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: Rewrite
Divide by $x$: $\dfrac{dy}{dx} = \dfrac{y}{x}\left(\log\dfrac{y}{x} + 1\right)$, a function of $\dfrac{y}{x}$ only.

Step 2: Substitution
With $y = vx$: $x\dfrac{dv}{dx} = v\log v$, so $\int \dfrac{dv}{v\log v} = \int \dfrac{dx}{x}$.

Step 3: Integrate
Put $t = \log v$: $\int \dfrac{dt}{t} = \log t$, so $\log(\log v) = \log x + \log c$ and $\log v = cx$.

Step 4: Compare the options
Since $v = \dfrac{y}{x}$, the result is $\log\dfrac{y}{x} = cx$.

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