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प्रश्न
If y = sin (m sin−1 x), then (1 − x2) y2 − xy1 is equal to
पर्याय
m2y
my
−m2y
none of these
उत्तर
(c)−m2y
Here,
\[y = \sin\left( m \sin^{- 1} x \right)\]
\[ \Rightarrow y_1 = \cos\left( m \sin^{- 1} x \right)\frac{m}{\sqrt{1 - x^2}}\]
\[ \Rightarrow y_2 = - \sin\left( m \sin^{- 1} x \right)\frac{m^2}{\left( 1 - x^2 \right)} + \frac{mx\cos\left( m \sin^{- 1} x \right)}{\left( 1 - x^2 \right)^{3/2}}\]
\[ \Rightarrow y_2 = - \sin\left( m \sin^{- 1} x \right)\frac{m^2}{\left( 1 - x^2 \right)} + \frac{xm\cos\left( m \sin^{- 1} x \right)}{\left( 1 - x^2 \right) \times \sqrt{1 - x^2}}\]
\[ \Rightarrow y_2 = - \sin\left( m \sin^{- 1} x \right)\frac{m^2}{\left( 1 - x^2 \right)} + \frac{x y_1}{\left( 1 - x^2 \right)}\]
\[ \Rightarrow \left( 1 - x^2 \right) y_2 = - y m^2 + x y_1 \]
\[ \Rightarrow \left( 1 - x^2 \right) y_2 - x y_1 = - m^2 y\]
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