Thursday, May 30, 2019

IGNOU SOLVED ASSIGNMENT 2019 MTE01 Q2(b)

We have

y = {\left( {{{\sin }^{ - 1}}\left( x \right)} \right)^2}

Differentiating

\frac{{dy}}{{dx}} = \frac{{2{{\sin }^{ - 1}}\left( x \right)}}{{\sqrt {1 - {x^2}} }}

{y^'}\left( {\sqrt {1 - {x^2}} } \right) = 2{\sin ^{ - 1}}\left( x \right)

Squaring both sides

{\left( {{y^'}} \right)^2}\left( {1 - {x^2}} \right) = 4y

Differentiating again

2{y^{''}}{y^'}\left( {1 - {x^2}} \right) - 2x{\left( {{y^'}} \right)^2} = 4{y^'}

The above equation gives us

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

Applying Leibniz theorem,

n{C_0}{y_{n + 2}}\left( {1 - {x^2}} \right) + n{C_1}{y_{n + 1}}\left( { - 2x} \right) + n{C_2}\left( { - 2} \right){y_n} - x{y_{n + 1}} - {y_n}n{C_1} = 0

{y_{n + 2}}\left( {1 - {x^2}} \right) - 2xn{y_{n + 1}} - \frac{{\left( 2 \right)\left( n \right)\left( {n - 1} \right)}}{2}{y_n} - x{y_{n + 1}} - {y_n}n = 0


{y_{n + 2}}\left( {1 - {x^2}} \right) - x\left( {2n + 1} \right){y_{n + 1}} - {n^2}{y_n} + n{y_n} - n{y_n} = 0


{y_{n + 2}}\left( {1 - {x^2}} \right) - x\left( {2n + 1} \right){y_{n + 1}} - {n^2}{y_n} = 0




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