# ode23 , matrix equation , 2nd order DE

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Gloria on 14 Jun 2019
Commented: darova on 19 Jun 2019
How I will solve this matrix in Matlab ?
w=10.02;
p= 7850; %%%Kg/m^3;
g=9.81;
G=77*10^9; %%% N/m^2
E=206*10^9; %%% N/m^2
L=9.6; %%%m
D=0.15 ; %%m
m=pi*(D/2)^2*L*p;
A=pi*D^2/4 ; %%m2
J=m*D^2/8; %%Kg*m^2
Ip=pi*D^4/32 ; %% m^4
I=pi*D^4/64 ;
cx=0.05*m*w;
cy=0.05*m*w;
ct=0.08*J*w;
kx=3*E*I/L;
ky=3*E*I/L;
kt=G*Ip/L; darova on 14 Jun 2019
You can find roots for x'', y'' and dtheta'' every iteration solving matrix equation
function du = func(t,u)
u1 = u(1:3)'; % [x y theta]'
du1 = u(4:6)'; % [dx dy dtheta]'
dth = u(6); % dtheta
C = [cx 0 0; 0 cy 0; 0 0 cz]; % C matrix
K = ... % K matrix
F = [me(w+dth)^2+Fx; ...] % force vector
A = [m 0 -mesin(wt)...] % mass matrix
B = -C*du1 -K*u1 + F;
du = zeros(6,1);
du(1:3) = u(4:6);
du(4:6) = A\B;
end
darova on 19 Jun 2019

### More Answers (1)

Torsten on 14 Jun 2019
Convert the system to a first order system and write it as
M*y' = f(t,y)
Then use the mass-matrix option of the ODE solvers to supply M, define f in a function file and use ODE45, ODE15S, ... to solve.
Jan on 16 Jun 2019
You can write s(6)', because ' is the Matlab operator for the complex conjugate transposition:
a = [1, 1i; ...
2, 2i]
a'
You provide real scalars. Then the ' operaotr does not change the value. So it is valid, but simply a confusing waste of time.