Chapter 6 Higher Order Derivatives-II

6.0.1 Leibnitz’s theorem

If \(y = uv\) where \(u\) and \(v\) are functions of \(x\) possessing \(\text{n}^{th}\) derivatives, then

\[\begin{equation*} \begin{split} y_n &= {}^{n}C_{0}u_nv + {}^{n}C_{1}u_{n-1}v_1 + {}^{n}C_{2}u_{n-2}v_2 + {}^{n}C_{3}u_{n-3}v_3 + \ldots + {}^{n}C_{r}u_{n-r}v_r + \ldots + {}^{n}C_{n}uv_n\\ &= u_nv + {}^{n}C_{1}u_{n-1}v_1 + {}^{n}C_{2}u_{n-2}v_2 + {}^{n}C_{3}u_{n-3}v_3 + \ldots + {}^{n}C_{r}u_{n-r}v_r + \ldots + uv_n\\ (uv)_{(n)} (x) &= \sum_{k=0}^{n} {}^{n}C_{k} u_{(n-k)}(x) v_{(k)} (x), \end{split} \end{equation*}\]

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 \(\frac{d^ny}{dx^n}\) if \(y\) is

• \(\boldsymbol{x^2e^{ax}}\)

Here \(u = e^{ax}\) and \(v = x^2\).

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

\[\begin{equation*} \begin{split} u_1 &= ae^{ax}\\ u_2 &= a^2 e^{ax}\\ u_3 &= a^3 e^{ax}\\ u_4 &= a^4 e^{ax}\\ \vdots\\ u_n &= a^n e^{ax} \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} v &= x^2 \\ v_1 &= 2x \\ v_2 &= 2\\ v_3 &= 0\\ \vdots\\ v_n &= 0 \end{split} \end{equation*}\]

From Leibnitz’s theorem,

\[\begin{equation*} \begin{split} y_n &= u_nv + {}^{n}C_{1}u_{n-1}v_1 + {}^{n}C_{2}u_{n-2}v_2 + {}^{n}C_{3}u_{n-3}v_3 + \ldots + {}^{n}C_{r}u_{n-r}v_r + \ldots + uv_n\\ &= a^n e^{ax}x^2 + {}^{n}C_{1}a^{n-1}e^{ax}.2x + {}^{n}C_{2}a^{n-2} e^{ax}.2 + \ldots + 0\\ &= a^n e^{ax}x^2 + 2n a^{n-1}e^{ax}x + \frac{n(n-1)}{2}a^{n-2}e^{ax}.2\\ y_n &= a^n e^{ax}x^2 + 2n a^{n-1}e^{ax}x + \frac{n(n-1)}{2}a^{n-2}e^{ax}.2\\ y_n &= a^n e^{ax}x^2 + 2n a^{n-1}e^{ax}x + n(n-1)a^{n-2}e^{ax} \end{split} \end{equation*}\]

• \(\boldsymbol{x^3\sin x}\)

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 \(\sin(ax+b)\).

\[\begin{equation*} \begin{split} u_1 &= \cos x\\ u_2 &= -\sin x\\ u_3 &= -\cos x\\ u_4 &= \sin x\\ \vdots\\ u_n &= 1^n\sin(x+ 0 +\frac{n\pi}{2})=\sin(x+\frac{n\pi}{2}) \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} v &= x^3 \\ v_1 &= 3x^2 \\ v_2 &= 6x\\ v_3 &= 6\\ v_4 &= 0\\ \vdots\\ v_n &= 0 \end{split} \end{equation*}\]

From Leibnitz’s theorem,

\[\begin{equation*} \begin{split} y_n &= u_nv + {}^{n}C_{1}u_{n-1}v_1 + {}^{n}C_{2}u_{n-2}v_2 + \\ {}^{n}C_{3}u_{n-3}v_3 + \ldots + {}^{n}C_{r}u_{n-r}v_r + \ldots + uv_n\\ &= \sin\left(x+\frac{n\pi}{2}\right).x^3 + \\ n. \sin \left(x + \frac{(n-1)\pi}{2}\right).3x^2 + \\ \frac{n(n-1)}{2}.\sin \left(x + \frac{(n-2)\pi}{2}\right).6x + \\ \frac{n(n-1)(n-2)}{6}\sin \left(x + \frac{(n-3)\pi}{2}\right).6 + \\ 0 + \ldots + 0\\ y_n &= x^3 \sin \left(x+\frac{n\pi}{2}\right) + \\ 3nx^2 \sin \left(x + \frac{(n-1)\pi}{2}\right) + \\ 3n(n-1)x \sin \left(x + \frac{(n-2)\pi}{2}\right) + \\ n(n-1)(n-2) \sin \left(x + \frac{(n-3)\pi}{2}\right) \end{split} \end{equation*}\]

