Question

If Y1/m + Y -1/m= 2x , prove that (x2 - 1) Yn + 2 +(2n + 1) xyn+1 +(n2 - m 2 ) Yn = 0 , where Yn denotes the nth derivative of Y.

Answer 1

SOLUTION

GIVEN

$\displaystyle \sf{ {y}^{ \frac{1}{m} } + {y}^{ - \frac{1}{m} } = 2x}$

TO PROVE

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

EVALUATION

Here it is given that

$\displaystyle \sf{ {y}^{ \frac{1}{m} } + {y}^{ - \frac{1}{m} } = 2x} \: \: - - - (1)$

Squaring both sides we get

$\displaystyle \sf{{( {y}^{ \frac{1}{m} } + {y}^{ - \frac{1}{m} } )}^{2} = 4 {x}^{2} }$

$\displaystyle \sf{ \implies \: {( {y}^{ \frac{1}{m} } - {y}^{ - \frac{1}{m} } )}^{2} + 4.{y}^{ \frac{1}{m} }.{y}^{ - \frac{1}{m} } = 4 {x}^{2} }$

$\displaystyle \sf{ \implies \: {( {y}^{ \frac{1}{m} } - {y}^{ - \frac{1}{m} } )}^{2} + 4= 4 {x}^{2} }$

$\displaystyle \sf{ \implies \: {( {y}^{ \frac{1}{m} } - {y}^{ - \frac{1}{m} } )}^{2} = 4 {x}^{2} - 4 }$

$\displaystyle \sf{ \implies \: {( {y}^{ \frac{1}{m} } - {y}^{ - \frac{1}{m} } )}^{2} = 4 ({x}^{2} - 1) }$

$\displaystyle \sf{ \implies \: {( {y}^{ \frac{1}{m} } - {y}^{ - \frac{1}{m} } )}^{} = 2 \sqrt{ ({x}^{2} - 1) }} \: \: - - - (1)$

Adding Equation 1 and Equation 2 we get

$\displaystyle \sf{2 {y}^{ \frac{1}{m} } = 2x + 2 \sqrt{ {x}^{2} - 1 } }$

$\displaystyle \sf{ \implies \: {y}^{ \frac{1}{m} } = x + \sqrt{ {x}^{2} - 1 } }$

$\displaystyle \sf{ \implies \: y = {( x + \sqrt{ {x}^{2} - 1 } )}^{m} }$

Differentiating both sides with respect to x we get

$\displaystyle \sf{ \implies \: y_1 = m{( x + \sqrt{ {x}^{2} - 1 } )}^{m - 1} \bigg(1 + \frac{x}{ \sqrt{ {x}^{2} - 1 } } \bigg) }$

$\displaystyle \sf{ \implies \: y_1 = m{( x + \sqrt{ {x}^{2} - 1 } )}^{m - 1} \bigg( \frac{x + \sqrt{ {x}^{2} - 1 } }{ \sqrt{ {x}^{2} - 1 } } \bigg) }$

$\displaystyle \sf{ \implies \: y_1 = m{( x + \sqrt{ {x}^{2} - 1 } )}^{m } \bigg( \frac{1 }{ \sqrt{ {x}^{2} - 1 } } \bigg) }$

$\displaystyle \sf{ \implies \: y_1 = \frac{my}{ \sqrt{ {x}^{2} - 1 } } }$

$\displaystyle \sf{ \implies \: (\sqrt{ {x}^{2} - 1}) \: y_1 = my }$

$\displaystyle \sf{ \implies \: ( {x}^{2} - 1) \: { y_1}^{2} = {m}^{2} {y}^{2} }$

Again Differentiating both sides with respect to x we get

$\displaystyle \sf{ \implies \: ( {x}^{2} - 1) 2y_1 y_2 + 2x { y_1}^{2} = 2 {m}^{2} yy_1 }$

$\displaystyle \sf{ \implies \: ( {x}^{2} - 1) y_2 + x y_1 = {m}^{2} y }$

Differentiating both sides n times using Leibniz Formula with respect to x we get

$\displaystyle \sf{ \implies \: ( {x}^{2} - 1) y_{n + 2} + 2nx y_{n + 1} + \frac{n(n - 1)}{2}2y_n + x y_{n + 1} +ny_n = {m}^{2} y_n }$

$\displaystyle \sf{ \implies \: ( {x}^{2} - 1) y_{n + 2} + 2nx y_{n + 1} + ( {n}^{2} - n) y_n + x y_{n + 1} +ny_n = {m}^{2} y_n }$

$\displaystyle \sf{ \implies \: ( {x}^{2} - 1)y_{n + 2} + (2n + 1)xy_{n + 1} + ( {n}^{2} - {m}^{2} )y_{n } = 0 }$

Hence proved

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Answer 2

Answer:

If y

1/m + y–1/m = 2x, then show that

(x

2 – 1)yn+2 + (2x + 1)xyn+1 + (n

2 + m2

)yn = 0.

Step-by-step explanation: