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D Y D X = X Log X

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Question

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

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

\[ \Rightarrow y = \log x\int x\ dx - \int\left[ \frac{d}{dx}\left( \log x \right)\int x\ dx \right]dx\]
\[ \Rightarrow y = \log x \times \frac{x^2}{2} - \int\left( \frac{1}{x} \times \frac{x^2}{2} \right)dx\]
\[ \Rightarrow y = \frac{1}{2} x^2 \log x - \int\frac{x}{2}dx\]
\[ \Rightarrow y = \frac{1}{2} x^2 \log x - \frac{x^2}{4} + C\]
\[ \text{ Hence, }y = \frac{1}{2} x^2 \log x - \frac{x^2}{4} + \text{C is the solution to the given differential equation.}\]

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Chapter 21: Differential Equations - Exercise 22.05 [Page 34]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 21 Differential Equations
Exercise 22.05 | Q 19 | Page 34

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