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प्रश्न
If y = (sec-1 x )2 , x > 0, show that
`x^2 (x^2 - 1) (d^2 y)/(dx^2) + (2x^3 - x ) dy/dx -2 = 0`
बेरीज
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उत्तर
y = ( sec-1 x)2
`dy/dx = 2 (sec^(-1) x) 1/ (xsqrt(x^2 - 1))`
`x sqrt(x^2 - 1 ) dy/dx = 2 sec^(-1) x`
Again differentiating both sides
`x sqrt(x^2 -1) (d^2 y )/(dx^2) + (dy)/(dx) [sqrt(x^2 - 1 ) +(x^2)/ sqrt(x^2 - 1) ] = (2 xx 1 ) /(x sqrt (x^2 - 1))`
`x sqrt(x^2 -1) (d^2 y )/(dx^2) + (dy)/(dx) ((x^2 - 1 + x^2)/ sqrt(x^2 - 1) ) = 2/(x sqrt (x^2 - 1))`
`[ x (x^2 -1) (d^2 y )/(dx^2) + (dy)/(dx)(2x^2 - 1)] 1/sqrt(x^2 - 1 ) = 2/( x sqrt(x^2 - 1))`
`x^2(x^2 - 1) (d^2y)/(dx^2) + x(2x^2 - 1 ) (dy)/(dx) = 2 `
`x^2(x^2 - 1) (d^2y)/(dx^2) + x(2x^3 - x ) (dy)/(dx) - 2 = 0`
Hence proved.
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