# Question #8826d

Jan 12, 2018

${y}_{n + 2} + \left(2 n x + 1\right) {y}_{n + 1} + n \left(n + 1\right) {y}_{n} = 0$.

#### Explanation:

$y = {\tan}^{-} 1 x$.

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = {y}_{1} = \frac{1}{1 + {x}^{2}}$.

$\therefore \left(1 + {x}^{2}\right) {y}_{1} = 1$.

Re-differentiating w.r.t. $x ,$ using the Product Rule,

$\left(1 + {x}^{2}\right) \left({y}_{1}\right) ' + {y}_{1} \left(1 + {x}^{2}\right) ' = 0$.

$\therefore \left(1 + {x}^{2}\right) {y}_{2} + 2 x {y}_{1} = 0$.

$\therefore {\left(\left(1 + {x}^{2}\right) {y}_{2}\right)}_{n} + {\left(2 x {y}_{1}\right)}_{n} = 0. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(\star\right)$.

Now, we use the following Leibnitz Rule (L) for the ${n}^{t h}$

derivative of product of functions :

${\left(u v\right)}_{n} = {u}_{n} + {\text{_nC_1u_(n-1)v_1+}}_{n} {C}_{2} {u}_{n - 2} {v}_{2} + \ldots + {v}_{n}$.

Applying L to $\left(1 + {x}^{2}\right) {y}_{2} , \text{ with, } u = {y}_{2} , \mathmr{and} v = 1 + {x}^{2}$,

${\left(\left(1 + {x}^{2}\right) {y}_{2}\right)}_{n} ,$

$= {\left({y}_{2}\right)}_{n} + n \cdot {\left({y}_{2}\right)}_{n - 1} \left(2 x\right) + \frac{n \left(n - 1\right)}{2} \cdot {\left({y}_{2}\right)}_{n - 2} \left(2\right) + 0$,

$= {y}_{n + 2} + 2 n x {y}_{n + 1} + n \left(n - 1\right) {y}_{n} \ldots \ldots \left({\star}^{1}\right)$.

Similarly, ${\left(2 x {y}_{1}\right)}_{n} ,$

$= {y}_{n + 1} + n \cdot {y}_{n} \cdot 2 + 0. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left({\star}^{2}\right)$.

$\left(\star\right) , \left({\star}^{1}\right) , \left({\star}^{2}\right) \Rightarrow {y}_{n + 2} + \left(2 n x + 1\right) {y}_{n + 1} + n \left(n + 1\right) {y}_{n} = 0$.