Question

(x - y) dx + x dy = 0. Please, solve by using an appropriate substitution.

Differential equations
ANSWERED
asked 2021-09-04
\(\displaystyle{\left({x}-{y}\right)}{\left.{d}{x}\right.}+{x}{\left.{d}{y}\right.}={0}\)
Please, solve by using an appropriate substitution.

Expert Answers (1)

2021-09-05
Given differential equation can be written as:
\(\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}={\frac{{{y}-{x}}}{{{x}}}}={\frac{{{y}}}{{{x}}}}-{1}\)
Let, \(\displaystyle{y}=\upsilon{x}\). Then \(\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}=\upsilon+{x}{d}\frac{\upsilon}{{\left.{d}{x}\right.}}\)
Substituting \(\displaystyle{y}=\upsilon{x}\) in the given equation we have:
\(\displaystyle\upsilon+{x}{d}\frac{\upsilon}{{\left.{d}{x}\right.}}=\upsilon-{1}\)
implies \(\displaystyle{x}{d}\frac{\upsilon}{{\left.{d}{x}\right.}}=-{1}\)
implies \(\displaystyle{d}\upsilon=-\frac{{\left.{d}{x}\right.}}{{x}}\)
implies \(\displaystyle\int{d}\upsilon=-\int\frac{{\left.{d}{x}\right.}}{{x}}\) (we take integration both side)
implies \(\displaystyle\upsilon=-{\ln{{x}}}+{c}\), (c is an integrating constant)
implies \(\displaystyle\frac{{y}}{{x}}=-{\ln{{x}}}+{c}\), (we put \(\displaystyle\upsilon=\frac{{y}}{{x}}\))
implies \(\displaystyle{y}={x}{\left({c}-{\ln{{x}}}\right)}\)
The solution of \(\displaystyle{\left({x}-{y}\right)}{\left.{d}{x}\right.}+{x}{\left.{d}{y}\right.}={0}{i}{s}{y}={x}{\left({c}-{\ln{{x}}}\right)}\), where c is an integrating constant
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