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प्रश्न
For the following differential equation, find the general solution:- \[\frac{dy}{dx} = \sqrt{4 - y^2}, - 2 < y < 2\]
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उत्तर
We have,
\[\frac{dy}{dx} = \sqrt{4 - y^2}\]
\[ \Rightarrow \frac{1}{\sqrt{4 - y^2}}dy = dx\]
Integrating both sides, we get
\[\int\frac{1}{\sqrt{4 - y^2}}dy = \int dx\]
\[ \Rightarrow \sin^{- 1} \frac{y}{2} = x + C\]
\[ \Rightarrow \frac{y}{2} = \sin \left( x + C \right)\]
\[ \Rightarrow y = 2\sin \left( x + C \right)\]
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