}^{\mathrm{v}}+\mathrm{v}\right)\\ \Rightarrow \log \left|\left({\mathrm{e}}^{\mathrm{v}}+\mathrm{v}\right)\mathrm{y}\right|=\mathrm{C}\Rightarrow \left|\left({\mathrm{e}}^{\mathrm{v}}+\mathrm{v}\right)\mathrm{y}\right|=\mathrm{C}\\ \Rightarrow \left({\mathrm{e}}^{\mathrm{v}}+\mathrm{v}\right)\mathrm{y}=\mathrm{C}\end{array}$

So, put $\mathrm{v}=\frac{\mathrm{x}}{\mathrm{y}}$, we get

${\left(e^{x/y}+\frac{x}{y}\right)}^{ }y=C\Rightarrow y{e}^{x/y}+x=C$

This is the required solution of the given differential equation.