COMEDK · Maths · 32. Differential Equations
The particular solution of \(\dfrac{d y}{d x}+\sqrt{\dfrac{1-y^2}{1-x^2}}=0\), when \(x=0, y=\dfrac{1}{2}\) is
- A \(\sin ^{-1} x+\sin ^{-1} y=\dfrac{\pi}{2}\)
- B \(\sin ^{-1} x+\sin ^{-1} y=\dfrac{\pi}{3}\)
- C \(\sin ^{-1} x-\sin ^{-1} y=\dfrac{\pi}{2}\)
- D \(\sin ^{-1} x+\sin ^{-1} y=\dfrac{\pi}{6}\)
Answer & Solution
Correct Answer
(D) \(\sin ^{-1} x+\sin ^{-1} y=\dfrac{\pi}{6}\)
Step-by-step Solution
Detailed explanation
The given differential equation is \(\dfrac{dy}{dx} + \sqrt{\dfrac{1-y^2}{1-x^2}} = 0\). Separating the variables, we have \(\dfrac{dy}{\sqrt{1-y^2}} = -\dfrac{dx}{\sqrt{1-x^2}}\). Integrating both sides, we get \(\int \dfrac{dy}{\sqrt{1-y^2}} = -\int \dfrac{dx}{\sqrt{1-x^2}}\).…
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