Question

求解联立微分方程组 x'(t)-x(t)-2y(t)=t ,y'(t)-y(t)...

求解联立微分方程组 x'(t)-x(t)-2y(t)=t ,y'(t)-y(t)-2x(t)=t ,其初始条件为x(0)=2,y(0)=4

Answer 1

两方程相减：x'-y'-x+y-2y+2x=0即x'-y'+x-y=0令h=x-y,则h'+h=0,得dh/dt=-h,dy/h=-dt,得h=Ce^(-t)当t=0时,x=2,y=4,h(0)=2-4=-2,则得C=-2即x-y=-2e^(-t)即y=x+2e^(-t)代入方程1得：x'-x-2x-4e^(-t)=t得：x'-3x=4e^(-t)+t特征根为3,通解为C1e^(3t),设特解x*=ae^(-t)+bt+c代入得：-ae^(-t)+b-3ae^(-t)-3bt-3c=4e^(-t)+t得：-2a=4,-3b=1,b-3c=0解得：a=-2,b=-1/3,c=-1/9即x=C1e^(3t)-2e^(-t)-t/3-1/9当t=0时,得x=C1-2-1/9=2,故C1=37/9因此x=37/9e^(3t)-2e^(-t)-t/3-1/9y=x+2e^(-t)=37/9e^(3t)-t/3-1/9