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Sum

If `y=e^(tan^(-1)x)`.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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#### Solution

`y=e^(tan^(-1)x)` ……..(1)

Diff. w.r.t x ,

`y=e^(tan^(-1)x)1/(x^2+1)`

`(x^2+1)y_1=e^(tan^(-1)x)` --------(from 1)

Again diff. w.r.t x,

`(x^2+1)y_2+2xy_1=y_1` ……………(1)

Now take n th derivative by applying Leibnitz theorem, Leibnitz theorem is :

`(uv)_n=u_nv+_1^ncu_(n-1)v_1+_2^ncu_(n-2)v_2+.....+uv_n`

`u=(x^2+1),v=y_2` …for first term in eqn (1)

`u=2x,v=y_1` …for second term in eqn (1)

`therefore(1+x^2)y_(n+2)+2(n+1)xy_(n+1)+n(n+1)y_n-y_(n+1)=0`

`therefore(1+x^2)y_(n+2)+[2(n+1)x-1]y_(n+1)+n(n+1)y_n=0`

Hence Proved.

Concept: Leibnitz’S Theorem (Without Proof) and Problems

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