• \(\boldsymbol{x^3 \log x}\)

\[\begin{equation*} \begin{split} u &= \log x\\ u_1 &= \frac{1}{x}\\ u_2 &= -\frac{1}{x^2}\\ u_3 &= \frac{2}{x^3}\\ u_4 &= -\frac{6}{x^4}\\ \vdots\\ u_n &= \frac{(-1)^{n-1}(n-1)!}{x^n} \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} v &= x^3 \\ v_1 &= 3x^2 \\ v_2 &= 6x\\ v_3 &= 6\\ v_4 &= 0\\ \vdots\\ v_n &= 0 \end{split} \end{equation*}\]

From Leibnitz’s theorem,

\[\begin{equation*} \begin{split} y_n &= u_nv + {}^{n}C_{1}u_{n-1}v_1 + {}^{n}C_{2}u_{n-2}v_2 + {}^{n}C_{3}u_{n-3}v_3 \\ + \ldots + {}^{n}C_{r}u_{n-r}v_r + \ldots + uv_n\\ y_n &= \frac{(-1)^{n-1}(n-1)!}{x^n}.x^3 + n.\frac{(-1)^{n-2}(n-2)!}{x^{n-1}}.3x^2 + \\ \frac{n(n-1)}{2}\frac{(-1)^{n-3}(n-3)!}{x^{n-2}}.6x + \\ \frac{n(n-1)(n-2)}{6}\frac{(-1)^{n-4}(n-4)!}{x^{n-3}}.6 + \\ 0 + \ldots + 0 \\ y_n &= \frac{(-1)^n}{x^{n-3}}\left[\frac{(n-1)!}{(-1)^1} + \frac{3n(n-2)!}{(-1)^2} + \\ \frac{3n(n-1)(n-3)!}{(-1)^3} + \frac{n(n-1)(n-2)(n-4)!}{(-1)^4}\right] \end{split} \end{equation*}\]

• \(\boldsymbol{x^ne^x}\)

\[\begin{equation*} \begin{split} u_1 &= e^x \\ u_2 &= e^x \\ u_3 &= e^x \\ u_4 &= e^x \\ \vdots\\ u_n &= e^x \\ \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} v &= x^n \\ v_1 &= nx^{n-1} \\ v_2 &= n(n-1)x^{n-2}\\ v_3 &= n(n-1)(n-2)x^{n-3}\\ v_4 &= n(n-1)(n-2)(n-3)x^{n-4}\\ \vdots\\ v_n &= n! \end{split} \end{equation*}\]

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,

\[\begin{equation*} \begin{split} y_n &= u_nv + {}^{n}C_{1}u_{n-1}v_1 + {}^{n}C_{2}u_{n-2}v_2 + {}^{n}C_{3}u_{n-3}v_3 \\ + \ldots + {}^{n}C_{r}u_{n-r}v_r + \ldots + uv_n\\ &= e^x x^n + n.e^x.nx^{n-1} + \\ \frac{n(n-1)}{2}.e^x.n(n-1)x^{n-2} + \\ \frac{n(n-1)(n-2)}{6}.e^x.n(n-1)(n-2)x^{n-3} + \\ \ldots + e^x.n! \\ \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} y_n &= e^x \left[x^n + \frac{n^2 x^{n-1}}{1!} + \frac{n^2 (n-1)^2 x^{n-2}}{2!} + \\ \frac{n^2 (n-1)^2 (n-2)^2 x^{n-3}}{3!} + \\ \ldots + n! \right]\\ &= e^x \left[x^n + \frac{n^2 x^{n-1}}{1!} + \frac{n^2 (n-1)^2 x^{n-2}}{2!} + \\ \frac{n^2 (n-1)^2 (n-2)^2 x^{n-3}}{3!} + \\ \ldots + \frac{(n!)^2}{n!} \right]\\ y_n &= e^x \left[x^n + \frac{n^2 x^{n-1}}{1!} + \frac{n^2 (n-1)^2 x^{n-2}}{2!} + \\ \frac{n^2 (n-1)^2 (n-2)^2 x^{n-3}}{3!} + \\ \ldots + \frac{n^2 (n-1)^2 (n-2)^2 (n-3)^2\ldots 1^2}{n!} \right]\\ \end{split} \end{equation*}\]

