Question

If y=(x+√1+x^2)^m then show that
(1+x^2)y2 +xy1-m^2y=0

Answer 1

$y = ( {x + \sqrt{1 + {x}^{2} } })^{m} ...(1) \\ \\ y_{1} = m( {x + \sqrt{1 + {x}^{2} } })^{m - 1} (1 + \frac{2x}{2 \sqrt{1 + {x}^{2} } } ) \\ \\ y_{1} = m \frac{(x + \sqrt{1 + {x}^{2} } ) ^{m - 1} ( \sqrt{1 + {x}^{2} } + x) }{ \sqrt{1 + {x}^{2} } } \\ \\ y_{1} = \frac{m( {x + \sqrt{1 + {x}^{2} } })^{m} }{ \sqrt{1 + {x}^{2} } } \\ \\ y_{1} = \frac{my}{ \sqrt{1 + {x}^{2} } } \: \: \: \: [from \: \: (1) \: \: ] \\ \\ \sqrt{1 + {x}^{2} } \: y_{1} = my \\ \\ (1 + {x}^{2} ) y_{1} ^{2} = {m}^{2} {y}^{2} \\ \\ again \: \: differentiate \: w. \: r .\: t \: \: x \\ \\ 2 y_{1} y_{2}(1 + {x}^{2} ) + { y_{1} }^{2} (2x) = {m}^{2} 2y y_{1} \\ \\ deviding \: \: both \: \: sides \: \: by \: \: 2 y_{1} \\ \\ (1 + {x}^{2} ) y_{2} + x y_{1} - {m}^{2} y = 0$