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

Carol Gates 2021-09-15 Answered
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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Expert Answer

SoosteethicU
Answered 2021-09-16 Author has 8391 answers
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 :
\(\displaystyle{\frac{{{1}}}{{{5}{\left({e}^{{x}}+{1}\right)}^{{5}}}}}=-{\frac{{{1}}}{{{\left({e}^{{y}}+{1}\right)}}}}+{c}\)
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