x''+40x-30y=0 y''-20x-20y=0 在t=0时,x=0.01,x'=0,y=0.02,y'=0.用m
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x''+40x-30y=0 y''-20x-20y=0 在t=0时,x=0.01,x'=0,y=0.02,y'=0.用matlab任何解出x和y阿?
我在写论文,遇到这个问题,自己也用matlab来解了一下,发现答案很奇怪,不知道是不是编程出了问题,麻烦哪位大侠帮帮忙啊~
我在写论文,遇到这个问题,自己也用matlab来解了一下,发现答案很奇怪,不知道是不是编程出了问题,麻烦哪位大侠帮帮忙啊~
[x,y]=dsolve('D2x+40*x-30*y=0','D2y-20*x-20*y=0','x(0)=0.01,Dx(0)=0,y(0)=0.02,Dy(0)=0')
x =
(-1/2000*15^(1/2)+1/400)*exp((-10-10*15^(1/2))^(1/2)*t)+(-1/2000*15^(1/2)+1/400)*exp(-(-10-10*15^(1/2))^(1/2)*t)+(1/2000*15^(1/2)+1/400)*exp((-10+10*15^(1/2))^(1/2)*t)+(1/2000*15^(1/2)+1/400)*exp(-(-10+10*15^(1/2))^(1/2)*t)
y =
(-1/2000*15^(1/2)+1/400)*exp((-10-10*15^(1/2))^(1/2)*t)-1/3*(-1/2000*15^(1/2)+1/400)*exp((-10-10*15^(1/2))^(1/2)*t)*15^(1/2)+(-1/2000*15^(1/2)+1/400)*exp(-(-10-10*15^(1/2))^(1/2)*t)-1/3*(-1/2000*15^(1/2)+1/400)*exp(-(-10-10*15^(1/2))^(1/2)*t)*15^(1/2)+(1/2000*15^(1/2)+1/400)*exp((-10+10*15^(1/2))^(1/2)*t)+1/3*(1/2000*15^(1/2)+1/400)*exp((-10+10*15^(1/2))^(1/2)*t)*15^(1/2)+(1/2000*15^(1/2)+1/400)*exp(-(-10+10*15^(1/2))^(1/2)*t)+1/3*(1/2000*15^(1/2)+1/400)*exp(-(-10+10*15^(1/2))^(1/2)*t)*15^(1/2)
x =
(-1/2000*15^(1/2)+1/400)*exp((-10-10*15^(1/2))^(1/2)*t)+(-1/2000*15^(1/2)+1/400)*exp(-(-10-10*15^(1/2))^(1/2)*t)+(1/2000*15^(1/2)+1/400)*exp((-10+10*15^(1/2))^(1/2)*t)+(1/2000*15^(1/2)+1/400)*exp(-(-10+10*15^(1/2))^(1/2)*t)
y =
(-1/2000*15^(1/2)+1/400)*exp((-10-10*15^(1/2))^(1/2)*t)-1/3*(-1/2000*15^(1/2)+1/400)*exp((-10-10*15^(1/2))^(1/2)*t)*15^(1/2)+(-1/2000*15^(1/2)+1/400)*exp(-(-10-10*15^(1/2))^(1/2)*t)-1/3*(-1/2000*15^(1/2)+1/400)*exp(-(-10-10*15^(1/2))^(1/2)*t)*15^(1/2)+(1/2000*15^(1/2)+1/400)*exp((-10+10*15^(1/2))^(1/2)*t)+1/3*(1/2000*15^(1/2)+1/400)*exp((-10+10*15^(1/2))^(1/2)*t)*15^(1/2)+(1/2000*15^(1/2)+1/400)*exp(-(-10+10*15^(1/2))^(1/2)*t)+1/3*(1/2000*15^(1/2)+1/400)*exp(-(-10+10*15^(1/2))^(1/2)*t)*15^(1/2)
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