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L e t y = e 2 sin − 1 x
Now , differentiating with respect to x , we get
y ′ = 1 − x 2 2 y ( 1 − x 2 ) y ′ = 2 y
Again differentiating with respect to x and simplifying the euation, we get
( 1 − x 2 ) y ′ ′ − x y ′ = 4 y ------------equation 1
Here, we use the Leibnitz's rule of finding n t h derivative of product of two functions:
( u v ) n = ( n 0 ) u v n + ( n 1 ) u 1 v n − 1 + ( n 2 ) u 2 v n − 2 + . . . . . . + ( n n ) u n v
where x n is the n t h derivative of x and ( a b ) = b ! ( a − b ) ! a !
Now, applying Leibnitz's rule to equation 1 , we get
( 1 − x 2 ) y n + 2 − 2 n x y n + 1 − n ( n − 1 ) y n − x y n + 1 − n y n = 4 y n
On simplifying, we get
( 1 − x 2 ) y n + 2 − ( 2 n + 1 ) x y n + 1 − ( 4 + n 2 ) y n = 0 -------------equation 2
where y n is the n t h derivative of y . Now, substituting value of x as 0 in equation 2 , we get
y n + 2 = ( 4 + n 2 ) y n ---------equation 3
We know that, y ( 0 ) = 1
y 1 ( 0 ) = 2
Using equation 3 , relation for y n ( 0 ) becomes ,
y n ( 0 ) = 2 ( 2 2 + 1 2 ) ( 2 2 + 3 2 ) . . . . . ( 2 2 + ( n − 2 ) 2 ) , n i s o d d y n ( 0 ) = 2 2 ( 2 2 + 2 2 ) ( 2 2 + 4 2 ) . . . . . ( 2 2 + ( n − 2 ) 2 ) , n i s e v e n
Substituting value of n = 1 0 , we get y 1 0 ( 0 ) = 1 7 4 0 8 0 0