Find the solution of the following Differential Equations by Exact

Tiffany Russell 2021-12-31 Answered
Find the solution of the following Differential Equations by Exact \(\displaystyle{\left({2}{y}+{x}{y}\right)}{\left.{d}{x}\right.}+{2}{x}{\left.{d}{y}\right.}={0},{y}{\left({3}\right)}=\sqrt{{{2}}}\)

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Expert Answer

esfloravaou
Answered 2022-01-01 Author has 904 answers

Given:
\(\displaystyle{\left({2}{x}+{x}{y}\right)}{\left.{d}{x}\right.}+{2}{x}{\left.{d}{y}\right.}={0},{y}{\left({3}\right)}=\sqrt{{{2}}}\)
this equation is form \(\displaystyle{M}{\left({x},{y}\right)}{\left.{d}{x}\right.}+{N}{\left({x},{y}\right)}{\left.{d}{y}\right.}={0}\)
At this equation is exact if and only it
\(\displaystyle{\frac{{\partial{M}}}{{\partial{y}}}}={\frac{{\partial{N}}}{{\partial{x}}}}\)
\(\displaystyle{M}={2}{y}+{x}{y}\ {\frac{{\partial{M}}}{{\partial{y}}}}={\frac{{\partial}}{{\partial{y}}}}{\left({2}{y}+{x}{y}\right)}={2}+{x}\)
\(\displaystyle{N}={2}{x}\ {\frac{{\partial{N}}}{{\partial{x}}}}={2}\)
Hence \(\displaystyle{\frac{{\partial{M}}}{{\partial{y}}}}\ne{\frac{{\partial{N}}}{{\partial{x}}}}\)
Now we find the integrating factor to made this equation exact.
\(\displaystyle{M}{y}-{N}{x}={2}+{x}-{2}={x}\)
\(\displaystyle{\frac{{{M}{y}-{N}{x}}}{{{N}}}}={\frac{{{x}}}{{{2}{x}}}}={\frac{{{1}}}{{{2}}}}\)
Hence \(\displaystyle{I}.{F}.={e}^{{\int{\frac{{{1}}}{{{2}}}}{\left.{d}{x}\right.}}}={e}^{{{\frac{{{x}}}{{{2}}}}}}\)
\(\displaystyle\Rightarrow{e}^{{\frac{{x}}{{2}}}}{\left({2}{y}+{x}{y}\right)}{\left.{d}{x}\right.}+{2}{e}^{{\frac{{x}}{{2}}}}{x}{\left.{d}{y}\right.}={0}\) (2)
\(=2e^{2/2}y+xe^{x/2}y\)
\(\displaystyle{N}={2}{e}^{{\frac{{x}}{{2}}}}{x}\)
\(\displaystyle{\frac{{\partial{M}}}{{\partial{y}}}}={2}{e}^{{\frac{{x}}{{2}}}}+{x}{e}^{{\frac{{x}}{{2}}}}\)
\(\displaystyle{\frac{{\partial{N}}}{{\partial{x}}}}={2}{e}^{{\frac{{x}}{{2}}}}{y}+{2}{x}\cdot{e}^{{\frac{{x}}{{2}}}}\cdot{\frac{{{1}}}{{{2}}}}={2}{e}^{{\frac{{x}}{{2}}}}+{x}{e}^{{\frac{{x}}{{2}}}}\)
\(\displaystyle\Rightarrow{\frac{{\partial{M}}}{{\partial{y}}}}={\frac{{\partial{M}}}{{\partial{x}}}}\)
equation (2) is exact equation:
Now solution of equation
\(\displaystyle\int{M}\ {\left.{d}{x}\right.}+\int{M}\ {\left.{d}{y}\right.}={c}\)
\(\displaystyle\int{\left({2}{y}{e}^{{\frac{{x}}{{2}}}}+{x}{y}{e}^{{\frac{{x}}{{2}}}}\right)}{\left.{d}{x}\right.}+\int{\left.{d}{y}\right.}={c}\)
\(\displaystyle{2}{y}\int{e}^{{\frac{{x}}{{2}}}}{\left.{d}{x}\right.}+{y}\int{x}{e}^{{\frac{{x}}{{2}}}}{\left.{d}{x}\right.}={c}\)
\(\displaystyle{2}{y}{\frac{{{e}^{{\frac{{x}}{{2}}}}}}{{\frac{{1}}{{2}}}}}+{y}{\left[{x}\cdot{\frac{{{e}^{{\frac{{x}}{{2}}}}}}{{\frac{{1}}{{2}}}}}-\int{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({x}\right)}\cdot{2}{e}^{{\frac{{x}}{{2}}}}{\left.{d}{x}\right.}\right]}={c}\)
\(\displaystyle{4}{y}{e}^{{\frac{{x}}{{2}}}}+{y}{\left[{2}{x}{e}^{{\frac{{x}}{{2}}}}-{2}{\frac{{{e}^{{\frac{{x}}{{2}}}}}}{{\frac{{1}}{{2}}}}}\right]}={c}\)
\(\displaystyle{4}{y}{e}^{{\frac{{x}}{{2}}}}+{2}{x}{y}{e}^{{\frac{{x}}{{2}}}}-{4}{y}{e}^{{\frac{{x}}{{2}}}}={c}\)
\(\displaystyle{2}{x}{y}{e}^{{\frac{{x}}{{2}}}}={c}\)
Here \(\displaystyle{x}={3},{y}=\sqrt{{{2}}}\)
\(\displaystyle{2}\times{3}\times\sqrt{{{2}}}{e}^{{\frac{{3}}{{2}}}}={c}\)
\(\displaystyle{c}={6}\sqrt{{{2}}}{e}^{{\frac{{3}}{{2}}}}\)
\(\displaystyle{2}{x}{y}{e}^{{\frac{{x}}{{2}}}}={6}\sqrt{{{2}}}{e}^{{\frac{{3}}{{2}}}}\)
\(\displaystyle{x}{y}{e}^{{\frac{{x}}{{2}}}}={3}\sqrt{{{2}}}{e}^{{\frac{{3}}{{2}}}}\)

