Question

If \(x \frac{\mathrm{~d} y}{\mathrm{~d} x}=y(\log y-\log x+1)\), then the solution of the equation is

• A. \(\mathrm{log}\frac{x}{y}=\mathrm{c}y\), where c is the constant of integration
• B. \(\mathrm{log}\frac{y}{x}=\mathrm{c}y\), where c is the constant of integration
• C. \(\mathrm{log}\frac{x}{y}=\mathrm{c}x\), where c is the constant of integration
• D. \(\mathrm{log}\frac{y}{x}=\mathrm{cx}\), where c is the constant of integration

Hint:

Put y = vx to reduce the equation to separable form and integrate.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The equation is homogeneous in \(x\) and \(y\), so substitute \(y = vx\) and \(\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}\).

Step 2: Substitute
\(x\dfrac{dy}{dx} = y(\log y - \log x + 1)\) becomes \(v + x\dfrac{dv}{dx} = v(\log v + 1)\).
So \(x\dfrac{dv}{dx} = v\log v\).

Step 3: Separate and integrate
\(\dfrac{dv}{v\log v} = \dfrac{dx}{x}\)
Integrating, \(\log(\log v) = \log x + \log c\), so \(\log v = cx\).

Step 4: Back-substitute
Replace \(v = \dfrac{y}{x}\) to get \(\log\dfrac{y}{x} = cx\), which matches option D.

Final Answer:
The solution is \(\log\dfrac{y}{x} = cx\). \[ \boxed{\text{(D) }\log\dfrac{y}{x} = cx} \]