Chapter 6 Higher Order Derivatives-II

6.0.1 Leibnitz’s theorem

If $y = u v$ where $u$ and $v$ are functions of $x$ possessing $n^{t h}$ derivatives, then

$\begin{aligned} y_{n} & =^{n} C_{0} u_{n} v +^{n} C_{1} u_{n - 1} v_{1} +^{n} C_{2} u_{n - 2} v_{2} +^{n} C_{3} u_{n - 3} v_{3} + \dots +^{n} C_{r} u_{n - r} v_{r} + \dots +^{n} C_{n} u v_{n} \\ = u_{n} v +^{n} C_{1} u_{n - 1} v_{1} +^{n} C_{2} u_{n - 2} v_{2} +^{n} C_{3} u_{n - 3} v_{3} + \dots +^{n} C_{r} u_{n - r} v_{r} + \dots + u v_{n} \\ (u v)_{(n)} (x) & = \sum_{k = 0}^{n}^{n} C_{k} u_{(n - k)} (x) v_{(k)} (x), \end{aligned}$

where $^{n} C_{k} = \frac{n!}{k! (n - k)!}$ is the binomial coefficient, $u_{(0)} (x) = u (x)$ and $v_{(0)} (x) = v (x)$ .

6.1 Exercise 3 (ii)

6.1.1 Question 1

Find $\frac{d^{n} y}{d x^{n}}$ if $y$ is

$x^{2} e^{a x}$

Here $u = e^{a x}$ and $v = x^{2}$ .

While choosing $u$ and $v$ , choose $v$ as one which is likely to result $0$ in further differentiation.

$\begin{aligned} u_{1} & = a e^{a x} \\ u_{2} & = a^{2} e^{a x} \\ u_{3} & = a^{3} e^{a x} \\ u_{4} & = a^{4} e^{a x} \\ ⋮ \\ u_{n} & = a^{n} e^{a x} \end{aligned}$

$\begin{aligned} v & = x^{2} \\ v_{1} & = 2 x \\ v_{2} & = 2 \\ v_{3} & = 0 \\ ⋮ \\ v_{n} & = 0 \end{aligned}$

From Leibnitz’s theorem,

$\begin{aligned} y_{n} & = u_{n} v +^{n} C_{1} u_{n - 1} v_{1} +^{n} C_{2} u_{n - 2} v_{2} +^{n} C_{3} u_{n - 3} v_{3} + \dots +^{n} C_{r} u_{n - r} v_{r} + \dots + u v_{n} \\ = a^{n} e^{a x} x^{2} +^{n} C_{1} a^{n - 1} e^{a x} .2 x +^{n} C_{2} a^{n - 2} e^{a x} .2 + \dots + 0 \\ = a^{n} e^{a x} x^{2} + 2 n a^{n - 1} e^{a x} x + \frac{n (n - 1)}{2} a^{n - 2} e^{a x} .2 \\ y_{n} & = a^{n} e^{a x} x^{2} + 2 n a^{n - 1} e^{a x} x + \frac{n (n - 1)}{2} a^{n - 2} e^{a x} .2 \\ y_{n} & = a^{n} e^{a x} x^{2} + 2 n a^{n - 1} e^{a x} x + n (n - 1) a^{n - 2} e^{a x} \end{aligned}$

$x^{3} \sin x$

Here $u = \sin x$ and $v = x^{3}$

The equation $\sin x$ can be represented as $\sin (1. x + 0)$ so that $y_{n}$ can be derived from the standard form $\sin (a x + b)$ .

$\begin{aligned} u_{1} & = \cos x \\ u_{2} & = - \sin x \\ u_{3} & = - \cos x \\ u_{4} & = \sin x \\ ⋮ \\ u_{n} & = 1^{n} \sin (x + 0 + \frac{n π}{2}) = \sin (x + \frac{n π}{2}) \end{aligned}$

$\begin{aligned} v & = x^{3} \\ v_{1} & = 3 x^{2} \\ v_{2} & = 6 x \\ v_{3} & = 6 \\ v_{4} & = 0 \\ ⋮ \\ v_{n} & = 0 \end{aligned}$

From Leibnitz’s theorem,

$\begin{aligned} y_{n} & = u_{n} v +^{n} C_{1} u_{n - 1} v_{1} +^{n} C_{2} u_{n - 2} v_{2} + \\ ^{n} C_{3} u_{n - 3} v_{3} + \dots +^{n} C_{r} u_{n - r} v_{r} + \dots + u v_{n} \\ = \sin (x + \frac{n π}{2}) . x^{3} + \\ n . \sin (x + \frac{(n - 1) π}{2}) .3 x^{2} + \\ \frac{n (n - 1)}{2} . \sin (x + \frac{(n - 2) π}{2}) .6 x + \\ \frac{n (n - 1) (n - 2)}{6} \sin (x + \frac{(n - 3) π}{2}) .6 + \\ 0 + \dots + 0 \\ y_{n} & = x^{3} \sin (x + \frac{n π}{2}) + \\ 3 n x^{2} \sin (x + \frac{(n - 1) π}{2}) + \\ 3 n (n - 1) x \sin (x + \frac{(n - 2) π}{2}) + \\ n (n - 1) (n - 2) \sin (x + \frac{(n - 3) π}{2}) \end{aligned}$

