differentiation(4)

y = xn sin x

y' = xn cos x + nxn-1 sin x

y" = - xn sin x + nxn-1 cos x + n(n - 1)xn-2 sin x + nxn-1 cos x

= - xn sin x + 2nxn-1 cos x + n(n - 1)xn-2 sin x

xy" - 2y' + xy = - xn+1 sin x + 2nxn cos x + n(n - 1)xn-1 sin x - 2xn cos x - 2nxn-1 sin x + xn+1 sin x

28xn-1 sin x + 2(n - 1)xn cos x = [n(n - 1)xn-1 - 2nxn-1] sin x + 2(n - 1)xn cos x

So we have:

n(n - 1) - 2n = 28

n2 - 3n - 28 = 0

(n - 7)(n + 4) = 0

n = 7 or -4 (rej.)

y = (x^n)sinx

y' = (x^n)cosx + nx^(n - 1)sinx

y''

= -(x^n)sinx + nx^(n - 1)cosx + nx^(n - 1)cosx + n(n - 1)x^(n - 2)sinx

= [n(n - 1) - x^2]x^(n - 2)sinx + 2nx^(n - 1)cosx

xy'' - 2y' + xy = [28x^(n - 1)]sinx + 2(n - 1)(x^n)cosx

[n(n - 1) - x^2]x^(n - 1)sinx + 2n(x^n)cosx - 2(x^n)cosx - 2nx^(n - 1)sinx +
x^(n + 1)sinx = [28x^(n - 1)]sinx + 2(n - 1)(x^n)cosx

{ [n(n - 1) - x^2 - 2n]x^(n - 1) + x^(n + 1) }sinx - 2(x^n)(n - 1)cosx
= [28x^(n - 1)]sinx + 2(n - 1)(x^n)cosx

{ [n^2 - 3n]x^(n - 1) }sinx + 2(x^n)(n - 1)cosx = [28x^(n - 1)]sinx + 2(n - 1)(x^n)cosx

Comparing the coefficient, n^2 - 3n = 28 => (n + 4)(n - 7) = 0

So, n = -4 (rejected) or n = 7