求解微分方程

显示全部楼层 · 2012-11-27 11:39:56

即需求A=010
-1 0 -1
010的特征值先|tI-A|= t -1 0
1 t1
0 -1 t=t(t^2+1)+t=t(t^2+2)=0t=0,±√2 i然后求特征向量即解(tI-A)x=0再标准化得t=0时，x=[√0.5,0,-√0.5]t=√2 i, x=[-0.5i,√0.5,-0.5i]t=-√2 i, x=[0.5i,√0.5,0.5i]令P=√0.5,0,-√0.5
-0.5i,√0.5,-0.5i
0.5i,√0.5,0.5i则PAP'=D=0 0 0...

千问 · 2012-11-27 11:39:56

x'=yy'=-x-zz'=yy''=-x'-z'=-2y解得：y=C1cos(√2)t+C2sin(√2)t, y(0)=1得：C1=1由x'=z'=y得：x=∫[cos(√2)t+C2sin(√2)t]dt=(1/√2)(sin(√2)t-C2cos(√2)t)+C3 x（0）=0得：0=-C2（1/√2)+C3z...

千问 · 2012-11-27 11:39:56

千问 · 2012-11-27 11:39:56

今天才看到，不好意思啊。楼上楼下高手都解答了...