IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

A.Vignesh

If y = ( x¹⁰/ 11) + (6 / x) , then value of n satisfying the equation

x dⁿy/dxⁿ + (n) d^n-1y/dx^n-1 = 0 is _________

kusum

hi,

i guess Vignesh u have to start workin on it as you 'll find the answer by just diff once , then twice and then check.there is no short cut to it.

edison

y = (x¹⁰)/11 + 6 / x

x dⁿy / dxⁿ + n d^n-1y / dx^n-1 = 0 .... (1)

To find the value of n

Solution) Since, y = (x¹⁰)/11 + 6 / x

Therefore,

dⁿy / dxⁿ = (10) (9)...[10 - (n - 1)] x^10-n / 11 + (-1)ⁿ n! 6 / xⁿ⁺¹

or dⁿy / dxⁿ = (10) (9)...(11 - n) x^10-n / 11 + (-1)ⁿ n! 6 / xⁿ⁺¹

Therefore,

x dⁿy / dxⁿ = (10) (9)...(11 - n) x^11-n / 11 + (-1)ⁿ n! 6 / xⁿ ...(2)

also,

n d^n-1y / dx^n-1 = n (10) (9)...(12 - n) x^11-n / 11 + n (-1)^n-1 (n-1)! 6 / xⁿ

or n d^n-1y / dx^n-1 = n (10) (9)...(12 - n) x^11-n / 11 + (-1)^n-1 n! 6 / xⁿ ...(3)

Substituting (2) and (3) in equation (1) we obtain

(10) (9)...(11 - n) x^11-n /11+ (-1)ⁿ n! 6 / xⁿ + n (10) (9)...(12 - n) x^11-n / 11

+ (-1)^n-1 n! 6 / xⁿ = 0

Here, (-1)ⁿ n! 6 / xⁿ + (-1)^n-1 n! 6 / xⁿ = 0

So,

(10) (9)...(11 - n) x^11-n /11 + n (10) (9)...(12 - n) x^11-n / 11 = 0

or, (10) (9)...(11 - n) + n (10) (9)...(12 - n) = 0

or, (10) (9)...(12 - n) [ 11 - n + n ] = 0

or, (10) (9)...(12 - n) 11 = 0

Which is possible if (12 - n ) = 0,

Hence, n = 12