matlab解二階微分方程,解出的結(jié)果如下,請問怎樣畫出y和時(shí)間t的函數(shù)圖像?
y=dsolve('101*D2y+100*Dy+1000*y=102*sin(20*pi*t)','y(0)=0,Dy(0)=0','t')
解出的結(jié)果是
y=exp(-(50*t)/101)*sin((10*985^(1/2)*t)/101)*((51*985^(1/2)*exp((50*t)/101)*((50*sin(t*(20*pi - (10*985^(1/2))/101)))/101 - cos(t*(20*pi - (10*985^(1/2))/101))*(20*pi - (10*985^(1/2))/101)))/(9850*((20*pi - (10*985^(1/2))/101)^2 + 2500/10201)) + (51*985^(1/2)*exp((50*t)/101)*((50*sin(t*(20*pi + (10*985^(1/2))/101)))/101 - cos(t*(20*pi + (10*985^(1/2))/101))*(20*pi + (10*985^(1/2))/101)))/(9850*((20*pi + (10*985^(1/2))/101)^2 + 2500/10201))) - (51*cos((10*985^(1/2)*t)/101)*(99485*pi^2*sin(20*pi*t
- (10*985^(1/2)*t)/101) - (4925*sin(20*pi*t + (10*985^(1/2)*t)/101))/2 - (4925*sin(20*pi*t - (10*985^(1/2)*t)/101))/2 + 99485*pi^2*sin(20*pi*t + (10*985^(1/2)*t)/101) + 4925*pi*cos(20*pi*t - (10*985^(1/2)*t)/101) + 4925*pi*cos(20*pi*t + (10*985^(1/2)*t)/101) + (25*985^(1/2)*cos(20*pi*t - (10*985^(1/2)*t)/101))/2 - (25*985^(1/2)*cos(20*pi*t +