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प्रश्न
If y = `e^(acos^(-1)x)`, −1 ≤ x ≤ 1, show that `(1- x^2) (d^2y)/(dx^2) -x dy/dx - a^2y = 0`.
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उत्तर
We have y = `e^(a cos^(-1)x)` ...(1)
Differentiating (1) both sides w.r.t. x, we get
`dy/dx = e^(a cos^(-1)x) d/dx (a cos^-1 x)`
`= e^(a cos^(-1)x) ((- a)/sqrt(1 - x^2))`
`= (- ay)/(sqrt(1 - x^2))` ...(2)
Differentiating (2) both sides w.r.t. x, we get
`(d^2y)/(dx^2) = -a[(sqrt(1-x^2) dy/dx - y d/dx sqrt(1 - x^2))/((1-x^2))]`
`(d^2y)/(dx^2) = -a[(sqrt(1-x^2)dy/dx - y/(2sqrt(1-x^2)) * (-2x))/((1-x^2))]`
`(1 - x^2) (d^2y)/dx^2 = -a[-ay + (xy)/sqrt(1-x^2)]` ....[from (2)]
`(1 - x^2) (d^2y)/dx^2 = -a[-ay + x * ((-1)/a * dy/dx)]`
`(1 - x^2) (d^2y)/(dx^2) = a^2y + x dy/dx`
`(1 - x^2) (d^2y)/(dx^2) - x dy/dx - a^2y = 0`
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