• \(\boldsymbol{e^{ax}\cos bx}\)

\[\begin{equation*} \begin{split} u &= e^{ax}\\ u_n &= a^n e^{ax}\\ \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} v &=\cos bx\\ v_n &= b^n \cos\left(bx+\frac{n\pi}{2}\right) \end{split} \end{equation*}\]

From Leibnitz’s theorem,

\[\begin{equation*} \begin{split} y_n &= u_nv + {}^{n}C_{1}u_{n-1}v_1 + {}^{n}C_{2}u_{n-2}v_2 + {}^{n}C_{3}u_{n-3}v_3 \\ + \ldots + {}^{n}C_{r}u_{n-r}v_r + \ldots + uv_n\\ &= a^n e^{ax} \cos bx + {}^{n}C_{1} a^{n-1}e^{ax}. b^1 \cos\left(bx+\frac{1\pi}{2}\right) + \\ {}^{n}C_{2}a^{n-2}e^{ax}.b^2 \cos\left(bx+\frac{2\pi}{2}\right) + \\ \ldots + e^{ax}. b^n \cos\left(bx+\frac{n\pi}{2}\right) \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} y_n &= e^{ax}\left[a^n \cos bx + {}^{n}C_{1} a^{n-1}b \cos\left(bx+\frac{\pi}{2}\right) + \\ {}^{n}C_{2}a^{n-2}b^2 \cos\left(bx+\frac{2\pi}{2}\right) + \\ \ldots + b^n \cos\left(bx+\frac{n\pi}{2}\right)\right] \\ \end{split} \end{equation*}\]

6.1.2 Question 2

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

Given equation is:

\[\begin{equation} \begin{split} y=x^{n-1} \log x \end{split} \tag{6.1} \end{equation}\]

Differentiating w.r.t \(x\),

\[\begin{equation*} \begin{split} y_1 &= \frac{x^{n-1}}{x} + \log x (n-1) x^{n-2}\\ &= \frac{x^{n-1}}{x} + \frac{\log x (n-1) x^{n-1}}{x} \end{split} \end{equation*}\]

From (6.1),

\[\begin{equation*} \begin{split} y_1 &= \frac{x^{n-1}}{x} + \frac{y(n-1)}{x}\\ xy_1 &= x^{n-1} + (n-1)y\\ xy_1 - (n-1)y &= x^{n-1} \end{split} \end{equation*}\]

Differentiating it \(n-1\) times using Leibnitz’s theorem,

\[\begin{equation*} \begin{split} {}^{n-1}C_{0} y_{1+(n-1)}x + {}^{n-1}C_{1}y_{n-1}\times 1 + 0 \\ - (n-1)y_{n-1} &= (n-1)!\\ \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} 1\times xy_n + \frac{(n-1)!}{1 !(n-2)!} y_{n-1} - (n-1)y_{n-1} &= (n-1)!\\ xy_n + (n-1) y_{n-1} - (n-1)y_{n-1} &= (n-1)!\\ xy_n &= (n-1)! \end{split} \end{equation*}\]

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 \(y_{n+1} - 2xy_n - 2ny_{n-1} = 0\).

Plotting reveals this graph:

Taking log and differentiating w.r.t \(x\),

\[\begin{equation*} \begin{split} \log y &= x^2 \log e\\ \frac{1}{y}y_1 &= 2x\\ y_1 &= 2xy \end{split} \end{equation*}\]

Differentiating \(n\)-times using Leibnitz’s theorem,

\[\begin{equation*} \begin{split} y_{n+1} &= 2[y_nx + {}^{n}C_{1}y_{n-1}\times 1 + 0]\\ y_{n+1} &= 2[y_nx + ny_{n-1}]\\ y_{n+1} - 2xy_n - 2ny_{n-1} &= 0 \end{split} \end{equation*}\]

Hence proved.

