$x = sin t y = cos m t$
Prove $(1 - x^{2}) y_{n + 2} - (2 n - 1) y_{n + 1} - (n^{2} - m^{2}) y_{n} = 0$

Open in App

Solution

$x = sin t y = cos m t$
$t = {sin}^{- 1} (x)$ $y = cos (m {sin}^{- 1} (x))$
$(1 - x^{2}) y_{n + 2} - (2 n + 1) x y_{n + 1} + (m^{2} - n^{2}) y_{n} = 0$
$\frac{d y}{d x} = - sin (m {sin}^{- 1} (x)) = \frac{m}{\sqrt{1 - x^{2}}}$
$\sqrt{1 - x^{2}} \frac{d y}{d x} = - m sin (m {sin}^{- 1} x)$
$(1 - x^{2}) {(\frac{d y}{d x})}^{2} = - m^{2} {[sin (m {sin}^{- 1} x))]}^{2}$
$= m^{2} [1 - {cos}^{2} (m {sin}^{- 1} x)]$
$(1 - x^{2}) {(\frac{d y}{d x})}^{2} = m^{2} [1 - y^{2}]$ $∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{d y}{d x} = y_{1} \frac{d^{2} y}{{d x}^{2}} = y_{2} \end{matrix}$
$(1 - x^{2}) 2 y_{1} y_{2} + y_{1}^{2} (- 2 x) = m^{2} (- 2 y) y_{1}$
$(1 - x^{2}) y_{2} - x y_{1} = - m^{2} y$
$∴ (1 - x^{2}) y_{2} - x y_{1} + m^{2} y = 0$
$(1 - x^{2}) y_{m + 2} + m (- 2 x)_{y m + 1} \neq \frac{m (n - 1)}{2} . (- 2) y_{m}$
$x y_{x + 1} - x y_{m} - m^{2} y_{n} = 0$
$(1 - x^{2}) y_{n + 2} - x (2 x + + 1) Y_{n + 1} + (n^{2} + n - n + m^{2}) y x = 0$
$(1 - x^{2}) y_{m + 2} - (2 n + 1) x y_{x + 1} + (m^{2} - n^{2}) y x = 0$

Suggest Corrections

Similar questions

x = sin t y = cos m t

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

y = e^{m s i n^{- 1} x}

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

y = e^{a s i n - 1} x \Rightarrow (1 - x^{2}) y_{n} + 2 - (2 n + 1) x y_{n} + 1

\frac{C_{0}}{1} - \frac{C_{1}}{4} + \frac{C_{2}}{9} - . . . + \frac{(- 1)^{n} C_{n}}{n^{2} + 2 n + 1}

= \frac{1}{n + 1} + \frac{1}{2 n + 2} + . . . + \frac{1}{n^{2} + 2 n + 1} .

y = e^{- x^{2}}

y_{n + 2} + 2 x . y_{n + 1} + 2 (n + 1) y_{n} =

Join BYJU'S Learning Program