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Exercise 6.3 · Q28

Q.Verify that y=(sin⁡−1x)2+cy=(\sin^{-1}x)^2+c is a solution of (1−x2)d2ydx2−xdydx=2(1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}=2.

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✓ Free question

Differentiate y=(sin⁡−1x)2+cy=(\sin^{-1}x)^2+c: dydx=2sin⁡−1x1−x2\dfrac{dy}{dx}=\dfrac{2\sin^{-1}x}{\sqrt{1-x^2}}. So 1−x2 dydx=2sin⁡−1x\sqrt{1-x^2}\,\dfrac{dy}{dx}=2\sin^{-1}x. Differentiate again: 1−x2 d2ydx2+−x1−x2dydx=21−x2\sqrt{1-x^2}\,\dfrac{d^2y}{dx^2}+\dfrac{-x}{\sqrt{1-x^2}}\dfrac{dy}{dx}=\dfrac{2}{\sqrt{1-x^2}}. Multiplying through by 1−x2\sqrt{1-x^2}: (1−x2)d2ydx2−xdydx=2(1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}=2, exactly as required.

✓Final answer

Verified: y=(sin⁡−1x)2+cy=(\sin^{-1}x)^2+c solves (1−x2)d2ydx2−xdydx=2(1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}=2

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