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D Y D X = X 2 + X − 1 X , X ≠ 0

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Question

\[\frac{dy}{dx} = x^2 + x - \frac{1}{x}, x \neq 0\]
Sum
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Solution

We have, 
\[\frac{dy}{dx} = x^2 + x - \frac{1}{x}\]
\[ \Rightarrow dy = \left( x^2 + x - \frac{1}{x} \right)dx\]
Integrating both sides, we get
\[ \Rightarrow \int dy = \int\left( x^2 + x - \frac{1}{x} \right)dx\]
\[ \Rightarrow y = \frac{x^3}{3} + \frac{x^2}{2} - \log\left| x \right| + C\]
\[\text{ Clearly, } y = \frac{x^3}{3} + \frac{x^2}{2} - \log\left| x \right| +\text{ C is defined for all } x \in\text{ R except }x = 0 . \]
\[\text{ Hence, } y = \frac{x^3}{3} + \frac{x^2}{2} - \log\left| x \right| + C,\text{ where }x \in R- \left\{ 0 \right\}, \text{ is the solution to the given differential equation }.\]
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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 1 | Page 34

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