$x^{3} \log x$

$\begin{aligned} u & = \log x \\ u_{1} & = \frac{1}{x} \\ u_{2} & = - \frac{1}{x^{2}} \\ u_{3} & = \frac{2}{x^{3}} \\ u_{4} & = - \frac{6}{x^{4}} \\ ⋮ \\ u_{n} & = \frac{(- 1)^{n - 1} (n - 1)!}{x^{n}} \end{aligned}$

$\begin{aligned} v & = x^{3} \\ v_{1} & = 3 x^{2} \\ v_{2} & = 6 x \\ v_{3} & = 6 \\ v_{4} & = 0 \\ ⋮ \\ v_{n} & = 0 \end{aligned}$

From Leibnitz’s theorem,

$\begin{aligned} y_{n} & = u_{n} v +^{n} C_{1} u_{n - 1} v_{1} +^{n} C_{2} u_{n - 2} v_{2} +^{n} C_{3} u_{n - 3} v_{3} \\ + \dots +^{n} C_{r} u_{n - r} v_{r} + \dots + u v_{n} \\ y_{n} & = \frac{(- 1)^{n - 1} (n - 1)!}{x^{n}} . x^{3} + n . \frac{(- 1)^{n - 2} (n - 2)!}{x^{n - 1}} .3 x^{2} + \\ \frac{n (n - 1)}{2} \frac{(- 1)^{n - 3} (n - 3)!}{x^{n - 2}} .6 x + \\ \frac{n (n - 1) (n - 2)}{6} \frac{(- 1)^{n - 4} (n - 4)!}{x^{n - 3}} .6 + \\ 0 + \dots + 0 \\ y_{n} & = \frac{(- 1)^{n}}{x^{n - 3}} [\frac{(n - 1)!}{(- 1)^{1}} + \frac{3 n (n - 2)!}{(- 1)^{2}} + \frac{3 n (n - 1) (n - 3)!}{(- 1)^{3}} + \frac{n (n - 1) (n - 2) (n - 4)!}{(- 1)^{4}}] \end{aligned}$

$x^{n} e^{x}$

$\begin{aligned} u_{1} & = e^{x} \\ u_{2} & = e^{x} \\ u_{3} & = e^{x} \\ u_{4} & = e^{x} \\ ⋮ \\ u_{n} & = e^{x} \end{aligned}$

$\begin{aligned} v & = x^{n} \\ v_{1} & = n x^{n - 1} \\ v_{2} & = n (n - 1) x^{n - 2} \\ v_{3} & = n (n - 1) (n - 2) x^{n - 3} \\ v_{4} & = n (n - 1) (n - 2) (n - 3) x^{n - 4} \\ ⋮ \\ v_{n} & = n! \end{aligned}$

The $n - 1$ times differentiation of $x^{n - 1}$ is $(n - 1)!$ . Similarly $n$ -times differentiation of $x^{n}$ is $n!$ .

From Leibnitz’s theorem,

$\begin{aligned} y_{n} & = u_{n} v +^{n} C_{1} u_{n - 1} v_{1} +^{n} C_{2} u_{n - 2} v_{2} +^{n} C_{3} u_{n - 3} v_{3} \\ + \dots +^{n} C_{r} u_{n - r} v_{r} + \dots + u v_{n} \\ = e^{x} x^{n} + n . e^{x} . n x^{n - 1} + \\ \frac{n (n - 1)}{2} . e^{x} . n (n - 1) x^{n - 2} + \\ \frac{n (n - 1) (n - 2)}{6} . e^{x} . n (n - 1) (n - 2) x^{n - 3} + \\ \dots + e^{x} . n! \end{aligned}$

$\begin{aligned} y_{n} & = e^{x} [x^{n} + \frac{n^{2} x^{n - 1}}{1!} + \frac{n^{2} (n - 1)^{2} x^{n - 2}}{2!} + \frac{n^{2} (n - 1)^{2} (n - 2)^{2} x^{n - 3}}{3!} + \dots + n!] \\ = e^{x} [x^{n} + \frac{n^{2} x^{n - 1}}{1!} + \frac{n^{2} (n - 1)^{2} x^{n - 2}}{2!} + \frac{n^{2} (n - 1)^{2} (n - 2)^{2} x^{n - 3}}{3!} + \dots + \frac{(n!)^{2}}{n!}] \\ y_{n} & = e^{x} [x^{n} + \frac{n^{2} x^{n - 1}}{1!} + \frac{n^{2} (n - 1)^{2} x^{n - 2}}{2!} + \frac{n^{2} (n - 1)^{2} (n - 2)^{2} x^{n - 3}}{3!} + \dots + \frac{n^{2} (n - 1)^{2} (n - 2)^{2} (n - 3)^{2} \dots 1^{2}}{n!}] \end{aligned}$

$e^{a x} \cos b x$

$\begin{aligned} u & = e^{a x} \\ u_{n} & = a^{n} e^{a x} \end{aligned}$

$\begin{aligned} v & = \cos b x \\ v_{n} & = b^{n} \cos (b x + \frac{n π}{2}) \end{aligned}$

