matlab 求解微分方程组已知u1如何求u2i1=C1*du1/dt;%%含有微分i5=C3*d(u2-u1)/dt;%%含有微分i3=i5+ii-i1;u5=u1-R1*i3-L1*di3/dt;%%含有微分u5=dfai/dt;%%含有微分fai=x;p1 = 1.367e-014; p2 = -2.585e-027; p3 = -8.37e-011;

Question

matlab 求解微分方程组

已知u1如何求u2

i1=C1*du1/dt;%%含有微分
i5=C3*d(u2-u1)/dt;%%含有微分
i3=i5+ii-i1;
u5=u1-R1*i3-L1*di3/dt;%%含有微分
u5=dfai/dt;%%含有微分
fai=x;
p1 = 1.367e-014;
p2 = -2.585e-027;
p3 = -8.37e-011;
p4 = 3.585e-023;
p5 = 3.2e-007;
p6 = -1.298e-019;
p7 = -0.0002556;
p8 = 1.143e-016;
ih= p1*x.^7 + p2*x.^6 + p3*x.^5 + p4*x.^4 + p5*x.^3 + p6*x.^2 + p7*x + p8;
i7=i3-ih;
i4=-100*i7;
u4=u5/100;
u2=u4+R2*i4+L2*di4/dt;%%含有微分
i2+i4+i5=0;

Answer 1

clear all;clc;
load 15.mat;
t=celiang(:,1);
u1=celiang(:,2);
syms u2 i5 ii x u5%定义u1,L1,L2,R1,C1,C3,ii 等
vU1=celiang(:,2);
k=length(u1);
C1=1;
h=5e-7;
C3=1;
R1=1;
L1=1;
R2=1;
L2=1;
C2=1;
i1(1)=0;i1(2)=0;%在开始的两个点用的是最简单的使之为零的方法.
for j=3:k-1
i1(j)=(vU1(j+1)-vU1(j))*12*C1/(23*h)+16*i1(j-1)/23-5*i1(j-2)/23;
end
i1(k)=(vU1(k)-vU1(k-1))*C1/h;
i1=i1';
i3=i5+ii-i1;
fai=x;
p1 = 1.367e-014;
p2 =-2.585e-027;
p3 = -8.37e-011;
p4 = 3.585e-023;
p5 = 3.2e-007;
p6 =-1.298e-019;
p7 = -0.0002556;
p8 = 1.143e-016;
ih= p1*x.^7 + p2*x.^6 + p3*x.^5 + p4*x.^4 + p5*x.^3 +p6*x.^2 + p7*x + p8;
i7=i3-ih;
i4=-100*i7;
u4=u5/100;
i2=-i4-i5;
u21=u2-u1;
%%%%%%%%%%%%%%%%%%%%%
s=dsolve(diff(u21,t)==i5/C3,diff(i3,t)==(u1-R1*i3-u5)/L1,diff(fai,t)==u5,diff(i4,t)==(u2-u4- R2*i4)/L2);
u2=s.u2