Question:

If the differential equation obtained by eliminating \(A, B\) from \(y = (\sin^{-1}x)^2 + A \cos^{-1} x + B\) is \((a-x^2)y'' - xy' = b\), then \(\frac{b+a}{b-a} =\)

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To eliminate constants \(A,B\) in function \(y(x)\), differentiate enough times and use linear combination to form DE independent of constants.
Updated On: Jul 18, 2026
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The Correct Option is C

Solution and Explanation

Step 1: Given function.
\(y = (\sin^{-1} x)^2 + A \cos^{-1} x + B\)

Step 2: Differentiate to eliminate constants.
\(y' = \frac{2 \sin^{-1}x}{\sqrt{1-x^2}} - \frac{A}{\sqrt{1-x^2}}\), \(y'' = \frac{2}{1-x^2} - \frac{2x \sin^{-1}x}{(1-x^2)^{3/2}} + \frac{A x}{(1-x^2)^{3/2}}\)

Step 3: Eliminate \(A\) and \(B\).
Using system of equations method, obtain differential equation independent of constants: \((a-x^2)y'' - x y' = b\)

Step 4: Compare coefficients.
Coefficient matching gives \(\frac{b+a}{b-a} = 3\)

Step 5: Check consistency.
Verify by substituting derivatives into DE to confirm ratio

Step 6: Final conclusion.
Hence, \[ \boxed{\frac{b+a}{b-a} = 3} \]
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