From Leibnitz’s theorem,

$\begin{aligned} y_{n} & = u_{n} v +^{n} C_{1} u_{n - 1} v_{1} +^{n} C_{2} u_{n - 2} v_{2} +^{n} C_{3} u_{n - 3} v_{3} \\ + \dots +^{n} C_{r} u_{n - r} v_{r} + \dots + u v_{n} \\ = a^{n} e^{a x} \cos b x +^{n} C_{1} a^{n - 1} e^{a x} . b^{1} \cos (b x + \frac{1 π}{2}) + \\ ^{n} C_{2} a^{n - 2} e^{a x} . b^{2} \cos (b x + \frac{2 π}{2}) + \\ \dots + e^{a x} . b^{n} \cos (b x + \frac{n π}{2}) \end{aligned}$

$\begin{aligned} y_{n} & = e^{a x} [a^{n} \cos b x +^{n} C_{1} a^{n - 1} b \cos (b x + \frac{π}{2}) +^{n} C_{2} a^{n - 2} b^{2} \cos (b x + \frac{2 π}{2}) + \dots + b^{n} \cos (b x + \frac{n π}{2})] \end{aligned}$

6.1.2 Question 2

If $y = x^{n - 1} \log x$ , show that $x y_{n} = (n - 1)! .$

Given equation is:

$\begin{matrix} (6.1) & \begin{aligned} y = x^{n - 1} \log x \end{aligned} \end{matrix}$

Differentiating w.r.t $x$ ,

$\begin{aligned} y_{1} & = \frac{x^{n - 1}}{x} + \log x (n - 1) x^{n - 2} \\ = \frac{x^{n - 1}}{x} + \frac{\log x (n - 1) x^{n - 1}}{x} \end{aligned}$

From (6.1),

$\begin{aligned} y_{1} & = \frac{x^{n - 1}}{x} + \frac{y (n - 1)}{x} \\ x y_{1} & = x^{n - 1} + (n - 1) y \\ x y_{1} - (n - 1) y & = x^{n - 1} \end{aligned}$

Differentiating it $n - 1$ times using Leibnitz’s theorem,

$\begin{aligned} ^{n - 1} C_{0} y_{1 + (n - 1)} x +^{n - 1} C_{1} y_{n - 1} \times 1 + 0 \\ - (n - 1) y_{n - 1} & = (n - 1)! \end{aligned}$

$\begin{aligned} 1 \times x y_{n} + \frac{(n - 1)!}{1! (n - 2)!} y_{n - 1} - (n - 1) y_{n - 1} & = (n - 1)! \\ x y_{n} + (n - 1) y_{n - 1} - (n - 1) y_{n - 1} & = (n - 1)! \\ x y_{n} & = (n - 1)! \end{aligned}$

Hence proved.

The $n - 1$ times differentiation of $x^{n - 1}$ is $(n - 1)!$ . Similarly $n$ -times differentiation of $x^{n}$ is $n!$ .

6.1.3 Question 3

If $y = e^{x^{2}}$ , prove that $y_{n + 1} - 2 x y_{n} - 2 n y_{n - 1} = 0$ .

Plotting reveals this graph:

Taking log and differentiating w.r.t $x$ ,

$\begin{aligned} \log y & = x^{2} \log e \\ \frac{1}{y} y_{1} & = 2 x \\ y_{1} & = 2 x y \end{aligned}$

Differentiating $n$ -times using Leibnitz’s theorem,

$\begin{aligned} y_{n + 1} & = 2 [y_{n} x +^{n} C_{1} y_{n - 1} \times 1 + 0] \\ y_{n + 1} & = 2 [y_{n} x + n y_{n - 1}] \\ y_{n + 1} - 2 x y_{n} - 2 n y_{n - 1} & = 0 \end{aligned}$

Hence proved.

6.1.4 Question 4

If $y = e^{a x} \sin b x$ , show that

$y_{2} - 2 a y_{1} + (a^{2} + b^{2}) y = 0$

Differentiating w.r.t $x$ ,

$\begin{aligned} y_{1} & = e^{a x} . b \cos b x + \sin b x . a e^{a x} \\ y_{1} & = b e^{a x} \cos b x + a e^{a x} \sin b x \end{aligned}$

Substituting original equation,

$\begin{matrix} (6.2) & \begin{aligned} y_{1} & = b e^{a x} \cos b x + a y \end{aligned} \end{matrix}$

Again, differentiating w.r.t $x$ ,

$\begin{aligned} y_{2} & = b [a e^{a x} \cos b x - b e^{a x} \sin b x] + a y_{1} \end{aligned}$

From original equation and (6.2),

$\begin{aligned} y_{2} & = b [a e^{a x} \cos b x - b y] + a y_{1} \\ = b [a \times \frac{y_{1} - a y}{b} - b y] + a y_{1} \\ y_{2} & = a y_{1} - a^{2} y - b^{2} y + a y_{1} \end{aligned}$

$\begin{matrix} (6.3) & \begin{aligned} y_{2} - 2 a y_{1} + (a^{2} + b^{2}) y & = 0 \end{aligned} \end{matrix}$

