To solve the given differential equation \( y \log x - y\,dx = x\,dy \), let's first rewrite it in a more familiar form for differential equations. This can be done by rearranging the terms as follows:
\(y \log x \,dx - y \,dx = x \,dy\)
This simplifies to:
\(y (\log x - 1) \,dx = x \,dy\)
We can now rewrite it in the standard form of a separable differential equation:
\(\frac{dy}{y} = \frac{\log x - 1}{x} \,dx\)
Now, we integrate both sides:
Integrating the left-hand side:
\(\int \frac{dy}{y} = \log |y| + C_1\)
Integrating the right-hand side:
\(\int \frac{\log x - 1}{x} \,dx = \int \frac{\log x}{x} \,dx - \int \frac{1}{x} \,dx\)
The integral of \(\frac{1}{x}\) is:
\(\int \frac{1}{x} \,dx = \log |x|\)
For the integral \(\int \frac{\log x}{x} \,dx\), we use integration by parts where:
Applying integration by parts:
\(\int u \,dv = uv - \int v \,du\)
This gives:
\(\int \frac{\log x}{x} \,dx = (\log x) \cdot x - \int 1 \,dx = x \log x - x\)
Therefore, the integration of the right-hand side is:
\(x \log x - x - \log |x| + C_2\)
Combining both sides, we have:
\(\log |y| = x \log x - x - \log |x| + C\)
where \(C = C_2 - C_1\)
This form does not match any of the given options, therefore the final answer is: None of these.