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CBSE Class 12 Math 2022 Term I Solved Paper

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Question : 33 of 50
Marks: +1, -0
If y=sin⁡(msin⁡−1x)y = \sin (m \sin^{-1} x), then which one of the following equations is true?
Solution:     👈: Video Solution
Given, y=sin⁡(m(sin⁡−1x))⋅⋅⋅⋅⋅⋅⋅(i)y = \sin (m (\sin^{-1} x)) \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)
Differentiating both sides w.r.t. xx , we get
dydx=cos⁡(msin⁡−1x)×m1−x2\frac{dy}{dx} = \cos (m \sin^{-1} x) \times \frac{m}{\sqrt{1-x^2}}
⇒dydx=mcos⁡(msin⁡−1x)1−x2⋅⋅⋅⋅⋅⋅⋅(ii)\Rightarrow \frac{dy}{dx} = \frac{m \cos (m \sin^{-1} x)}{\sqrt{1-x^2}} \cdot \cdot \cdot \cdot \cdot \cdot \cdot (ii)
⇒(1−x2)y′=mcos⁡(msin⁡−1x)1−x2⋅⋅⋅⋅⋅⋅⋅(iii)\Rightarrow (\sqrt{1-x^2}) y' = \frac{m \cos (m \sin^{-1} x)}{\sqrt{1-x^2}} \cdot \cdot \cdot \cdot \cdot \cdot \cdot (iii)
⇒(1−x2)y′=mcos⁡(msin⁡−1x)\Rightarrow (\sqrt{1-x^2}) y' = m \cos (m \sin^{-1} x)
Differentiating again w.r.t. ' xx ', we get
y′′(1−x2)+y′−2x21−x2y'' (\sqrt{1-x^2}) + y' \frac{-2x}{2\sqrt{1-x^2}} =−m2sin⁡(msin⁡−1x)11−x2= -m^2 \sin (m \sin^{-1} x) \frac{1}{\sqrt{1-x^2}}
⇒y′′(1−x2)−xy′=−m2y\Rightarrow y'' (1-x^2) - x y' = -m^2 y
⇒y′′(1−x2)−xy′+m2y=0\Rightarrow y'' (1-x^2) - x y' + m^2 y = 0
 or, (1−x2)d2ydx2−xdydx+m2y=0\text{ or, } (1-x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} + m^2 y = 0
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