Q.5 (c) If y=e^tan⁻¹x then prove that (1+x²)y_(n+2)+[2(n+1)x-1]y_(n+1)+n(n+1)y_n=0 .

y₁=e^tan⁻¹x·frac11+x²=frac y1+x²⇒(1+x²)y₁=y . Differentiate again: (1+x²)y₂+2xy₁=y₁⇒(1+x²)y₂+(2x-1)y₁=0 . Leibnitz's theorem (differentiate n times): Dⁿ[(1+x²)y₂]=(1+x²)y_(n+2)+n(2x)y_(n+1)+n(n-1)/2(2)y_n Dⁿ[(2x-1)y₁]=(2x-1)y_(n+1)+n(2)y_n Adding: (1+x²)y_(n+2)+(2nx+2x-1)y_(n+1)+(n²-n+2n)y_n=0 (1+x²)y_(n+2)+[2(n+1)x-1]y_(n+1)+n(n+1)y_n=0 (proved).

If \(y=e^{\tan^{-1}x}\) then prove that \((1+x^2)y_{n+2}+[2(n+1)x-1]y_{n+1}+n(n+1)y_n=0\).

Q.5 (c) If \(y=e^{\tan^{-1}x}\) then prove that \((1+x^2)y_{n+2}+[2(n+1)x-1]y_{n+1}+n(n+1)y_n=0\).

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\(y_1=e^{\tan^{-1}x}\cdot\frac1{1+x^2}=\frac y{1+x^2}\Rightarrow(1+x^2)y_1=y\).

Differentiate again: \((1+x^2)y_2+2xy_1=y_1\Rightarrow(1+x^2)y_2+(2x-1)y_1=0\).

Leibnitz's theorem (differentiate n times):

\(D^n[(1+x^2)y_2]=(1+x^2)y_{n+2}+n(2x)y_{n+1}+\frac{n(n-1)}{2}(2)y_n\)

\(D^n[(2x-1)y_1]=(2x-1)y_{n+1}+n(2)y_n\)

Adding: \((1+x^2)y_{n+2}+(2nx+2x-1)y_{n+1}+(n^2-n+2n)y_n=0\)

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

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If \(y=e^{\tan^{-1}x}\) then prove that \((1+x^2)y_{n+2}+[2(n+1)x-1]y_{n+1}+n(n+1)y_n=0\).

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