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(Y2 + 1) Dx − (X2 + 1) Dy = 0 - Mathematics

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प्रश्न

(y2 + 1) dx − (x2 + 1) dy = 0

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उत्तर

We have,
\[\left( y^2 + 1 \right) dx - \left( x^2 + 1 \right) dy = 0\]
\[ \Rightarrow \left( y^2 + 1 \right) dx = \left( x^2 + 1 \right) dy\]
\[ \Rightarrow \frac{1}{x^2 + 1}dx = \frac{1}{y^2 + 1}dy\]
Integrating both sides, we get
\[\int\frac{1}{x^2 + 1}dx = \int\frac{1}{y^2 + 1}dy\]
\[ \Rightarrow \tan^{- 1} x = \tan^{- 1} y + C\]
\[ \Rightarrow \tan^{- 1} x - \tan^{- 1} y = C\]
\[\text{ Hence, } \tan^{- 1} x - \tan^{- 1} y = \text{ C is the required solution .}\]

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पाठ 22: Differential Equations - Exercise 22.07 [पृष्ठ ५५]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 22 Differential Equations
Exercise 22.07 | Q 31 | पृष्ठ ५५

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