If y=(x + rt (x^2 +1))^m, prove that (x^2 +1) y2 + xy1 - m^2 y =0 - Maths - Continuity and Differentiability

Question

If y=(x + rt (x^2 +1))^m, prove that (x^2 +1) y2 + xy1 - m^2 y =0

R S Prasobh Sankar · Accepted Answer

Here , y = ( x + √(x2 + 1)
)m
So y1 = m (x + √(x2 + 1)
)m-1 d/dx( x + √(x2 + 1) )
= m ( x + √(x2 + 1) ) m-1 ( 1 +
2x/2√(x2 + 1))
=m( x+ √(x2 + 1))m-1( x +
√(x2 + 1)) / √(x2 + 1)
=m( x + √(x2 + 1)
)m/√(x2 + 1)
Cross multiplying ,
√(x2 + 1)y1 = m ( x +
√(x2 + 1) )m
Similarly , differentiating both sides w r t x ,
√(x2 + 1)y2 + y1
(d/dx(√(x2 + 1)) = m2 (x +
√(x2 + 1) )m-1 d/dx( x +
√(x2 + 1) )
√(x2 + 1)y2 + y1
x/√(x2 + 1) = m2 ( x +
√(x2 + 1) ) m-1 ( 1 +
2x/2√(x2 + 1))
=m2( x+ √(x2 + 1))m-1( x
+ √(x2 + 1)) / √(x2 + 1)
=m2( x + √(x2 + 1)
)m/√(x2 + 1)
Cross multiplying ,
We get , ( x2 + 1)y2 + xy1 =
m2( x + √(x2 + 1)
)m
= m2y [ y = ( x + √(x2 + 1) )m
]
Re arranging , ( x2+ 1)y2 +
xy1 – m2y = 0
Hope that helps