Math, asked by as24hour, 7 months ago

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.

Answers

Answered by pulakmath007
10

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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Answered by naitishsahu263
1

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:

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