Question

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

Integrals
ANSWERED
asked 2021-01-05
Give the correct answer and solve the given equation
\(\displaystyle{\left({x}+{y}\right)}{\left.{d}{x}\right.}+{\left({x}-{y}\right)}{\left.{d}{y}\right.}={0}\)

Answers (1)

2021-01-06
First rewrite it as \(\displaystyle{\left({x}-{y}\right)}{\left.{d}{y}\right.}=-{\left({x}+{y}\right)}{\left.{d}{x}\right.}\) (1)
Suppose that \(\displaystyle{x}-{y}={0},\ \text{so}\ {y}={x}\). Then
\(\displaystyle{0}=-{2}{x}\),
so \(\displaystyle{y}={x}\) is not a solution of the starding equation. Therefore, \(\displaystyle{y}\ne{x}\), and we can divide (1) by \(\displaystyle{x}-{y}\):
\(\displaystyle{\left.{d}{y}\right.}=\frac{{-{\left({x}+{y}\right)}}}{{{x}-{y}}}{\left.{d}{x}\right.}=\frac{{{x}+{y}}}{{{y}-{x}}}{\left.{d}{x}\right.}\)
We can also write this as
\(\displaystyle{y}'=\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}=\frac{{{x}+{y}}}{{{y}-{x}}}=\frac{{{y}-{x}+{2}{x}}}{{{y}-{x}}}={1}+{2}\frac{x}{{{y}-{x}}}={1}+\frac{2}{{\frac{y}{{x}}-{1}}}\) (2)
Now use the substitution
\(\displaystyle{u}=\frac{y}{{x}}\),
which we can also write as
\(\displaystyle{y}-{u}{x},\)
to get that
\(\displaystyle{y}'={u}'{x}+{u}\) (3)
From(2),
\(\displaystyle{y}'={1}+\frac{2}{{\frac{y}{{x}}-{1}}}={1}+\frac{2}{{{u}-{1}}}\)
Combining this and (3) we get
\(\displaystyle{u}'{x}+{u}={1}+\frac{2}{{{u}-{1}}}\)
Furthermore,
\(\displaystyle{u}'{x}={1}-{u}+\frac{2}{{{u}-{1}}}=\frac{{{\left({1}-{u}\right)}{\left({u}-{1}\right)}+{2}}}{{{u}-{1}}}=\frac{{-{\left({u}-{1}\right)}^{2}+{2}}}{{{u}-{1}}}=-\frac{{{\left({u}-{1}\right)}^{2}+{2}}}{{{u}-{1}}}\)
Now write \(\displaystyle{u}'={d}\frac{u}{{\left.{d}{x}\right.}}:\)
\(\displaystyle{d}\frac{u}{{\left.{d}{x}\right.}}\cdot{x}=-\frac{{{\left({u}-{1}\right)}^{2}+{2}}}{{{u}-{1}}}\)
Whis is a separable equation, because we can write it in the form h
\(\displaystyle{h}{\left({u}\right)}{d}{u}= g{{\left({x}\right)}}{\left.{d}{x}\right.}\)
To get this, first divide the equation by \(\displaystyle\frac{{{\left({u}-{1}\right)}^{2}+{2}}}{{{u}-{1}}}\) (notice that this is always nonzero!):
\(\displaystyle\frac{{{u}-{1}}}{{{\left({u}-{1}\right)}^{2}+{2}}}\cdot{d}\frac{u}{{\left.{d}{x}\right.}}\cdot{x}=-{1}\)
Now divide the equation by x:
\(\displaystyle\frac{{{u}-{1}}}{{{\left({u}-{1}\right)}^{2}+{2}}}\cdot\frac{{{d}{u}}}{{\left.{d}{x}\right.}}=-\frac{1}{{x}}\)
Finally, "multiply" thq equation by dx:
\(\displaystyle\frac{{{u}-{1}}}{{{\left({u}-{1}\right)}^{2}+{2}}}\cdot{d}{u}=-\frac{{\left.{d}{x}\right.}}{{x}}\)
All that is left is to integrate the two sides:
\(\displaystyle\int\frac{{{u}-{1}}}{{{\left({u}-{1}\right)}^{2}+{2}}}{d}{u}=\int-\frac{{\left.{d}{x}\right.}}{{x}}\) (4)
The RHS of (4) is easy:
\(\displaystyle\int-\frac{{\left.{d}{x}\right.}}{{x}}=-\int\frac{{\left.{d}{x}\right.}}{{x}}=-{\ln}{\left|{x}\right|}+{C}_{{2}}=\frac{ \ln{{1}}}{{{\left|{x}\right|}}}+{C}_{{2}}\),
where \(\displaystyle{C}_{{2}}\) is some constant.
Now onto the LHS of (4):
\(\displaystyle\int\frac{{{u}-{1}}}{{{\left({u}-{1}\right)}^{2}+{2}}}{d}{u}={\left\lbrace{\left({t}+{\left({u}-{1}\right)}^{2}\right)},{\left({\left.{d}{t}\right.}={2}{\left({u}-{1}\right)}{d}{u}\right)}\right\rbrace}\)
\(\displaystyle=\frac{1}{{2}}\int\frac{{\left.{d}{t}\right.}}{{{t}+{2}}}\)
\(\displaystyle=\frac{1}{{2}}{\ln}{\left|{t}+{2}\right|}+{C}_{{1}}\)
\(\displaystyle= \ln{\sqrt{{{\left|{t}+{2}\right|}}}}+{C}_{{1}}\)
\(\displaystyle= \ln{\sqrt{{{\left|{\left({u}-{1}\right)}^{2}+{2}\right|}}}}+{C}_{{1}}\)
\(\displaystyle=^{{{\left(\ast\right)}}} \ln{\sqrt{{{\left({u}-{1}\right)}^{2}+{2}}}}+{C}_{{1}},\)
where \(\displaystyle{C}_{{1}}\) is some constant, and in \(\displaystyle{\left(\ast\right)}\) we used that \(\displaystyle{\left({u}-{1}\right)}^{2}+{2}\ge{0}.\)
So, (4) becomes
\(\displaystyle \ln{\sqrt{{{\left({u}-{1}\right)}^{2}+{2}}}}+{C}_{{1}}=\frac{ \ln{{1}}}{{\left|{x}\right|}}+{C}_{{2}}\Rightarrow \ln{\sqrt{{{\left({u}-{1}\right)}^{2}+{2}}}}=\frac{ \ln{{1}}}{{\left|{x}\right|}}+{C}\), (5)
where \(\displaystyle{C}={C}_{{2}}-{C}_{{1}}\) is a constant.
Now recall that \(\displaystyle{u}=\frac{y}{{x}}\), so plug this into (5) to obtain the result:
\(\displaystyle \ln{\sqrt{{{\left(\frac{y}{{x}}-{1}\right)}^{2}+{2}}}}=\frac{ \ln{{1}}}{{\left|{x}\right|}}+{C}\)
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