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(D) Given the differential equation $\frac{dy}{dx} = x \log x$. Integrating both sides with respect to $x$: $y = \int x \log x \, dx$. Using integrati

Vedclass · Accepted Answer

None of these

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(D) Given the differential equation $\frac{dy}{dx} = x \log x$. Integrating both sides with respect to $x$: $y = \int x \log x \, dx$. Using integration by parts,where $u = \log x$ and $dv = x \, dx$: $du = \frac{1}{x} \, dx$ and $v = \frac{x^2}{2}$. Applying the formula $\int u \, dv = uv - \int v \, du$: $y = (\log x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx$. $y = \frac{x^2}{2} \log x - \frac{1}{2} \int x \, dx$. $y = \frac{x^2}{2} \log x - \frac{1}{2} \cdot \frac{x^2}{2} + c$. $y = \frac{x^2}{2} \log x - \frac{x^2}{4} + c$. Comparing this with the given options,none of the options match the correct result. Therefore,the correct option is $D$.

The solution of $\frac{dy}{dx} = x \log x$ is

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