Give the correct answer and solve the given equation displaystyle{left({x}-{y}right)}{left.{d}{x}right.}+{left({3}{x}+{y}right)}{left.{d}{y}right.}={0},text{when} {x}={3},{y}=-{2}

Question
Integrals
asked 2020-11-24
Give the correct answer and solve the given equation \(\displaystyle{\left({x}-{y}\right)}{\left.{d}{x}\right.}+{\left({3}{x}+{y}\right)}{\left.{d}{y}\right.}={0},\text{when}\ {x}={3},{y}=-{2}\)

Answers (1)

2020-11-25
The given diferential equation is
\(\displaystyle{\left({x}-{y}\right)}{\left.{d}{x}\right.}+{\left({3}{x}+{y}\right)}{\left.{d}{y}\right.}={0},\)
with \(\displaystyle{x}={3},{y}={2}\) Now
\(\displaystyle{\left({x}-{y}\right)}{\left.{d}{x}\right.}+{\left({3}{x}+{y}\right)}{\left.{d}{y}\right.}={0},\Rightarrow\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}=\frac{{{y}-{x}}}{{{3}{x}={y}}}\)
since the given differential equation is a homogenuous equation, putting \(\displaystyle{y}={v}{x}\), we have
\(\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}={v}+{x}{d}\frac{v}{{\left.{d}{x}\right.}}\)
So the equation becomes.
\(\displaystyle{v}+{x}{d}\frac{v}{{\left.{d}{x}\right.}}=\frac{{{v}-{1}}}{{{v}+{3}}}\)
\(\displaystyle\Rightarrow{x}{d}\frac{v}{{\left.{d}{x}\right.}}=\frac{{{v}-{1}}}{{{v}+{3}}}-{v}=-\frac{{{\left({v}+{1}\right)}^{2}}}{{{v}+{3}}}\)
\(\displaystyle\Rightarrow\frac{{{v}+{3}}}{{\left({v}+{1}\right)}^{2}}{d}{v}+\frac{{\left.{d}{x}\right.}}{{x}}={0}\)
\(\displaystyle\Rightarrow\frac{{{v}+{1}+{2}}}{{{\left({v}+{1}\right)}^{2}}}{d}{v}+\frac{{\left.{d}{x}\right.}}{{x}}={0}\)
\(\displaystyle\Rightarrow{d}\frac{v}{{{\left({v}+{1}\right)}}}+\frac{{{2}{d}{v}}}{{{\left({v}+{1}\right)}^{2}}}+\frac{{\left.{d}{x}\right.}}{{x}}={0}\)
\(\displaystyle\Rightarrow\int\frac{{\left.{d}{x}\right.}}{{{\left({v}+{1}\right)}}}+\int\frac{{{2}{d}{v}}}{{\left({v}+{1}\right)}^{2}}+\int\frac{{\left.{d}{x}\right.}}{{x}}={C}\) where C is a integrating constant
\(\displaystyle\Rightarrow \ln{{\left({v}+{1}\right)}}=\frac{2}{{{\left({v}+{1}\right)}}}+ \ln{{\left({x}\right)}}={C}\)
\(\displaystyle\Rightarrow \ln{{\left({x}{\left({v}+{1}\right)}\right)}}-\frac{2}{{{\left({v}+{1}\right)}}}={C}\)
putting \(\displaystyle{v}=\frac{y}{{x}}\) we have
\(\displaystyle\Rightarrow \ln{{\left({y}+{x}\right)}}-{2}\frac{x}{{{\left({x}+{y}\right)}}}={C}\)
Since \(\displaystyle{x}+{3},{y}+{2}\), we have \(\displaystyle{C}= \ln{{5}}-\frac{6}{{5}}\)
Hence the solution of the equation is
\(\displaystyle\Rightarrow \ln{{\left({y}+{x}\right)}}-{2}\frac{x}{{{\left({x}+{y}\right)}}}= \ln{{5}}-\frac{6}{{5}}\)
0

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