Question

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

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

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