If `x=sint ,y=sinp t ,"p r o v et h a t"` `(1-x^2)(d^2y)/(dx^2)-x(dy)/(dx)+p^2y=0.`

Answer

N/a

Explanation

We have, `x = sint` and `y = sin pt`
`:. (dx)/(dt) = cost` and `(dy)/(dt) = cos pt.p`
` rArr (dy)/(dx) = (dt//dt)/(dx//dt) = (p.cospt)/(cost)`
Again differentiating both sides w.r.t. x, we get
`(d^(2)y)/(dx^(2)) = (cost.(d)/(dt)(p.cospt)(dt)/(dx)-pcospt.(d)/(dt)cost.(dt)/(dx))/(cos^(2)t)`
` =([cost.p.(-sinpt).p-pcospt.(-sint)](dt)/(dx))/(cos^(2)t)`
`= ([-p^(2)sinpt.cost+p sint.cospt].(1)/(cost))/(cos^(2)t)`
`rArr (d^(2)y)/(dx^(2)) = (-p^(2)sin pt. cost + p cospt.sint)/(cos^(3)t)"......"(ii)`
Since, we have to prove
`(1-x^(2))(d^(2)y)/(dx^(2))-x'(dy)/(dx)+p^(2)y=0`
`:. LHS = (1-sin^(2)t)'([-p^(2)sin pt. cost+p cos pt. sint])/(cos^(3)t) - sint. (p cos pt)/(cos t) + p^(2) sin pt`
`(1)/(cos^(3)t)[{:((1-sin^(2)t)(-p^(2)sin pt . cost+pcospt.sint)),(-p cos pt. sint. cos^(2)t+p^(2) sinpt.cos^(3)t):}]`
`= (1)/(cos^(3)t)[{:(-p^(2)sinpt.cos^(3)t+pcospt.sint.cos^(2)t),(-pcospt.sint.cos^(2)t+p^(2)sinpt.cos^(3)t):}],[:' 1-sin^(2)t=cos^(2)t]`
`= (1)/(cos^(3)t).0`
`= 0`