解答:(a)L{e2t}=∫∞0e2te−stdt=∫∞0e(2−s)tdt=[12−se(2−s)t]|∞0=1s−2(b)y′=[2−2−13]y+[42]e2t,y(0)=0⇒{y′1=2y1−2y2+4e2t,y1(0)=0y′2=−y1+3y2+2e2t,y2(0)=0⇒{L{y′1}=L{2y1−2y2+4e2t}L{y′2}=L{−y1+3y2+2e2t}⇒{sY1(s)=2Y1(s)−2Y2(s)+4s−2⋯(1)sY2(s)=−Y1(s)+3Y2(s)+2s−2⋯(2)由(1)⇒Y2(s)=2−s2Y1(s)+2s−2代入(2)⇒Y1(s)=4(s−1)(s−2)⇒Y2=2s−2−2s−1⇒y1(t)=L−1{Y1(s)}=4L−1{1s−2−1s−1}=4(e2t−et)y2(t)=L−1{Y2(s)}=2L−1{1s−2−1s−1}=2(e2t−et)⇒{y1(t)=4(e2t−et)y2(t)=2(e2t−et)(c){limt→∞y1(t)>>y1(0)limt→∞y2(t)>>y2(0)⇒unstable
解答:(a){A=[10−1213]B=[210034]⇒A+B=[31−1247]⇒(A+B)T=[3214−17](b){2A−B=[0−1−24−12]C+D=[−132422]⇒(2A−B)(C+D)=[−6−8−212](c)A−B=[−1−1−12−2−1]⇒X=−32(A−B)=[3/23/23/2−333/2]

解答:(a){→A=(1,−3,2)→B=(3,−2,3)⇒→A⋅→B=1×3+(−3)×(−2)+2×3=15(b)→A×→B=|→i→j→k1−323−23|=−5→i+3→j+7→k(c)(→A⋅→B‖→B‖)→B‖→B‖=15√22(3√22,−2√22,3√22)=(4522,−3022,4522)=4522→i−1511→j+4522→k
解答:A=[201010203]⇒det
================= END ==============
解題僅供參考,其他歷年試題及詳解
沒有留言:
張貼留言