6.1.4 Question 4

If \(y=e^{ax}\sin bx\), show that

• \(\boldsymbol{y_2 - 2ay_1 + (a^2 +b^2)y = 0}\)

Differentiating w.r.t \(x\),

\[\begin{equation*} \begin{split} y_1 &= e^{ax}.b\cos bx + \sin bx. ae^{ax}\\ y_1 &= b e^{ax} \cos bx + a e^{ax}\sin bx\\ \end{split} \end{equation*}\]

Substituting original equation,

\[\begin{equation*} \begin{split} y_1 &= b e^{ax} \cos bx + ay \\ \end{split} \tag{6.2} \end{equation*}\]

Again, differentiating w.r.t \(x\),

\[\begin{equation*} \begin{split} y_2 &= b[a e^{ax}\cos bx - b e^{ax}\sin bx] + ay_1\\ \end{split} \end{equation*}\]

From original equation and (6.2),

\[\begin{equation*} \begin{split} y_2 &= b[a e^{ax}\cos bx - by] +ay_1\\ &= b\left[a \times \frac{y_1-ay}{b}-by\right] + ay_1\\ y_2 &= ay_1 - a^2y - b^2y +ay_1\\ \end{split} \end{equation*}\]

\[\begin{equation} \begin{split} y_2 - 2ay_1 + (a^2 + b^2)y &= 0 \end{split} \tag{6.3} \end{equation}\]

• \(\boldsymbol{y_{n+1} = 2ay_n - (a^2 + b^2)y_{n-1}}\)

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

\[\begin{equation*} \begin{split} y_{(n-1)+2} -2ay_{(n-1)+1} + (a^2 + b^2)y_{n-1} &= 0\\ y_{n+1} -2ay_n + (a^2 + b^2)y_{n-1} &= 0\\ y_{n+1} &= 2ay_n - (a^2 + b^2)y_{n-1} \end{split} \end{equation*}\]

6.1.5 Question 5

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

• \(\boldsymbol{x^2y_2 + xy_1 + y = 0}\)

Given equation is,

\[\begin{equation} \begin{split} y &= a \cos (\log x) + b \sin (\log x) \end{split} \tag{6.4} \end{equation}\]

Differentiating w.r.t \(x\),

\[\begin{equation} \begin{split} y_1 &= -\frac{a\sin (\log x)}{x} + \frac{b \cos(\log x)}{x}\\ \end{split} \end{equation}\]

which can also be written as,

\[\begin{equation} \begin{split} xy_1 &= b\cos(\log x) - a\sin (\log x) \end{split} \tag{6.5} \end{equation}\]

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

\[\begin{equation*} \begin{split} xy_2 + y_1 &= \dfrac{-b\sin(\log x)}{x} - \dfrac{a\cos \log x}{x}\\ x^2y_2 + xy_1 &= -(a \cos (\log x) + b \sin (\log x))\\ \end{split} \end{equation*}\]

From original equation (6.4),

\[\begin{equation*} \begin{split} x^2y_2 + xy_1 &= -(y)\\ \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} x^2y_2 + xy_1 + y &= 0 \end{split} \tag{6.6} \end{equation*}\]

Hence proved.

• \(\boldsymbol{x^2y_{n+2} + (2n+1)xy_{n+1} + (n^2 +1)y_n = 0}\)

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

\[\begin{equation*} \begin{split} y_{n+2}.x^2 + n.y_{n+1}.2x + \frac{n(n-1)}{2}y_n.2 + 0 + \\ y_{n+1}.x + n y_n \times 1 + 0 +y_n &=0 \\ \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} x^2y_{n+2} + (2n+1)xy_{n+1} + (n^2 -n + n + 1)y_n &= 0 \\ x^2y_{n+2} + (2n+1)xy_{n+1} + (n^2 + 1)y_n &= 0 \end{split} \end{equation*}\]

Hence proved.

6.1.6 Question 6

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

• \(\boldsymbol{(a^2 + x^2)y_2 + xy_1 = 0}\)

Differentiating w.r.t \(x\),

\[\begin{equation*} \begin{split} y_1 &= \frac{1}{x + \sqrt{a^2+x^2}} \times \left[1 + \frac{2x}{2\sqrt{a^2 + x^2}}\right]\\ &= \frac{1}{x + \sqrt{a^2+x^2}} \times \left[1 + \frac{x}{\sqrt{a^2 + x^2}}\right]\\ &= \frac{1}{x + \sqrt{a^2+x^2}} \times \left[\frac{x + \sqrt{a^2+x^2}}{\sqrt{a^2+x^2}}\right]\\ y_1 &= \frac{1}{\sqrt{a^2 + x^2}} \end{split} \end{equation*}\]

\[\begin{equation} \begin{split} y_1 &= \frac{1}{\sqrt{a^2 + x^2}} \end{split} \tag{6.7} \end{equation}\]

Differentiating again the equation (6.7),

\[\begin{equation*} \begin{split} y_2 &= -\frac{1}{2}(a^2 + x^2)^{-3/2} \times 2x \\ y_2 &= -\frac{x}{(a^2 + x^2)^{3/2}}\\ y_2 &= -\frac{x}{\sqrt{a^2 + x^2} (a^2 + x^2)}\\ \end{split} \end{equation*}\]

From (6.7),

\[\begin{equation*} \begin{split} y_2 &= -\frac{xy_1}{a^2 + x^2}\\ \end{split} \end{equation*}\]

\[\begin{equation} \begin{split} (a^2 + x^2)y_2 + xy_1 &= 0 \end{split} \tag{6.8} \end{equation}\]

Hence proved.

