求解一題微分方程組的問題

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(1) GivenA=[.a b]
..[-b a]x={x1}
..{x2}x(0)={1}
.....{0}Solve the equation system:x'=Ax => dx/x=Adtln(x)=At*c=[.at bt]*{c1}
.[-bt at].{c2}={.c1*at+c2*bt}
.{-c1*bt+c2*at}x={e^(c1*at).+e^(c2*bt)}
..{e^(-c1*bt)+e^(c2*at)}={k1*e^(at)+k2*e^(bt)}
.{k1*e^(-bt)+k2*e^(at)}Initial coditions:x(0)={1}
.....{0} =>1=k1+k2 => k1=k2-10=k1+k2 => k2=-k1=>x(t)={k*e^(at)+(k-1)*e^(bt)}
.....{k*e^(-bt)-k*e^(at)}
(2) GivenA=[.3 .1 .0]
..[-4 -1 .0]
..[.4 -8 -2]x={x1}
..{x2}
..{x3}x(0)={1}
.....{0}
.....{0}Solve the equation system:dx/x=Adt => ln(x)=At*cx=e^(At*c)=e^(At)*k{x}={k1*e^3t+k2*e^t}
....{k1*e^(-4t)+k2*e^(-t)}
....{k1*e^4t+k2*e^(-8t)+k3*e^(-2t)}Initial conditions:x(0)={k1+k2...}={1}
.....{k1+k2...}.{0}
.....{k1+k2+k3}.{0} =>k2=1-k1k2=-k1k3=-(k1+k2){x}={k*e^3t+(1-k)*e^t}
....{k*e^(-4t)-k*e^(-t)}
....{k1*e^4t+k2*e^(-8t)-(k1+k2)*e^(-2t)}