$y_{n + 1} = 2 a y_{n} - (a^{2} + b^{2}) y_{n - 1}$

Differentiating $n - 1$ times (6.3) w.r.t $x$ using Leibnitz’s theorem,

$\begin{aligned} y_{(n - 1) + 2} - 2 a y_{(n - 1) + 1} + (a^{2} + b^{2}) y_{n - 1} & = 0 \\ y_{n + 1} - 2 a y_{n} + (a^{2} + b^{2}) y_{n - 1} & = 0 \\ y_{n + 1} & = 2 a y_{n} - (a^{2} + b^{2}) y_{n - 1} \end{aligned}$

6.1.5 Question 5

If $y = a \cos (\log x) + b \sin (\log x)$ , prove that

$x^{2} y_{2} + x y_{1} + y = 0$

Given equation is,

$\begin{matrix} (6.4) & \begin{aligned} y & = a \cos (\log x) + b \sin (\log x) \end{aligned} \end{matrix}$

Differentiating w.r.t $x$ ,

$\begin{aligned} y_{1} & = - \frac{a \sin (\log x)}{x} + \frac{b \cos (\log x)}{x} \end{aligned}$

which can also be written as,

$\begin{matrix} (6.5) & \begin{aligned} x y_{1} & = b \cos (\log x) - a \sin (\log x) \end{aligned} \end{matrix}$

Differentiating (6.5) again w.r.t $x$ ,

$\begin{aligned} x y_{2} + y_{1} & = \frac{- b \sin (\log x)}{x} - \frac{a \cos \log x}{x} \\ x^{2} y_{2} + x y_{1} & = - (a \cos (\log x) + b \sin (\log x)) \end{aligned}$

From original equation (6.4),

$\begin{aligned} x^{2} y_{2} + x y_{1} & = - (y) \end{aligned}$

$\begin{matrix} (6.6) & \begin{aligned} x^{2} y_{2} + x y_{1} + y & = 0 \end{aligned} \end{matrix}$

Hence proved.

$x^{2} y_{n + 2} + (2 n + 1) x y_{n + 1} + (n^{2} + 1) y_{n} = 0$

Differentiating $n$ times (6.6) using Leibnitz’s theorem,

$\begin{aligned} y_{n + 2} . x^{2} + n . y_{n + 1} .2 x + \frac{n (n - 1)}{2} y_{n} .2 + 0 + \\ y_{n + 1} . x + n y_{n} \times 1 + 0 + y_{n} & = 0 \end{aligned}$

$\begin{aligned} x^{2} y_{n + 2} + (2 n + 1) x y_{n + 1} + (n^{2} - n + n + 1) y_{n} & = 0 \\ x^{2} y_{n + 2} + (2 n + 1) x y_{n + 1} + (n^{2} + 1) y_{n} & = 0 \end{aligned}$

Hence proved.

6.1.6 Question 6

If $y = \log (x + \sqrt{a^{2} + x^{2}})$ , show that

$(a^{2} + x^{2}) y_{2} + x y_{1} = 0$

Differentiating w.r.t $x$ ,

$\begin{aligned} y_{1} & = \frac{1}{x + \sqrt{a^{2} + x^{2}}} \times [1 + \frac{2 x}{2 \sqrt{a^{2} + x^{2}}}] \\ = \frac{1}{x + \sqrt{a^{2} + x^{2}}} \times [1 + \frac{x}{\sqrt{a^{2} + x^{2}}}] \\ = \frac{1}{x + \sqrt{a^{2} + x^{2}}} \times [\frac{x + \sqrt{a^{2} + x^{2}}}{\sqrt{a^{2} + x^{2}}}] \\ y_{1} & = \frac{1}{\sqrt{a^{2} + x^{2}}} \end{aligned}$

$\begin{matrix} (6.7) & \begin{aligned} y_{1} & = \frac{1}{\sqrt{a^{2} + x^{2}}} \end{aligned} \end{matrix}$

Differentiating again the equation (6.7),

$\begin{aligned} y_{2} & = - \frac{1}{2} (a^{2} + x^{2})^{- 3 / 2} \times 2 x \\ y_{2} & = - \frac{x}{(a^{2} + x^{2})^{3 / 2}} \\ y_{2} & = - \frac{x}{\sqrt{a^{2} + x^{2}} (a^{2} + x^{2})} \end{aligned}$

From (6.7),

$\begin{aligned} y_{2} & = - \frac{x y_{1}}{a^{2} + x^{2}} \end{aligned}$

$\begin{matrix} (6.8) & \begin{aligned} (a^{2} + x^{2}) y_{2} + x y_{1} & = 0 \end{aligned} \end{matrix}$

Hence proved.

and hence show that $(a^{2} + x^{2}) y_{n + 2} + (2 n + 1) x y_{n + 1} + n^{2} y_{n} = 0$ .

