Question

设x(t)是微分方程5x''+10x'+6x=0的解,证明函数f(t)=[x(t)]^2/[1+x(t)^4]有最大值,

Answer 1

解得x(t)=e^(-x)·(C1·cos((√5/5)x) + C2·sin((√5/5)x) )

当x→∞（正或负）时,x(t)→∞.

f(t)=[x(t)]^2/[1+x(t)^4]

=1 / { 1/[x(t)]^2 + [x(t)]^2 }

1/[x(t)]^2 + [x(t)]^2 ≥ 2,

因此 f(t)≤ 1/2