Question

The solution of differential equation \(y\log x - y\,dx = x\,dy\) is

• A. \(y(\log e^x + Cx) = 1\)
• B. \(\left(\log \frac{x}{e} + Cx\right)x = y\)
• C. \((\log Cx^2 + ex^2)y = x\)
• D. None of these

Hint:

For equations like \(y(\log x -1)dx = xdy\), separate variables and integrate carefully.

Answer 1

The Correct Option is D

Step 1: Rewrite equation.
\[ y\log x\, dx - y\, dx = x\, dy \] \[ y(\log x -1)\, dx = x\, dy \]

Step 2: Separate variables.
\[ \frac{dy}{y} = \frac{\log x -1}{x}\, dx \]

Step 3: Integrate.
\[ \int \frac{dy}{y} = \int \frac{\log x}{x}dx - \int \frac{1}{x}dx \] \[ \log y = \frac{(\log x)^2}{2} - \log x + C \]

Step 4: Simplify.
\[ y = e^{\frac{(\log x)^2}{2} - \log x + C} \] \[ y = C \cdot \frac{e^{(\log x)^2/2}}{x} \]

Step 5: Compare with options.
None of the given options match this form. Conclusion: \[ {\text{None of these}} \]