# Solve the given differential equation by separation of variables. (e^y+1)^2e^(-y)dx+(e^x+1)^6e^(-x)dy=0

Solve the given differential equation by separation of variables.
$$\displaystyle{\left({e}^{{y}}+{1}\right)}^{{2}}{e}^{{-{y}}}{\left.{d}{x}\right.}+{\left({e}^{{x}}+{1}\right)}^{{6}}{e}^{{-{x}}}{\left.{d}{y}\right.}={0}$$

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SoosteethicU
Step 1
Rearranging the given equation:
$$\displaystyle{\left({e}^{{y}}+{1}\right)}^{{2}}{e}^{{-{y}}}{\left.{d}{x}\right.}+{\left({e}^{{x}}+{1}\right)}^{{6}}{e}^{{-{x}}}{\left.{d}{y}\right.}={0}$$
Rearranging, $$\displaystyle{\left({e}^{{y}}+{1}\right)}^{{2}}{e}^{{-{y}}}{\left.{d}{x}\right.}=-{\left({e}^{{x}}+{1}\right)}^{{6}}{e}^{{-{x}}}{\left.{d}{y}\right.}$$
Separating the variables ,we get
$$\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{-{\left({e}^{{x}}+{1}\right)}^{{6}}{e}^{{-{x}}}}}}={\frac{{{\left.{d}{y}\right.}}}{{{\left({e}^{{y}}+{1}\right)}^{{2}}{e}^{{-{y}}}}}}$$
or, $$\displaystyle{\frac{{{e}^{{x}}{\left.{d}{x}\right.}}}{{-{\left({e}^{{x}}+{1}\right)}^{{6}}}}}={\frac{{{e}^{{y}}{\left.{d}{y}\right.}}}{{{\left({e}^{{y}}+{1}\right)}^{{2}}}}}$$
Step 2
Now, integrating both sides, we get :
$$\displaystyle\int{\frac{{{e}^{{x}}{\left.{d}{x}\right.}}}{{-{\left({e}^{{x}}+{1}\right)}^{{6}}}}}=\int{\frac{{{e}^{{y}}{\left.{d}{y}\right.}}}{{{\left({e}^{{y}}+{1}\right)}^{{2}}}}}$$
Let us take, $$\displaystyle{\left({e}^{{x}}+{1}\right)}={u}\ \text{ and }\ {\left({e}^{{y}}+{1}\right)}={v}$$
Then, $$\displaystyle{e}^{{x}}{\left.{d}{x}\right.}={d}{u}\ \text{ and }\ {e}^{{y}}{\left.{d}{y}\right.}={d}{v}$$
Therefore, $$\displaystyle\int{\frac{{{d}{u}}}{{-{\left({u}\right)}^{{6}}}}}=\int{\frac{{{d}{v}}}{{{\left({v}\right)}^{{2}}}}}$$
$$\displaystyle\text{or }\ -{\frac{{{u}^{{-{6}+{1}}}}}{{-{6}+{1}}}}={\frac{{{v}^{{-{2}+{1}}}}}{{-{2}+{1}}}}+{c}$$
$$\displaystyle\text{or }\ -{\frac{{{u}^{{-{5}}}}}{{-{5}}}}={\frac{{{v}^{{-{1}}}}}{{-{1}}}}+{c}$$
$$\displaystyle\text{or }\ {\frac{{{1}}}{{{5}{u}^{{5}}}}}=-{\frac{{{1}}}{{{v}}}}+{c}$$
$$\displaystyle\text{or }\ {\frac{{{1}}}{{{5}{\left({e}^{{x}}+{1}\right)}^{{5}}}}}=-{\frac{{{1}}}{{{\left({e}^{{y}}+{1}\right)}}}}+{c}$$
Step 3
Thus, solution of the given differential equation by separation of variables is :
$$\displaystyle{\frac{{{1}}}{{{5}{\left({e}^{{x}}+{1}\right)}^{{5}}}}}=-{\frac{{{1}}}{{{\left({e}^{{y}}+{1}\right)}}}}+{c}$$