If the function y(x) reprsented by x = sin t ,
`y = ae^(tsqrt(2)) + be^(tsqrt(2)) , t in (-(pi)/(2),(pi)/(2)) ` satisfiles the equation
` (1-x^(2)) y''- xy'= ky` , then k is equal to

Options

A. 1
B. `-2`
C. 2
D. None of these

Answer

C

Explanation

We have , ` (dx)/(dt) = cos t " and " (dy)/(dx) = sqrt(2) (a + b) e^(t sqrt(2))`
` therefore (dy)/(dx) = ((dy)/(dt))/((dx)/(dt)) = (sqrt( 2)(a + b)e^(tsqrt(2)))/(cos t) = (sqrt(2)*y)/( sqrt( 1 - x^(2)))`
` rArr (1 - x^(2)) ((dy)/(dx))^(2) = 2y^(2)`
On differentiating w.r.t.w, we get
`(1 - x^(2)) 2y'y'' - 2x (y')^(2) = 4yy'`
` rArr (1 - x^(2) )y''- xy' = 2y " "` [dividing by 2y]
` therefore k = 2 ` .