• and hence show that \(\boldsymbol{(a^2 + x^2)y_{n+2} + (2n+1)xy_{n+1} + n^2y_n = 0}\).

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

\[\begin{equation*} \begin{split} y_{n+2}(a^2 + x^2) + ny_{n+1}.2x + \frac{n(n-1)}{2}y_n.2 + 0 + \\ y_{n+1}.x + ny_n \times 1 &= 0\\ \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} (a^2 + x^2)y_{n+2} + (2n+1)xy_{n+1} + (n^2 - n + n )y_n &= 0\\ (a^2 + x^2)y_{n+2} + (2n+1)xy_{n+1} + n^2 y_n &= 0 \end{split} \end{equation*}\]

6.1.7 Question 7

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

• \(\boldsymbol{(1-x^2)y_2 -xy_1 = 0}\)

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

\[\begin{equation} \begin{split} y_1 &= \frac{1}{\sqrt{1-x^2}} \end{split} \tag{6.9} \end{equation}\]

Differentiating again w.r.t \(x\),

\[\begin{equation*} \begin{split} y_2 &= -\frac{1}{2}\times \frac{-2x}{\sqrt{1-x^2}(1-x^2)}\\ y_2 &= \frac{x}{\sqrt{1-x^2}(1-x^2)} \end{split} \end{equation*}\]

Substituting from (6.9),

\[\begin{equation*} \begin{split} y_2 &= \frac{xy_1}{1-x^2}\\ (1-x^2)y_2 - xy_1 &= 0 \end{split} \end{equation*}\]

\[\begin{equation} \begin{split} (1-x^2)y_2 - xy_1 &= 0 \end{split} \tag{6.10} \end{equation}\]

Thus proved.

• \(\boldsymbol{(1-x^2)y_{n+2} - (2n+1)xy_{n+1} - n^2 y_n = 0}\)

Differentiating \(n\)-times (6.10),

\[\begin{equation*} \begin{split} y_{n+2}(1-x^2) -ny_{n+1}.2x - \frac{n(n-1)}{2}y_n.2 + 0 \\ - y_{n+1}x -ny_n \times 1 + 0 &= 0\\ (1-x^2)y_{n+2} - (2n+1)xy_{n+1} - (n^2 -n + n) y_n &= 0 \\ \end{split} \end{equation*}\]

\[\begin{equation} \begin{split} (1-x^2)y_{n+2} - (2n+1)xy_{n+1} - n^2 y_n &= 0 \end{split} \tag{6.11} \end{equation}\]

• \(\boldsymbol{(y_{n+2})_0 = n^2(y_n)_0}\). Find also the value of \(\boldsymbol{(y_n)_0}\).

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

\[\begin{equation} \begin{split} (y_{n+2})_0 &= n^2(y_n)_0 \end{split} \tag{6.12} \end{equation}\]

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 \((n-2)^2(n-4)^2(n-6)^2\ldots 5^2.3^2.1^2\).

6.1.8 Question 8

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

Given equation is

\[\begin{equation} \begin{split} y &= e^{a\sin^{-1}x} \end{split} \tag{6.13} \end{equation}\]

Taking log and differentiating both sides,

\[\begin{equation*} \begin{split} \log y &= a\sin^{-1}x\\ \frac{1}{y}y_1 &= \frac{a}{\sqrt{1-x^2}}\\ y_1 \sqrt{1-x^2} &= ay \end{split} \end{equation*}\]

Squaring both sides,

\[\begin{equation} \begin{split} {y_1}^2(1-x^2) &= a^2y^2 \end{split} \tag{6.14} \end{equation}\]

Differentiating w.r.t \(x\),

\[\begin{equation*} \begin{split} (1-x^2)2y_1 \times y_2 - {y_1}^2 \times 2x &= a^2 \times 2y \times y_1\\ \end{split} \end{equation*}\]

