A.Math微分2條數

37.
y= (x/k)sin(k/x)
y' =(1/k) sin(k/x) + (x/k)cos(k/x)[-k/x^2]
= (1/k)sin(k/x) - cos(k/x)/x
y" = (1/k)cos(k/x)[-k/x^2] - [-cos(k/x)/x^2 -(1/x) sin(k/x)(-k/x^2)]
= - cos(k/x)/x^2 + cos(k/x)/x^2 - ksin(k/x)/x^3
= -ksin(k/x)/x^3
= -(k/x^3)(ky/x)
x^4y" = -k^2y
so x^4y" + k^2y = 0.
38.
y = Ax^n
y' = Anx^(n-1)
y" = An(n-1)x^(n-2)
so x^2y" = An(n-1)x^n and
xy' = Anx^n
Therefore,
x^2y" - 2xy' - 4y
= An(n-1)x^n - 2Anx^n - 4Ax^n = 0
n(n-1) - 2n - 4 = 0
n^2 - n - 2n - 4 = 0
n^2 - 3n - 4 = 0
(n+1)(n-4) = 0
n = - 1 or 4.
For n = -1. y' = A(-1)x^(-1-1) = -A/x^2.
When x = 1, y' = 2, therefore, 2 = -A, A = -2.
That is y = -2/x.
For n = 4, y' = 4Ax^(4-1) = 4Ax^3.
When x = 1, y' = 2, therefore, 2 = 4A, A = 1/2.
So y = x^4/2.

2008-11-04 22:35:18 補充：
Q37 is using product rule to find y' and y".