Question

If \(y=e^{m\sin^{-1}x}\) and \(y_n\) indicates \(n^{th}\) derivative of \(y\) with respect to \(x\), then which of the following relation is true?

• A. \((1+x^2)y_{n+2}-(2n+1)xy_{n+1}-(n^2+m^2)y_n=0\)
• B. \((1-x^2)y_{n+2}+(2n+1)xy_{n+1}-(n^2+m^2)y_n=0\)
• C. \((1-x^2)y_{n+2}+(2n+1)xy_{n+1}+(n^2+m^2)y_n=0\)
• D. \((1-x^2)y_{n+2}-(2n+1)xy_{n+1}-(n^2+m^2)y_n=0\)

Hint:

For successive derivative questions, first form the differential equation satisfied by \(y\), then apply Leibniz theorem to differentiate \(n\) times.

Answer 1

The Correct Option is D

Concept:
This question is based on successive differentiation. First we form a differential equation satisfied by \(y\), and then differentiate it \(n\) times.

Step 1: Given function. \[ y=e^{m\sin^{-1}x} \] Taking logarithm: \[ \log y=m\sin^{-1}x \]

Step 2: Differentiate once. \[ \frac{1}{y}\frac{dy}{dx}=\frac{m}{\sqrt{1-x^2}} \] \[ y'=\frac{my}{\sqrt{1-x^2}} \]

Step 3: Form the differential equation.
On differentiating again and simplifying, the standard differential equation obtained is: \[ (1-x^2)y''-xy'-m^2y=0 \]

Step 4: Differentiate \(n\) times. \[ D^n\left[(1-x^2)y''\right]-D^n[xy']-m^2D^n[y]=0 \] Using Leibniz theorem: \[ D^n\left[(1-x^2)y''\right] =(1-x^2)y_{n+2}-2nxy_{n+1}-n(n-1)y_n \] Also, \[ D^n[xy']=xy_{n+1}+ny_n \]

Step 5: Substitute these values. \[ (1-x^2)y_{n+2}-2nxy_{n+1}-n(n-1)y_n-xy_{n+1}-ny_n-m^2y_n=0 \] \[ (1-x^2)y_{n+2}-(2n+1)xy_{n+1}-(n^2+m^2)y_n=0 \] \[ \therefore \text{Correct Answer is (D)} \]