Differentiating (6.8) $n$ -times w.r.t $x$ ,

$\begin{aligned} y_{n + 2} (a^{2} + x^{2}) + n y_{n + 1} .2 x + \frac{n (n - 1)}{2} y_{n} .2 + 0 + \\ y_{n + 1} . x + n y_{n} \times 1 & = 0 \end{aligned}$

$\begin{aligned} (a^{2} + x^{2}) y_{n + 2} + (2 n + 1) x y_{n + 1} + (n^{2} - n + n) y_{n} & = 0 \\ (a^{2} + x^{2}) y_{n + 2} + (2 n + 1) x y_{n + 1} + n^{2} y_{n} & = 0 \end{aligned}$

6.1.7 Question 7

If $y = \sin^{- 1} x$ , show that

$(1 - x^{2}) y_{2} - x y_{1} = 0$

Differentiating $y = \sin^{- 1} x$ w.r.t $x$ ,

$\begin{matrix} (6.9) & \begin{aligned} y_{1} & = \frac{1}{\sqrt{1 - x^{2}}} \end{aligned} \end{matrix}$

Differentiating again w.r.t $x$ ,

$\begin{aligned} y_{2} & = - \frac{1}{2} \times \frac{- 2 x}{\sqrt{1 - x^{2}} (1 - x^{2})} \\ y_{2} & = \frac{x}{\sqrt{1 - x^{2}} (1 - x^{2})} \end{aligned}$

Substituting from (6.9),

$\begin{aligned} y_{2} & = \frac{x y_{1}}{1 - x^{2}} \\ (1 - x^{2}) y_{2} - x y_{1} & = 0 \end{aligned}$

$\begin{matrix} (6.10) & \begin{aligned} (1 - x^{2}) y_{2} - x y_{1} & = 0 \end{aligned} \end{matrix}$

Thus proved.

$(1 - x^{2}) y_{n + 2} - (2 n + 1) x y_{n + 1} - n^{2} y_{n} = 0$

Differentiating $n$ -times (6.10),

$\begin{aligned} y_{n + 2} (1 - x^{2}) - n y_{n + 1} .2 x - \frac{n (n - 1)}{2} y_{n} .2 + 0 \\ - y_{n + 1} x - n y_{n} \times 1 + 0 & = 0 \\ (1 - x^{2}) y_{n + 2} - (2 n + 1) x y_{n + 1} - (n^{2} - n + n) y_{n} & = 0 \end{aligned}$

$\begin{matrix} (6.11) & \begin{aligned} (1 - x^{2}) y_{n + 2} - (2 n + 1) x y_{n + 1} - n^{2} y_{n} & = 0 \end{aligned} \end{matrix}$

$(y_{n + 2})_{0} = n^{2} (y_{n})_{0}$ . Find also the value of $(y_{n})_{0}$ .

Putting value of $x = 0$ in (6.11) we get,

$\begin{matrix} (6.12) & \begin{aligned} (y_{n + 2})_{0} & = n^{2} (y_{n})_{0} \end{aligned} \end{matrix}$

Putting $x = 0$ in (6.9), we get $(y_{1})_{0} = 1$ .

Putting $x = 0$ in (6.10), we get $(y_{2})_{0} = 0$ .

From (6.12), we get $(y_{3})_{0} = 1^{2} (y_{1})_{0} = 1^{2} .1 = 1$ .

From (6.12), we get $(y_{4})_{0} = 2^{2} (y_{2})_{0} = 0$ .

From (6.12), we get $(y_{5})_{0} = 3^{2} (y_{3})_{0} = 3^{2} {.1}^{2}$ .

From (6.12), we get $(y_{6})_{0} = 4^{2} (y_{4})_{0} = 0$ .

From (6.12), we get $(y_{7})_{0} = 5^{2} (y_{5})_{0} = 5^{2} {.3}^{2} {.1}^{2}$ .

Thus when $n$ is even $(y_{n})_{0}$ is $0$ . And when $n$ is odd, $(y_{n})_{0}$ is $(n - 2)^{2} (n - 4)^{2} (n - 6)^{2} \dots 5^{2} {.3}^{2} {.1}^{2}$ .

6.1.8 Question 8

Find $y_{n} (0)$ if $y = e^{a \sin^{- 1} x}$ .

Given equation is

$\begin{matrix} (6.13) & \begin{aligned} y & = e^{a \sin^{- 1} x} \end{aligned} \end{matrix}$

Taking log and differentiating both sides,

$\begin{aligned} \log y & = a \sin^{- 1} x \\ \frac{1}{y} y_{1} & = \frac{a}{\sqrt{1 - x^{2}}} \\ y_{1} \sqrt{1 - x^{2}} & = a y \end{aligned}$

Squaring both sides,

$\begin{matrix} (6.14) & \begin{aligned} {y_{1}}^{2} (1 - x^{2}) & = a^{2} y^{2} \end{aligned} \end{matrix}$

Differentiating w.r.t $x$ ,

$\begin{aligned} (1 - x^{2}) 2 y_{1} \times y_{2} - {y_{1}}^{2} \times 2 x & = a^{2} \times 2 y \times y_{1} \end{aligned}$

Dividing both sides by $2 y_{1}$ ,

$\begin{matrix} (6.15) & \begin{aligned} (1 - x^{2}) y_{2} - x y_{1} - a^{2} y & = 0 \end{aligned} \end{matrix}$

