Question

solve the equation using the substitution
$(x - y)dx + xdy = 0$
$xdy = -(x - y)dx$
$dy = \frac{y}{x}dx$
$\frac{dy}{dx} = \frac{y}{x}$
$v = \frac{y}{x}$
$y = vx$
$v + x\frac{dv}{dx} = \frac{vx}{x}$
$v + x\frac{dv}{dx} = v$
$x\frac{dv}{dx} = 0$
$\int dv = \int \frac{1}{x}dx$

Explanation:

Step1: Rewrite the given equation

Given the differential equation \((x - y)dx + xdy = 0\), we can rewrite it as \(xdy=(y - x)dx\), then \(\frac{dy}{dx}=\frac{y - x}{x}=\frac{y}{x}-1\).

Step2: Use substitution \(v = \frac{y}{x}\) (i.e., \(y=vx\))

Differentiate \(y = vx\) with respect to \(x\) using the product rule. We get \(\frac{dy}{dx}=v + x\frac{dv}{dx}\).

Substitute \(\frac{dy}{dx}\) and \(y = vx\) (so \(\frac{y}{x}=v\)) into the equation \(\frac{dy}{dx}=\frac{y}{x}-1\):
\(v + x\frac{dv}{dx}=v - 1\)

Step3: Simplify the equation

Subtract \(v\) from both sides of the equation \(v + x\frac{dv}{dx}=v - 1\), we obtain:
\(x\frac{dv}{dx}=- 1\)

Step4: Separate variables and integrate

Rewrite the equation as \(dv=-\frac{1}{x}dx\).
Integrate both sides: \(\int dv=\int-\frac{1}{x}dx\)
The left - hand side integral \(\int dv=v + C_1\), and the right - hand side integral \(\int-\frac{1}{x}dx=-\ln|x|+C_2\).

Combining the constants \(C = C_2 - C_1\), we have \(v=-\ln|x|+C\)

Step5: Substitute back \(v=\frac{y}{x}\)

Since \(v = \frac{y}{x}\), we substitute it back into the equation \(v=-\ln|x|+C\):
\(\frac{y}{x}=-\ln|x|+C\)
Multiply both sides by \(x\) to get the general solution of the differential equation:
\(y=-x\ln|x|+Cx\)

Answer:

The general solution of the differential equation \((x - y)dx + xdy = 0\) is \(y = Cx-x\ln|x|\) (where \(C\) is an arbitrary constant).