Correct option is (b) \(lx+\frac{l(1-l^2)}{3!} x^3+\frac{l(1-l^2)(9-l^2)}{5!} x^5+…\)
For explanation: Given, y = f(x) = Sin(lSin^-1(x))
Now, differentiating,
\(\frac{dy}{dx}=Cos(lSin^{-1} (x))(\frac{l}{\sqrt{1-x^2}})\)
Hence,
\((1-x^2)(\frac{dy}{dx})^2=lCos(lSin^{-1}(x))^2\)
\((1-x^2)(y_1)^2=l^2 Cos(lSin^{-1}(x))^2=l^2 [1-y^2]\)
Hence, differentiating again we get,
\((1-x^2)2y_1 y_2-2xy_1^2=-2l^2 yy_1\)
\((1-x^2) y_2-xy_1+l^2 y=0\)
Hence by Leibniz theorem,
\((1-x^2) y_{(n+2)}-(2n+1)xy_{(n+1)}-(n^2-m^2) y_n=0\)
Therefore by putting x=0, we get,
\(y_{(n+2)}(0)=(n^2-l^2)y_n (0)\)
putting ,n=1,2,3,4,…..
\(y_3 (0)=(1-l^2) y_1 (0)=l(1-l^2)\)
\(y_4 (0)=(4-l^2) y_2 (0)=0\)
\(y_5 (0)=(9-l^2)3(0)=l(1-l^2)(9-l^2)\)
Hence,\(y=Sin(lSin^{-1}(x))=lx+\frac{l(1-l^2)}{3!}x^3+\frac{l(1-l^2)(9-l^2)}{5!} x^5+…\).