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eskalopit
Answered 2022-01-02 Author has 350 answers
\(\displaystyle{\left({2}{y}+{x}{y}\right)}{\left.{d}{x}\right.}+{2}{x}{\left.{d}{y}\right.}={0}\)
rearranging,
\(\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}=-\frac{{{2}{y}+{x}{y}}}{{{2}{x}}}=-\frac{{{2}+{x}}}{{{2}{x}}}.{y}=-{g{{\left({x}\right)}}}.{y}\)
\(\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}+{g{{\left({x}\right)}}}.{y}={0}\)
Integrating Factor
\(\displaystyle{I}{F}={\exp{{\left(\int{g{{\left({x}\right)}}}{\left.{d}{x}\right.}\right)}}}\)
\(\displaystyle\int{g{{\left({x}\right)}}}{\left.{d}{x}\right.}=\int\frac{{{2}+{x}}}{{{2}{x}}}{\left.{d}{x}\right.}=\int{\left\lbrace\frac{{1}}{{x}}+\frac{{1}}{{2}}\right\rbrace}{\left.{d}{x}\right.}\)
\(\displaystyle\int{g{{\left({x}\right)}}}{\left.{d}{x}\right.}={\ln{{\left({x}\right)}}}+\frac{{x}}{{2}}\)
Hence,
\(\displaystyle{I}{F}={\exp{{\left({\ln{{\left({x}\right)}}}+\frac{{x}}{{2}}\right)}}}={x}\cdot{e}^{{\frac{{x}}{{2}}}}\)
The DE now becomes,
\(\displaystyle{d}{\left({I}{F}\cdot{y}\right)}={0}\)
\(\displaystyle{d}{\left({x}\cdot{e}^{{\frac{{x}}{{2}}}}.{y}\right)}={0}\)
Integrating,
\(\displaystyle{x}\cdot{e}^{{\frac{{x}}{{2}}}}.{y}={c}{o}{n}{s}{t}\)
Answer: \(\displaystyle{x}{y}.{e}^{{\frac{{x}}{{2}}}}={c}\)
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karton
Answered 2022-01-09 Author has 8454 answers

\(\begin{array}{}(2y+xy)dx+2ydy=0 \\\Rightarrow 2y+xy+2y \frac{dy}{dx}=0 \\\text{Substitute}\ \frac{dy}{dx}\ \text{with y'} \\\Rightarrow 2y+xy+2yy'=0 \\\Rightarrow 2+x+2y'=0 \\\text{If}\ f'(x)=g(x)\ \text{then}\ f(x)=\int g(x)dx \\\Rightarrow y=\int-\frac{x+2}{2}dx \\\Rightarrow y=-\frac{x^{2}}{4}-x+c_{1} \\\text{Now, using}\ y(3)=\sqrt{2} \\\text{So,}\ \sqrt{2}=-\frac{9}{4}-3+c_{1} \\\Rightarrow c_{1}=\sqrt{2}+\frac{21}{4} \\\text{Answer: Thus,}\ y=-\frac{x^{2}}{4}-x+\sqrt{2}+\frac{21}{4} \end{array}\)

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