Find the solution of the following Differential Equations (2x+e^{y})dx + xe^{y}

reproacht3 2021-12-31 Answered
Find the solution of the following Differential Equations
\(\displaystyle{\left({2}{x}+{e}^{{{y}}}\right)}{\left.{d}{x}\right.}+{x}{e}^{{{y}}}{\left.{d}{y}\right.}={0}\)

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Expert Answer

nghodlokl
Answered 2022-01-01 Author has 2021 answers
Step 1
\(\displaystyle{\left({2}{x}+{e}^{{{y}}}\right)}{\left.{d}{x}\right.}+{x}{e}^{{{y}}}{\left.{d}{y}\right.}={0}\)
\(\displaystyle\Rightarrow{2}{x}{\left.{d}{x}\right.}+{e}^{{{y}}}{\left.{d}{x}\right.}+{x}{e}^{{{y}}}{\left.{d}{y}\right.}={0}\)
\(\displaystyle\Rightarrow{2}{x}{\left.{d}{x}\right.}+{d}{\left({x}{e}^{{{y}}}\right)}={0}\)
\(\displaystyle\therefore{d}{\left({f}\cdot{g}\right)}={f}{d}{\left({g}\right)}+{g}{d}{\left({f}\right)}\)
Step 2
\(\displaystyle\Rightarrow\int{2}{x}{\left.{d}{x}\right.}+\int{d}{\left({x}{e}^{{{y}}}\right)}={0}\)
\(\displaystyle\Rightarrow{2}{\frac{{{x}^{{{2}}}}}{{{2}}}}+{x}{e}^{{{y}}}={c}\)
\(\displaystyle\Rightarrow{x}^{{{2}}}+{x}{e}^{{{y}}}={c}\)
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Pademagk71
Answered 2022-01-02 Author has 4935 answers
Solution:
\(\displaystyle{\left({2}{x}+{e}^{{{y}}}\right)}{\left.{d}{x}\right.}+{x}{e}^{{{y}}}{\left.{d}{y}\right.}={0},{y}{\left({1}\right)}={0}\)
Compare above with \(\displaystyle{M}{\left.{d}{x}\right.}+{N}{\left.{d}{y}\right.}={0}\)
\(\displaystyle{M}={2}{x}+{e}^{{{y}}},{N}={x}{e}^{{{y}}}\)
\(\displaystyle{\frac{{\partial{M}}}{{\partial{y}}}}={e}^{{{y}}},{\frac{{\partial{N}}}{{\partial{x}}}}={e}^{{{y}}}\)
Since \(\displaystyle{\frac{{\partial{M}}}{{\partial{y}}}}={\frac{{\partial{N}}}{{\partial{y}}}}={\frac{{\partial{N}}}{{\partial{x}}}}\), therefore differential equation is exact
\(\displaystyle{\frac{{\partial{f}}}{{\partial{x}}}}={2}{x}+{e}^{{{y}}}\) (i)
\(\displaystyle{\frac{{\partial{f}}}{{\partial{y}}}}={x}{e}^{{{y}}}\) (ii)
integrate (i) wrt x, we get
\(\displaystyle{f{{\left({x},{y}\right)}}}=\int{\left({2}{x}+{e}^{{{y}}}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{{2}}}+{x}{e}^{{{y}}}+{h}{\left({y}\right)}\)
Differentiate above wrt y, we get
\(\displaystyle{\frac{{\partial{f}}}{{\partial{y}}}}={x}{e}^{{{y}}}+{h}'{\left({y}\right)}\)
compare above and (ii), we get
\(\displaystyle{x}{e}^{{{y}}}+{h}'{\left({y}\right)}={x}{e}^{{{y}}}\)
\(\displaystyle{h}'{\left({y}\right)}={0}\)
integrate both sides wrt y, we get \(\displaystyle{h}{\left({y}\right)}={C}\)
therefore, we get
\(\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{{2}}}+{x}{e}^{{{y}}}+{C}\)
or we can write as \(\displaystyle{x}^{{{2}}}+{x}{e}^{{{y}}}+{C}={0}\)
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karton
Answered 2022-01-09 Author has 8454 answers

Consider the differential equation
\((2x+e^{y})dx+xe^{y}dy=0\)
Compare it with \(M(x,y)dx+N(x,y)dy=0\), get
\(M(x,y)=2x+e^{y}\ \text{and}\ N(x,y)=xe^{y}\)
\(\frac{\partial M}{\partial y}=e^{y}, \frac{\partial N}{\partial x}=e^{y}\)
Observe that, \(\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\)
So, the given differential equation is in exact form
\((2x+e^{y})dx+xe^{y}dy=0\)
\(2xdx+e^{y}dx+xe^{y}dy=0\)
\(2xdx+d(xe^{y})=0\)
on taking integration get
\(x^{2}+xe^{y}=C\)
which is the required solution.

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