解一个方程

不断变形再多次两边同时积分就可以了。
原式 => d²U/dz²=-k²U
=> d(dU/dz)/dz=-k²U
=> d(dU/dz)=-k²U/dz
=> ∫d(dU/dz)=∫-k²Udz
=> dU/dz=-k²Uz
=> dU=-k²Uzdz
=> ∫dU=∫-k²Uzdz
=> U=-k²Uz²/2
=> 2=-k²z²

U=a sin(mx), a=常数
U'=am cos(mx)
U"=-a(m*m)sin(mx)
U"+(k*k)U=0
-a(m*m)sin(mx)+(k*k)a*sin(mx)=0 ==>
k*k=m*m ==>m=±k,sin(-kx)=sin(kx) ==>
U=a sin(kx), a=任意常数

“d*z的平方”为分母，则d≠0，z≠0,[d^2/(d*z^2)+k^2]≠0
又∵[d^2/（d*z^2）+k^2]*U = 0,
∴U = 0

d2(U)/d2(z)+K2U＝0
d2U/d2z=-UK2
(dz/dU)2=-1/UK2
dz/dU=±(√-1/U)/K
z=±(3√-ln|U|)/2K+ C