Q.6 (b) If y=e^msin⁻¹x , prove that (1-x²)y_(n+2)-(2n+1)xy_(n+1)-(n²+m²)y_n=0

y₁=e^msin⁻¹x·frac m√(1-x²)⇒√(1-x²) y₁=my⇒(1-x²)y₁²=m²y² . Differentiate: (1-x²)2y₁y₂-2xy₁²=2m²yy₁ ; divide by 2y₁ : (1-x²)y₂-xy₁-m²y=0 . Leibnitz's theorem (n times): (1-x²)y_(n+2)+n(-2x)y_(n+1)+fracn(n-1)2(-2)y_n-[xy_(n+1)+ny_n]-m²y_n=0 (1-x²)y_(n+2)-(2n+1)xy_(n+1)-(n²+m²)y_n=0 (proved).

If \(y=e^{m\sin^{-1}x}\), prove that \((1-x^2)y_{n+2}-(2n+1)xy_{n+1}-(n^2+m^2)y_n=0\)

Q.6 (b) If \(y=e^{m\sin^{-1}x}\), prove that \((1-x^2)y_{n+2}-(2n+1)xy_{n+1}-(n^2+m^2)y_n=0\)

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\(y_1=e^{m\sin^{-1}x}\cdot\frac m{\sqrt{1-x^2}}\Rightarrow\sqrt{1-x^2}\,y_1=my\Rightarrow(1-x^2)y_1^2=m^2y^2\).

Differentiate: \((1-x^2)2y_1y_2-2xy_1^2=2m^2yy_1\); divide by \(2y_1\): \((1-x^2)y_2-xy_1-m^2y=0\).

Leibnitz's theorem (n times):

\((1-x^2)y_{n+2}+n(-2x)y_{n+1}+\frac{n(n-1)}2(-2)y_n-[xy_{n+1}+ny_n]-m^2y_n=0\)

\((1-x^2)y_{n+2}-(2n+1)xy_{n+1}-(n^2+m^2)y_n=0\) (proved).

Doubtly AI · 25 Oct 2026 AI-verified · Oct 2026 Report

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If \(y=e^{m\sin^{-1}x}\), prove that \((1-x^2)y_{n+2}-(2n+1)xy_{n+1}-(n^2+m^2)y_n=0\)

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