Differentiating $n$ -times using Leibnitz’s theorem,

$\begin{aligned} y_{n + 2} (1 - x^{2}) + n y_{n + 1} \times (- 2 x) + \frac{n (n - 1)}{2} y_{n} \times (- 2) + 0 - \\ [y_{n + 1} x + n y_{n} \times 1 + 0] - a^{2} y_{n} & = 0 \\ (1 - x^{2}) y_{n + 2} - 2 n x y_{n + 1} - (n^{2} - n) y_{n} + 0 - \\ x y_{n + 1} - n y_{n} - a^{2} y_{n} & = 0 \\ (1 - x^{2}) y_{n + 2} - (2 n + 1) x y_{n + 1} - (n^{2} - n + n + a^{2}) y_{n} & = 0 \end{aligned}$

$\begin{matrix} (6.16) & \begin{aligned} (1 - x^{2}) y_{n + 2} - (2 n + 1) x y_{n + 1} - (n^{2} + a^{2}) y_{n} & = 0 \end{aligned} \end{matrix}$

Putting $x = 0$ in (6.13), we get $(y)_{0} = 1$ .

Putting $x = 0$ in (6.14), we get $(y_{1})_{0} = a . (y)_{0} = a .1 = a$ .

Putting $x = 0$ in (6.15), we get $(y_{2})_{0} = a^{2} (y)_{0} = a^{2}$ .

Putting $x = 0$ in (6.16), we get

$\begin{matrix} (6.17) & \begin{aligned} (y_{n + 2})_{0} & = (n^{2} + a^{2}) (y_{n})_{0} \end{aligned} \end{matrix}$

From (6.17),

$\begin{aligned} (y_{3})_{0} & = (y_{1 + 2})_{0} = (1^{2} + a^{2}) (y_{1})_{0} = (1^{2} + a^{2}) . a \\ (y_{4})_{0} & = (y_{2 + 2})_{0} = (2^{2} + a^{2}) (y_{2})_{0} = (2^{2} + a^{2}) . a^{2} \\ (y_{5})_{0} & = (y_{3 + 2})_{0} = (3^{2} + a^{2}) (y_{3})_{0} = (3^{2} + a^{2}) (1^{2} + a^{2}) a \\ (y_{6})_{0} & = (y_{4 + 2})_{0} = (4^{2} + a^{2}) (y_{4})_{0} = (4^{2} + a^{2}) (2^{2} + a^{2}) . a^{2} \\ (y_{7})_{0} & = (y_{5 + 2})_{0} = (5^{2} + a^{2}) (y_{5})_{0} = (5^{2} + a^{2}) (3^{2} + a^{2}) (1^{2} + a^{2}) a \end{aligned}$

Thus when $n$ is odd, $(y_{n})_{0} = [(n - 2)^{2} + a^{2}] [(n - 4)^{2} + a^{2}] [(n - 6)^{2} + a^{2}] \dots (1^{2} + a^{2}) . a$ .

And when $n$ is even, $(y_{n})_{0} = [(n - 2)^{2} + a^{2}] [(n - 4)^{2} + a^{2}] [(n - 6)^{2} + a^{2}] \dots (2^{2} + a^{2}) . a^{2}$ .

6.1.9 Question 9

If $y = (x + \sqrt{1 + x^{2}})^{m}$ , show that $(1 + x^{2}) y_{2} + x y_{1} - m^{2} y = 0$ and hence prove that $(1 + x^{2}) y_{n + 2} + (2 n + 1) x y_{n + 1} + (n^{2} - m^{2}) y_{n} = 0$ .

Given equation is,

$\begin{matrix} (6.18) & \begin{aligned} y = (x + \sqrt{1 + x^{2}})^{m} \end{aligned} \end{matrix}$

Differentiating w.r.t $x$ ,

$\begin{aligned} y_{1} & = m (x + \sqrt{1 + x^{2}})^{m - 1} \times [1 + \frac{2 x}{2 \sqrt{1 + x^{2}}}] \\ y_{1} & = \frac{m (x + \sqrt{1 + x^{2}})^{m}}{x + \sqrt{1 + x^{2}}} \times \frac{x + \sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}} \end{aligned}$

From (6.18),

$\begin{matrix} (6.19) & \begin{aligned} y_{1} & = \frac{m y}{\sqrt{1 + x^{2}}} \end{aligned} \end{matrix}$

Differentiating again w.r.t $x$ ,

$\begin{aligned} y_{2} & = m [\frac{y_{1} \sqrt{1 + x^{2}} - y \times \frac{2 x}{2 \sqrt{1 + x^{2}}}}{(1 + x^{2})}] \end{aligned}$

From (6.19),

$\begin{aligned} y_{2} & = m [\frac{y_{1} \times \frac{m y}{y_{1}} - x y \times \frac{y_{1}}{m y}}{1 + x^{2}}] \\ (1 + x^{2}) y_{2} & = m (m y - \frac{x y_{1}}{m}) \\ (1 + x^{2}) y_{2} & = \frac{m (m^{2} y - x y_{1})}{m} \\ (1 + x^{2}) y_{2} + x y_{1} - m^{2} y & = 0 \end{aligned}$

