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प्रश्न
For the following differential equation, find the general solution:- \[\frac{dy}{dx} = \sin^{- 1} x\]
बेरीज
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उत्तर
We have,
\[\frac{dy}{dx} = \sin^{- 1} x\]
\[ \Rightarrow dy = \left( \sin^{- 1} x \right)dx\]
Integrating both sides, we get
\[\int dy = \int\left( \sin^{- 1} x \right)dx\]
\[ \Rightarrow \int dy = \sin^{- 1} x\int1 dx - \int\left[ \frac{d}{dx}\left( \sin^{- 1} x \right)\int1 dx \right]dx\]
\[ \Rightarrow y = x \sin^{- 1} x - \int\frac{x}{\sqrt{1 - x^2}}dx\]
\[\text{Putting }t^2 = 1 - x^2,\text{ we get}\]
\[2t\ dt = - 2x\ dx\]
\[ \Rightarrow - t\ dt = x\ dx\]
\[ \therefore y = x \sin^{- 1} x + \int dt\]
\[ \Rightarrow y = x \sin^{- 1} x + t + C\]
\[ \Rightarrow y = x \sin^{- 1} x + \sqrt{1 - x^2} + C\]
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