The general solution of the differential equation \(\displaystyle\frac{dy}{dx} = y\log x\), where \(x \gt 0\) and \(y \gt 0\), is (C is an arbitrary constant)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\log y = x\log x - x + C\)
Separating the variables:
\( \displaystyle \frac{dy}{y} = \log x\,dx \)
The left side integrates to \(\log y\). The right side is done by parts, taking \(\log x\) as the first function and 1 as the second:
\( \displaystyle \int \log x\,dx = x\log x - \int x \cdot \frac{1}{x}\,dx \)
\( \displaystyle \int \log x\,dx = x\log x - x \)
So
\( \log y = x\log x - x + C \)
Why the other options are incorrect
Option (1) \(\log y = x\log x + x + C\) — Incorrect. The sign of the x term is wrong. Differentiating gives \(\frac{1}{y}\frac{dy}{dx} = \log x + 2\).
Option (2) \(\log y = \displaystyle\frac{(\log x)^2}{2} + C\) — Incorrect. This is the integral of \(\frac{\log x}{x}\), not of \(\log x\). Differentiating gives \(\frac{1}{y}\frac{dy}{dx} = \frac{\log x}{x}\).
Option (3) \(y = x\log x - x + C\) — Incorrect. This solves \(\frac{dy}{dx} = \log x\); the factor y was not separated.