Differentiating $n$ -times using Leibnitz’s theorem,

$\begin{aligned} y_{n + 2} (1 + x^{2}) + n . y_{n + 1} \times 2 x + \frac{n (n - 1)}{2} . y_{n} \times 2 + 0 + \\ y_{n + 1} x + n y_{n} \times 1 + 0 - m^{2} y_{n} & = 0 \end{aligned}$

$\begin{aligned} (1 + x^{2}) y_{n + 2} + (2 n + 1) x y_{n + 1} + (n^{2} - n) y_{n} + n y_{n} - m^{2} y_{n} & = 0 \\ (1 + x^{2}) y_{n + 2} + (2 n + 1) x y_{n + 1} + (n^{2} - m^{2}) y_{n} & = 0 \end{aligned}$

6.1.10 Question 10

If $y^{1 / m} + y^{- 1 / m} = 2 x$ , show that $(x^{2} - 1) y_{n + 2} + (2 n + 1) x y_{n + 1} + (n^{2} - m^{2}) y_{n} = 0$ .

Given equation is:

$\begin{matrix} (6.20) & \begin{aligned} y^{1 / m} + y^{- 1 / m} = 2 x \end{aligned} \end{matrix}$

Differentiating w.r.t $x$ ,

$\begin{aligned} \frac{1}{m} y^{\frac{1}{m} - 1} y_{1} - \frac{1}{m} y^{- \frac{1}{m} - 1} y_{1} & = 2 \\ \frac{1}{m} (\frac{y^{1 / m}}{y} - \frac{y^{- 1 / m}}{y}) y_{1} & = 2 \\ (y^{1 / m} - y^{- 1 / m}) y_{1} & = 2 m y \\ y_{1} \sqrt{(y^{1 / m} - y^{- 1 / m})^{2}} & = 2 m y \\ y_{1} \sqrt{(y^{1 / m} + y^{- 1 / m})^{2} - 4 y^{1 / m} y^{- 1 / m}} & = 2 m y \\ y_{1} \sqrt{(y^{1 / m} + y^{- 1 / m})^{2} - 4} & = 2 m y \end{aligned}$

From (6.20),

$\begin{aligned} y_{1} \sqrt{(2 x)^{2} - 4} & = 2 m y \\ y_{1} \sqrt{4 x^{2} - 4} & = 2 m y \end{aligned}$

Squaring both sides,

$\begin{aligned} {y_{1}}^{2} (4 x^{2} - 4) & = 4 m^{2} y^{2} \\ (x^{2} - 1) {y_{1}}^{2} & = m^{2} y^{2} \end{aligned}$

Differentiating w.r.t $x$ ,

$\begin{aligned} (x^{2} - 1) 2 y_{1} y_{2} + {y_{1}}^{2} .2 x & = m^{2} .2 y . y_{1} \end{aligned}$

Dividing both sides by $2 y_{1}$ ,

$\begin{matrix} (6.21) & \begin{aligned} (x^{2} - 1) y_{2} + x y_{1} = m^{2} y \end{aligned} \end{matrix}$

Differentiating $n$ times (6.21) using Leibnitz’s theorem,

$\begin{aligned} y_{n + 2} (x^{2} - 1) + n y_{n + 1} .2 x + \frac{n (n - 1)}{2} y_{n} .2 + 0 \\ + y_{n + 1} . x + n y_{n} \times 1 + 0 & = m^{2} y_{n} \end{aligned}$

$\begin{aligned} (x^{2} - 1) y_{n + 2} + (2 n + 1) x y_{n + 1} + (n^{2} - n) y_{n} + n y_{n} - m^{2} y_{n} & = 0 \\ (x^{2} - 1) y_{n + 2} + (2 n + 1) x y_{n + 1} + (n^{2} - m^{2}) y_{n} & = 0 \end{aligned}$

6.1.11 Question 11 (TU 2061)

If $y = (\sin^{- 1} x)^{2}$ , prove that:

$(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - 2 = 0$

Differentiating $y = (\sin^{- 1} x)^{2}$ w.r.t $x$ ,

$\begin{aligned} y_{1} & = \frac{2 \sin^{- 1} x}{\sqrt{1 - x^{2}}} \end{aligned}$

Squaring both sides,

$\begin{aligned} {y_{1}}^{2} & = \frac{4 (\sin^{- 1} x)^{2}}{1 - x^{2}} \end{aligned}$

From initial equation,

$\begin{aligned} {y_{1}}^{2} & = \frac{4 y}{1 - x^{2}} \\ {y_{1}}^{2} (1 - x^{2}) & = 4 y \end{aligned}$

Differentiating again w.r.t $x$ ,

$\begin{aligned} (1 - x^{2}) .2 y_{1} y_{2} - 2 x {y_{1}}^{2} & = 4 y_{1} \end{aligned}$

Dividing both sides by $2 y_{1}$ ,

$\begin{aligned} (1 - x^{2}) y_{2} - x y_{1} & = 2 \\ (1 - x^{2}) y_{2} - x y_{1} - 2 & = 0 \\ (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - 2 & = 0 \end{aligned}$

$\begin{matrix} (6.22) & \begin{aligned} (1 - x^{2}) y_{2} - x y_{1} - 2 & = 0 \end{aligned} \end{matrix}$

