Question:
The general solution of the DE \(\dfrac{dy}{dx} = (\sqrt{1-x^2})(\sqrt{1-y^2})\) is
अवकल समीकरण \(\dfrac{dy}{dx} = (\sqrt{1-x^2})(\sqrt{1-y^2})\) का व्यापक हल क्या है?
Options:
- \(2\sin^{-1} y – \sin^{-1} x = C\)
- \(3\sin^{-1} y + \sin^{-1} x = C\)
- \(\sin^{-1} y – \sin^{-1} x = x\sqrt{1-x^2} + C\)
- \(2\sin^{-1} y – \sin^{-1} x = x\sqrt{1-x^2} + C\)
- \(2\sin^{-1} y – \sin^{-1} x = C\)
- \(3\sin^{-1} y + \sin^{-1} x = C\)
- \(\sin^{-1} y – \sin^{-1} x = x\sqrt{1-x^2} + C\)
- \(2\sin^{-1} y – \sin^{-1} x = x\sqrt{1-x^2} + C\)
Correct Answer: c)
Explanation:
Separating variables: \(\dfrac{dy}{\sqrt{1-y^2}} = \sqrt{1-x^2}\,dx\).
Integrating both sides: \(\sin^{-1} y = \dfrac{x}{2}\sqrt{1-x^2} + \dfrac{1}{2}\sin^{-1} x + C_1\).
Multiplying by 2: \(2\sin^{-1} y – \sin^{-1} x = x\sqrt{1-x^2} + C\).
चरों को पृथक करने पर: \(\dfrac{dy}{\sqrt{1-y^2}} = \sqrt{1-x^2}\,dx\)।
समाकलन करने पर: \(2\sin^{-1} y – \sin^{-1} x = x\sqrt{1-x^2} + C\)।