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 \(\frac{b+a}{b-a} =\)

• A. 2
• B. -2
• C. 3
• D. -3

Hint:

To eliminate constants \(A,B\) in function \(y(x)\), differentiate enough times and use linear combination to form DE independent of constants.

Answer 1

The Correct Option is C

Step 1: Differentiate once and isolate the awkward constant.
Given \(y=(\sin^{-1}x)^2+A\cos^{-1}x+B\), differentiating gives
\[ y'=\frac{2\sin^{-1}x}{\sqrt{1-x^2}}-\frac{A}{\sqrt{1-x^2}} \implies \sqrt{1-x^2}\,y' = 2\sin^{-1}x - A \]
Notice B has already vanished, and this form isolates A cleanly on the right.

Step 2: Differentiate again to remove A too.
Differentiating \(\sqrt{1-x^2}\,y'=2\sin^{-1}x-A\) with respect to x (the right side's derivative is simply \(\frac{2}{\sqrt{1-x^2}}\), since A is a constant),
\[ -\frac{x}{\sqrt{1-x^2}}y'+\sqrt{1-x^2}\,y''=\frac{2}{\sqrt{1-x^2}} \]

Step 3: Clear the square roots.
Multiplying through by \(\sqrt{1-x^2}\),
\[ (1-x^2)y''-xy'=2 \]

Step 4: Compare with the given form.
Matching \((1-x^2)y''-xy'=2\) with \((a-x^2)y''-xy'=b\), we read off \(a=1, b=2\).

Step 5: Final conclusion.
\[ \frac{b+a}{b-a}=\frac{2+1}{2-1}=\boxed{3} \]