Question

If $y=sin(m~sin^-1x)$, then $(1-x²)y^\prime\prime-xy^\prime}$ is equal to

• A. $m^{2}y$
• B. $my$
• C. $-m^{2}y$
• D. None of these

Hint:

If $y=sin(m~sinx)$, then $(1-x)y$ is equal to

Answer 1

The Correct Option is C

Step 1: Concept
Differentiate $y$ with respect to $x$.
Step 2: Analysis
$y' = cos(m~sin^{-1}x) \cdot \frac{m}{\sqrt{1-x^2}}$. Rearranging: $\sqrt{1-x^2} y' = m~cos(m~sin^{-1}x)$.
Step 3: Evaluation
Squaring both sides: $(1-x^2)(y')^2 = m^2 cos^2(m~sin^{-1}x) = m^2(1 - y^2)$. Now, differentiate again.
Step 4: Conclusion
$(1-x^2)2y'y'' + (y')^2(-2x) = -2m^2yy'$. Dividing by $2y'$: $(1-x^2)y'' - xy' = -m^2y$.
Final Answer: (c)