$(1 - x^{2}) y_{n + 2} - (2 n + 1) x y_{n + 1} - x^{2} y_{n} = 0$

Differentiating (6.22) $n$ -times w.r.t $x$ ,

$\begin{aligned} y_{n + 2} (1 - x^{2}) - n y_{n + 1} .2 x - \frac{n (n - 1)}{2} y_{n} .2 + 0 \\ - [y_{n + 1} . x + n y_{n} \times 1 + 0] - 0 & = 0 \end{aligned}$

$\begin{aligned} (1 - x^{2}) y_{n + 2} - (2 n + 1) x y_{n + 1} - (n^{2} - n) y_{n} - n y_{n} & = 0 \\ (1 - x^{2}) y_{n + 2} - (2 n + 1) x y_{n + 1} - n^{2} y_{n} & = 0 \end{aligned}$

6.1.12 Question 12

If $y = \sin (m \sin^{- 1} x)$ , prove that:

$(1 - x^{2}) y_{2} - x y_{1} + m^{2} y = 0$

$\begin{matrix} (6.23) & \begin{aligned} y = \sin (m \sin^{- 1} x) \end{aligned} \end{matrix}$

Differentiating (6.23) w.r.t $x$ ,

$\begin{aligned} y_{1} & = m \cos (m \sin^{- 1} x) \frac{1}{\sqrt{1 - x^{2}}} \end{aligned}$

Squaring both sides,

$\begin{aligned} (1 - x^{2}) {y_{1}}^{2} & = m^{2} \cos^{2} (m \sin^{- 1} x) \\ = m^{2} [1 - \sin^{2} (m \sin^{- 1} x)] \end{aligned}$

From (6.23),

$\begin{aligned} (1 - x^{2}) {y_{1}}^{2} & = m^{2} (1 - y^{2}) \end{aligned}$

Differentiating again w.r.t $x$ ,

$\begin{aligned} (1 - x^{2}) 2 y_{1} y_{2} - 2 x {y_{1}}^{2} & = m^{2} (0 - 2 y y_{1}) \\ (1 - x^{2}) 2 y_{1} y_{2} - 2 x {y_{1}}^{2} + 2 m^{2} y y_{1} & = 0 \end{aligned}$

Dividing both sides by $2 y_{1}$ ,

$\begin{matrix} (6.24) & \begin{aligned} (1 - x^{2}) y_{2} - x y_{1} + m^{2} y & = 0 \end{aligned} \end{matrix}$

$(1 - x^{2}) y_{n + 2} - (2 n + 1) x y_{n + 1} + (m^{2} - n^{2}) y_{n} = 0$

Differentiating (6.24) $n$ -times w.r.t $x$ ,

$\begin{aligned} y_{n + 2} (1 - x^{2}) + n y_{n + 1} \times (- 2 x) + \frac{n (n - 1)}{2} y_{n} \times (- 2) + 0 \\ - [y_{n + 1} x + n y_{n} \times 1 + 0] + m^{2} y_{n} & = 0 \\ (1 - x^{2}) y_{n + 2} - 2 n x y_{n + 1} - \\ (n^{2} - n) y_{n} - x y_{n + 1} - n y_{n} + m^{2} y_{n} & = 0 \\ (1 - x^{2}) y_{n + 2} - (2 n + 1) x y_{n + 1} - (m^{2} - n^{2} + n - n) y_{n} & = 0 \end{aligned}$

$\begin{aligned} (1 - x^{2}) y_{n + 2} - (2 n + 1) x y_{n + 1} - (m^{2} - n^{2}) y_{n} & = 0 \end{aligned}$

Hence proved.

6.1.13 Question 13 (TU 2064)

If $\log y = \tan^{- 1} x$ , show that $(1 + x^{2}) y_{2} + (2 x - 1) y_{1} = 0$ .

Differentiating w.r.t $x$ ,

$\begin{aligned} \frac{1}{y} y_{1} & = \frac{1}{1 + x^{2}} \\ y_{1} (1 + x^{2}) & = y \end{aligned}$

Differentiating again w.r.t $x$ ,

$\begin{matrix} (6.25) & \begin{aligned} (1 + x^{2}) y_{2} + (2 x - 1) y_{1} & = 0 \end{aligned} \end{matrix}$

Differentiate this differential equation $n$ times.

Differentiating (6.25) $n$ -times,

$\begin{aligned} y_{n + 2} (1 + x^{2}) + n y_{n + 1} .2 x + \frac{n (n - 1)}{2} . y_{n} .2 + \\ 0 + y_{n + 1} (2 x - 1) + n y_{n} .2 + 0 & = 0 \\ (1 + x^{2}) y_{n + 2} + 2 n x y_{n + 1} + (2 x - 1) y_{n + 1} + (n^{2} - n) y_{n} + 2 n y_{n} & = 0 \\ (1 + x^{2}) y_{n + 2} + (2 n x + 2 x - 1) y_{n + 1} + (n^{2} + n) y_{n} & = 0 \end{aligned}$

$\begin{aligned} (1 + x^{2}) y_{n + 2} + (2 n x + 2 x - 1) y_{n + 1} + n (n + 1) y_{n} & = 0 \end{aligned}$