Dividing both sides by \(2y_1\),

\[\begin{equation} \begin{split} (1-x^2)y_2 - xy_1 - a^2y &= 0 \end{split} \tag{6.15} \end{equation}\]

Differentiating \(n\)-times using Leibnitz’s theorem,

\[\begin{equation*} \begin{split} y_{n+2}(1-x^2) + ny_{n+1}\times (-2x) + \frac{n(n-1)}{2}y_n \times (-2) + 0 - \\ [y_{n+1}x + ny_n \times 1 + 0] -a^2y_n &= 0\\ (1-x^2)y_{n+2} -2nxy_{n+1} - (n^2 -n)y_n + 0 - \\ xy_{n+1} -ny_n - a^2y_n &= 0\\ (1-x^2)y_{n+2} - (2n+1)xy_{n+1} - (n^2 -n + n + a^2)y_n &= 0\\ \end{split} \end{equation*}\]

\[\begin{equation} \begin{split} (1-x^2)y_{n+2} - (2n+1)xy_{n+1} - (n^2 + a^2)y_n &= 0\\ \end{split} \tag{6.16} \end{equation}\]

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

\[\begin{equation} \begin{split} (y_{n+2})_0 &= (n^2 + a^2)(y_n)_0 \end{split} \tag{6.17} \end{equation}\]

From (6.17),

\[\begin{equation*} \begin{split} (y_3)_0 &= (y_{1+2})_0 = (1^2 + a^2)(y_1)_0 = (1^2 + a^2).a \\ (y_4)_0 &= (y_{2+2})_0 = (2^2 + a^2)(y_2)_0 = (2^2 + a^2).a^2\\ (y_5)_0 &= (y_{3+2})_0 = (3^2 + a^2)(y_3)_0 = (3^2 + a^2)(1^2 + a^2)a\\ (y_6)_0 &= (y_{4+2})_0 = (4^2 + a^2)(y_4)_0 = (4^2 + a^2)(2^2 + a^2).a^2\\ (y_7)_0 &= (y_{5+2})_0 = (5^2 + a^2)(y_5)_0 = (5^2 + a^2)(3^2 + a^2)(1^2 + a^2)a \end{split} \end{equation*}\]

Thus when \(n\) is odd, \((y_n)_0 = [(n-2)^2 + a^2][(n-4)^2 + a^2][(n-6)^2 + a^2]\ldots (1^2+a^2).a\).

And when \(n\) is even, \((y_n)_0 = [(n-2)^2 + a^2][(n-4)^2 + a^2][(n-6)^2 + a^2]\ldots (2^2 + a^2).a^2\).

6.1.9 Question 9

If \(y=(x+\sqrt{1+x^2})^m\), show that \((1+x^2)y_2 + xy_1 -m^2y = 0\) and hence prove that \((1+x^2)y_{n+2} + (2n+1)xy_{n+1} + (n^2 - m^2)y_n = 0\).

Given equation is,

\[\begin{equation} \begin{split} y=(x+\sqrt{1+x^2})^m \end{split} \tag{6.18} \end{equation}\]

Differentiating w.r.t \(x\),

\[\begin{equation*} \begin{split} y_1 &= m(x+\sqrt{1+x^2})^{m-1} \times \left[1 + \frac{2x}{2\sqrt{1+x^2}}\right]\\ y_1 &= \frac{m(x+\sqrt{1+x^2})^m}{x+\sqrt{1+x^2}} \times \frac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}}\\ \end{split} \end{equation*}\]

From (6.18),

\[\begin{equation} \begin{split} y_1 &= \frac{my}{\sqrt{1+x^2}} \end{split} \tag{6.19} \end{equation}\]

Differentiating again w.r.t \(x\),

\[\begin{equation*} \begin{split} y_2 &= m\left[\frac{y_1\sqrt{1+x^2}- y\times \frac{2x}{2\sqrt{1+x^2}}}{(1+x^2)}\right]\\ \end{split} \end{equation*}\]

From (6.19),

\[\begin{equation*} \begin{split} y_2 &= m\left[\frac{y_1 \times \frac{my}{y_1} - xy\times \frac{y_1}{my}}{1+x^2}\right]\\ (1+x^2)y_2 &= m\left(my -\frac{xy_1}{m}\right)\\ (1+x^2)y_2 &= \frac{m(m^2y - xy_1)}{m}\\ (1+x^2)y_2 + xy_1 - m^2y &= 0 \end{split} \end{equation*}\]

