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प्रश्न
If 2y = `sqrt(x + 1) + sqrt(x - 1)`, show that 4(x2 – 1)y2 + 4xy1 – y = 0.
उत्तर
2y = `sqrt(x + 1) + sqrt(x - 1)` ...[Given] (1)
Differentiating both sides w.r.t. x, we get
∴ `2 dy/dx = d/dx (sqrt(x + 1)) + d/dx (sqrt(x - 1))`
∴ `2 dy/dx = (1)/(2sqrt(x + 1))(1 + 0) + (1)/(2sqrt(x - 1))(1 - 0)`
∴ `2 dy/dx = (1)/(2sqrt(x + 1)) + (1)/(2sqrt(x - 1)`
∴ `2 dy/dx = (sqrt(x - 1) + sqrt(x + 1))/(2sqrt(x + 1).sqrt(x - 1)`
∴ `2 dy/dx = (cancel2y)/(cancel2sqrt(x^2 - 1)` ...[By (1)]
∴ `2sqrt(x^2 - 1) dy/dx` = y
Taking square both the sides,
∴ `4(x^2 - 1).(dy/dx)^2` = y2
Differentiating both sides w.r.t. x, we get,
`4(x^2 - 1) d/dx (dy/dx)^2 + (dy/dx)^2. d/dx [4(x^2 - 1)] = 2y dy/dx`
∴ `4(x^2 - 1).2 dy/dx.(d^2y)/(dx^2) + (dy/dx)^2 . 4(2x) = 2y(dy/dx)`
Cancelling `2 dy/dx` on both sides, we get,
`4(x^2 - 1)(d^2y)/(dx^2) + 4x dy/dx` = y
∴ `4(x^2 - 1)(d^2y)/(dx^2) + 4x dy/dx - y` = 0
∴ 4(x2 – 1)y2 + 4xy1 – y = 0.
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