If `y=sin(m\ sin^(-1)x)` , then `(1-x^2)y_2-x y_1` is equal to `m^2y` (b) `m y` (c) `-m^2y` (d) none of these

Explanation

Answer is c

`y=sin (m sin ^{-1} x)`
`
frac{d y}{d x}=m cos (m sin ^{-1} x) frac{1}{sqrt{(1-x^{2})}}`
`
x frac{d y}{d x}=cos (m sin ^{-1} x) frac{m x}{sqrt{(1-x^{2})}}`
`
frac{d^{2} y}{d x^{2}}`
`
=m{frac{-m sin (m sin ^{-1} x) sqrt{1-x^{2}} frac{1}{sqrt{1-x^{2}}}-frac{1}{2 sqrt{(1-x^{2})}(-2 x) cos (m sin ^{-1} x)}}{(sqrt{.(1-x^{2}))^{2}}.}}`
`
=frac{m}{(1-x^{2})}{-m sin (m sin ^{-1} x)+frac{x}{sqrt{(1-x^{2}.}} cos (m sin ^{-1} x)} `
`
(1-x^{2}) y_{2}=m{-m sin (m sin ^{-1} x)+frac{x}{sqrt{(1-x^{2}.}} cos (m sin ^{-1} x)} `
`
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