Differentiating \(n\)-times using Leibnitz’s theorem,

\[\begin{equation*} \begin{split} y_{n+2}(1+x^2) + n.y_{n+1}\times 2x + \frac{n(n-1)}{2}.y_n \times 2 + 0 + \\ y_{n+1}x + ny_n \times 1 + 0 -m^2y_n &= 0\\ \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} (1+x^2)y_{n+2} + (2n+1)xy_{n+1} + (n^2 -n)y_n + ny_n -m^2y_n &= 0\\ (1+x^2)y_{n+2} + (2n+1)xy_{n+1} + (n^2 - m^2)y_n &= 0 \end{split} \end{equation*}\]

6.1.10 Question 10

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

Given equation is:

\[\begin{equation} \begin{split} y^{1/m} + y^{-1/m} = 2x \end{split} \tag{6.20} \end{equation}\]

Differentiating w.r.t \(x\),

\[\begin{equation*} \begin{split} \frac{1}{m}y^{\frac{1}{m} -1}y_1 - \frac{1}{m}y^{-\frac{1}{m} -1}y_1 &= 2\\ \frac{1}{m}\left(\frac{y^{1/m}}{y} - \frac{y^{-1/m}}{y} \right)y_1 &= 2\\ (y^{1/m} - y^{-1/m})y_1 &=2my\\ y_1\sqrt{(y^{1/m} - y^{-1/m})^2} &= 2my\\ y_1\sqrt{(y^{1/m} + y^{-1/m})^2 - 4y^{1/m}y^{-1/m}} &= 2my\\ y_1\sqrt{(y^{1/m} + y^{-1/m})^2 - 4} &= 2my\\ \end{split} \end{equation*}\]

From (6.20),

\[\begin{equation*} \begin{split} y_1\sqrt{(2x)^2 - 4} &= 2my\\ y_1\sqrt{4x^2 - 4} &= 2my\\ \end{split} \end{equation*}\]

Squaring both sides,

\[\begin{equation*} \begin{split} {y_1}^2 (4x^2 -4) &= 4m^2y^2\\ (x^2 -1){y_1}^2 &= m^2y^2 \end{split} \end{equation*}\]

Differentiating w.r.t \(x\),

\[\begin{equation*} \begin{split} (x^2-1)2y_1y_2 + {y_1}^2.2x &= m^2.2y.y_1\\ \end{split} \end{equation*}\]

Dividing both sides by \(2y_1\),

\[\begin{equation*} \begin{split} (x^2-1)y_2 + xy_1 = m^2y \end{split} \tag{6.21} \end{equation*}\]

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

\[\begin{equation*} \begin{split} y_{n+2}(x^2-1) + ny_{n+1}.2x + \frac{n(n-1)}{2}y_n.2 + 0 \\ + y_{n+1}.x + ny_n\times 1 + 0 &= m^2y_n\\ \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} (x^2-1)y_{n+2} + (2n+1)xy_{n+1} + (n^2 -n)y_n + ny_n -m^2y_n &= 0\\ (x^2-1)y_{n+2} + (2n+1)xy_{n+1} + (n^2 -m^2)y_n &= 0 \end{split} \end{equation*}\]

6.1.11 Question 11 (TU 2061)

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

• \(\boldsymbol{(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx}-2 = 0}\)

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

\[\begin{equation*} \begin{split} y_1 &= \frac{2\sin^{-1}x}{\sqrt{1-x^2}}\\ \end{split} \end{equation*}\]

Squaring both sides,

\[\begin{equation*} \begin{split} {y_1}^2 &= \frac{4 (\sin^{-1}x)^2}{1-x^2}\\ \end{split} \end{equation*}\]

From initial equation,

\[\begin{equation*} \begin{split} {y_1}^2 &= \frac{4y}{1-x^2}\\ {y_1}^2 (1-x^2) &= 4y\\ \end{split} \end{equation*}\]

Differentiating again w.r.t \(x\),

\[\begin{equation*} \begin{split} (1-x^2).2y_1y_2 - 2x{y_1}^2 &= 4y_1\\ \end{split} \end{equation*}\]

Dividing both sides by \(2y_1\),

\[\begin{equation*} \begin{split} (1-x^2)y_2 - xy_1 &= 2\\ (1-x^2)y_2 - xy_1 - 2 &= 0\\ (1-x^2)\frac{d^2y}{dx^2} -x\frac{dy}{dx} - 2 &= 0 \end{split} \end{equation*}\]

\[\begin{equation} \begin{split} (1-x^2)y_2 - xy_1 - 2 &= 0 \end{split} \tag{6.22} \end{equation}\]

