A.maths 高手請答

1)y= x( A cos5X + B sin 5X) (given)
y'=Acos5x+Bsin5x-5Axsin5x+5Bxcos5x
y''-25Axcos5x-10Asin5x+25Bxsin5x+10Bcos5x
As y' ' + 25 y = 8 cos
5X-25Axcos5x-10Asin5x+25Bxsin5x+10Bcos5x-25Axcos5x-25Bxsin5x= 8 cos 5X
-50Axcos5x-10Asin5x+10Bcos5x= 8 cos 5X
By comparing ,as there is no sine function
A=0 i.e. 10Bcos5x = 8cos5x
10B=8
B=4/5

2) Let P(n) be the proposition that [d^n sin X] / dx^n = sin (x+ nπ/2)
where n is true for all positive integer.
Put n=1, L.H.S.=dsinx/dx=cosx=sin(x+π/2)=R.H.S.
therefore , P(1) is true
Assume that P(k) is also true where k is any integer
i.e.[d^k sin X] / dx^k = sin (x+ kπ/2)
for n = k+1,
=[d^k+1 sin X] / dx^k+1
=dsin (x+ kπ/2)/dx (by assumption)
=cos(x+kπ/2)
=sin(x+(k+1)kπ/2)
therefore P(k+1)is true
By the principle M.I., P(n) is true for all positive integers n.

3) √x+ √y =4 ------(*)
diff.w.r.t.x
1/2x^-1/2+dy/dx(1/2y^-1/2)=0
dy/dx = -x^-1/2/y^-1/2
from (*),√y=4- √x
i.e.dy/dx = -x^-1/2(4- √x)
=-4x^-1/2
therefore d^2y/dx^2 = 2x^-3/2
=2/x^3/2
=2 /( x√x)

4)b^2x^2+ a^2y^2 =a^2b^2 ---- (*)
diff.w.r.t.x.
2b^2x+2a^2ydy/dx=0
dy/dx=-b^2x/a^2y
from(*),y=(b^2-b^2x^2/a^2)^1/2
i.e.dy/dx=-b^2x/a^2(b^2-b^2x^2/a^2)^1/2
=-b^2x/a(b^2-b^2x^2)^1/2
d^2/dx^2=- b^4 /( a^2 y^3 )

2007-09-01 07:25:13 補充：
By comparing L.H.S. and R.H.S.,-50Axcos5x-10Asin5x 10Bcos5x= 8 cos 5X It is obviously to see that there is just the COSINE function in R.H.S.So,the problem of sine functions should not be there in the L.H.S. and it can induce that the sine term -10Asin5x=0,i.e. A=0

2007-09-02 00:51:09 補充：
哦,通常泥d題目都系就住出.tan ge 話我都唔識計

2007-09-02 00:53:35 補充：
哦,通常泥d題目都系就住出,如果tan ge 話我都唔識計.