t**********4
发帖数: 2115 | 1 【 以下文字转载自 Dreamer 讨论区 】
发信人: Dreamer (不要问我从哪里来), 信区: Dreamer
标 题: 麻烦好心人转帖
发信站: BBS 未名空间站 (Tue Sep 27 18:40:47 2011, 美东)
帮忙转Fashion或NewYork
菜鸟一枚，请懂的姐妹帮忙看看～～谢谢！
ZALET家的钻石
1.25ct, E, VV2, Triple EX, GIA
含税$21,500，可能还可以再低一点，值得买不？ |
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f*****k
发帖数: 353 | 3 depends on A's eigenvalues,
if A has two real non-equal eigenvalues r1, r2, and the eigenvalue is v1, v2
, then
u = C*(e^(r1 t) v1 + e^(r2 t) v2);
if A has two conjugate complex eigenvalue r1, r2, then
then pick up the real and the imaginary part of e^(r1 t) v1, the linear
combination is the solution.
if A has double eigenvalue r, then find the two linear independent solution
v1, v2 of
(A - r I )^2 v = 0,
vv1 = (A - rI)*v1, vv2 = (A - rI)*v2;
solution u = C (e^(rt)*(v1 + vv1*x) + e^(rt)*(v2 + v |
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