※ 引述《chienluen (小捻)》之銘言：
想問問下面兩題證明題~
Q2
Show that, if y"=-y, y'=1, and y=0 when x=0, then y=sinx.
pf:
今將證明改為若 y"(x) + β^2*y(x) = 0 ， 求其解。
y(x)
令 z(x) = -------- ， 即 y(x) = z(x)*cosβx
cosβx
則 y'(x) = z'(x)*cosβx - β*z(x)*sinβx
→ y"(x) = z"(x)*cosβx - 2β*z'(x)*sinβx - β^2*z(x)*cosβx
= z"(x)*cosβx - 2β*z'(x)*sinβx - β^2*y(x)
→ 0 = y"(x) + β^2*y(x) = z"(x)*cosβx - 2β*z'(x)*sinβx
→ z"(x)*cos^2βx - 2β*z'(x)*sinβx*cosβx = 0
→ [ z'(x)*cos^2βx ]' = 0
→ z'(x)*cos^2βx = C2*β (C2為常數)
→ z'(x) = C2*β*sec^2βx
y(t)
→ z(x) = C1 + C2*tanβx = -------- (C1,C2為常數)
cosβx
所以 y(t) = C1*cosβx + C2*sinβx (C1,C2為常數)
今將題目代入此公式解: y"(t) + y(t) = 0 ( β=1 )
y(t) = C1*cosx + C2*sinx (C1,C2為常數)
又 y(0) = 0 → C1 = 0
y'(0) = 1 → C2 = 0
所以 y(t) = sinx ，由此得證
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