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D Y D X = ( 1 + X 2 ) ( 1 + Y 2 )

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प्रश्न

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

We have,
\[\frac{dy}{dx} = \left( 1 + x^2 \right)\left( 1 + y^2 \right)\]
\[ \Rightarrow \frac{1}{\left( 1 + y^2 \right)}dy = \left( 1 + x^2 \right) dx \]
Integrating both sides, we get
\[\int\frac{1}{\left( 1 + y^2 \right)}dy = \int\left( 1 + x^2 \right) dx \]
\[ \Rightarrow \tan^{- 1} y = x + \frac{x^3}{3} + C\]
\[\text{ Hence, }\tan {}^{- 1} y = x + \frac{x^3}{3} +\text{ C is the required solution }.\]

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अध्याय 21: Differential Equations - Exercise 22.07 [पृष्ठ ५५]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 21 Differential Equations
Exercise 22.07 | Q 33 | पृष्ठ ५५

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