Question

\[\frac{dy}{dx} = \log x\]

Answer 1

We have,
\[\frac{dy}{dx} = \log x\]
\[ \Rightarrow dy = \left( \log x \right)dx\]
Integrating both sides, we get
\[\int dy = \int\left( \log x \right)dx\]

\[ \Rightarrow \int dy = \log x\int1 dx - \int\left[ \frac{d}{dx}\left( \log x \right)\int1 dx \right]dx\]
\[ \Rightarrow y = x\log x - \int\frac{x}{x}dx\]
\[ \Rightarrow y = x\log x - \int1dx\]
\[ \Rightarrow y = x\log x - x\]
\[ \Rightarrow y = x\left( \log x - 1 \right) + C\]
\[ \Rightarrow y = x\left( \log x - 1 \right) + C\]
\[\text{ So, } y = x\left( \log x - 1 \right) +\text{ C is defined for all }x \in R\text{ except }x = 0.\]
\[\text{ Hence, } y = x\left( \log x - 1 \right) +\text{ C, where } x \in R - \left\{ 0 \right\},\text{ is the solution to the given differential equation.}\]