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D Y D X = 1 X 2 + 4 X + 5 - Mathematics

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प्रश्न

\[\frac{dy}{dx} = \frac{1}{x^2 + 4x + 5}\]

बेरीज
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उत्तर

We have,

\[\frac{dy}{dx} = \frac{1}{x^2 + 4x + 5}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{1}{x^2 + 4x + 4 + 1}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{1}{\left( x + 2 \right)^2 + \left( 1 \right)^2}\]
\[ \Rightarrow dy = \frac{1}{\left( x + 2 \right)^2 + \left( 1 \right)^2}dx\]

Integrating both sides, we get

\[\int y = \int\frac{1}{\left( x + 2 \right)^2 + \left( 1 \right)^2}dx\]

\[ \Rightarrow y = \tan^{- 1} \frac{x + 2}{1} + C\]

\[ \Rightarrow y = \tan^{- 1} \left( x + 2 \right) + C\]

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पाठ 22: Differential Equations - Revision Exercise [पृष्ठ १४५]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 22 Differential Equations
Revision Exercise | Q 19 | पृष्ठ १४५

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