• \(\boldsymbol{(1-x^2)y_{n+2} - (2n+1)xy_{n+1} - x^2y_n = 0}\)

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

\[\begin{equation*} \begin{split} y_{n+2}(1-x^2) - ny_{n+1}.2x - \frac{n(n-1)}{2}y_n.2 + 0 \\ - [y_{n+1}.x + ny_n\times 1 + 0] - 0 &= 0\\ \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} (1-x^2)y_{n+2} - (2n+1)xy_{n+1} - (n^2 -n)y_n - ny_n &= 0\\ (1-x^2)y_{n+2} - (2n+1)xy_{n+1} - n^2y_n &= 0\\ \end{split} \end{equation*}\]

6.1.12 Question 12

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

• \(\boldsymbol{(1-x^2)y_2 - xy_1 + m^2y =0}\)

\[\begin{equation} \begin{split} y=\sin(\text{m}\sin^{-1}x) \end{split} \tag{6.23} \end{equation}\]

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

\[\begin{equation*} \begin{split} y_1 &= m\cos(\text{m}\sin^{-1}x)\frac{1}{\sqrt{1-x^2}} \end{split} \end{equation*}\]

Squaring both sides,

\[\begin{equation*} \begin{split} (1-x^2){y_1}^2 &= m^2 \cos^2(\text{m}\sin^{-1}x)\\ &= m^2[1- \sin^2(\text{m}\sin^{-1}x)] \end{split} \end{equation*}\]

From (6.23),

\[\begin{equation*} \begin{split} (1-x^2){y_1}^2 &= m^2(1-y^2) \end{split} \end{equation*}\]

Differentiating again w.r.t \(x\),

\[\begin{equation*} \begin{split} (1-x^2)2y_1y_2 - 2x{y_1}^2 &= m^2(0-2yy_1)\\ (1-x^2)2y_1y_2 - 2x{y_1}^2 + 2m^2yy_1 &= 0 \end{split} \end{equation*}\]

Dividing both sides by \(2y_1\),

\[\begin{equation} \begin{split} (1-x^2)y_2 - xy_1 + m^2y &= 0 \end{split} \tag{6.24} \end{equation}\]

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

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

\[\begin{equation*} \begin{split} y_{n+2}(1-x^2) + ny_{n+1}\times(-2x) + \frac{n(n-1)}{2}y_n \times(-2) + 0 \\ - [y_{n+1}x + ny_n\times 1 + 0] + m^2y_n &= 0\\ (1-x^2)y_{n+2} -2nxy_{n+1} - \\ (n^2 -n)y_n - xy_{n+1} -ny_n + m^2y_n &= 0\\ (1-x^2)y_{n+2} -(2n+1)xy_{n+1} -(m^2 - n^2 + n -n)y_n &= 0\\ \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} (1-x^2)y_{n+2} -(2n+1)xy_{n+1} -(m^2 - n^2)y_n &= 0\\ \end{split} \end{equation*}\]

Hence proved.

6.1.13 Question 13 (TU 2064)

If \(\log y = \tan^{-1} x\), show that \((1+x^2)y_2 + (2x-1)y_1 = 0\).

Differentiating w.r.t \(x\),

\[\begin{equation*} \begin{split} \frac{1}{y}y_1 &= \frac{1}{1+x^2}\\ y_1(1+x^2) &= y\\ \end{split} \end{equation*}\]

Differentiating again w.r.t \(x\),

\[\begin{equation} \begin{split} (1+x^2)y_2 + (2x -1)y_1 &= 0 \end{split} \tag{6.25} \end{equation}\]

Differentiate this differential equation \(n\) times.

Differentiating (6.25) \(n\)-times,

\[\begin{equation*} \begin{split} y_{n+2}(1+x^2) + ny_{n+1}.2x + \frac{n(n-1)}{2}.y_n.2 + \\ 0 + y_{n+1}(2x-1) + ny_n.2 + 0 &= 0\\ (1+x^2)y_{n+2} + 2nxy_{n+1} + (2x-1)y_{n+1} + (n^2 -n)y_n + 2ny_n &= 0\\ (1+x^2)y_{n+2} + (2nx + 2x-1)y_{n+1} + (n^2 + n)y_n &= 0\\ \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} (1+x^2)y_{n+2} + (2nx + 2x-1)y_{n+1} + n(n + 1)y_n &= 0\\ \end{split} \end{equation*}\]