solving system of equations

Tried to get this to work but have had no luck.

Main Code:

syms x K M c1 c2 c3 c4
x0 = [0,0,0,0];
y0 = [1,200,-80,900];
coeff = [1,-0.02,12.5,-0.058,27.77];
rx = -0.05*exp(0.005*x)*cos(11.7*x) + 0.07*exp(0.005*x)*sin(11.7*x);
y_new = Examprac(x0,y0,coeff,rx);

g = 

g = 

C = struct with fields:
    M: -(40000000*cos((117*x)/10) - 56000000*sin((117*x)/10))/(1254221280*cos((117*x)/10) + 27288860249597*sin((117*x)/10))
    K: 0

Output argument "y" (and possibly others) not assigned a value in the execution with "solution>Examprac" function.

Function:

function y = Examprac(x0,y0,coeff,rx)
syms x c1 c2 c3 c4 K M 
yp = 2*K*cos((117*x)/10)*exp(x/200) + 2*M*sin((117*x)/10)*exp(x/200);
yp_der1 = diff(yp);
yp_der2 = diff(yp_der1);
yp_der3 = diff(yp_der2);
yp_der4 = diff(yp_der3);
yp_der1= simplify(yp_der1);
yp_der2= simplify(yp_der2);
yp_der3= simplify(yp_der3);
yp_der4= simplify(yp_der4);
g = simplify(coeff(1)*yp_der4 + coeff(2)*yp_der3 + coeff(3)*yp_der2 + coeff(4)*yp_der1 + coeff(5)*yp == rx)
g = coeff(1)*yp_der4 + coeff(2)*yp_der3 + coeff(3)*yp_der2 + coeff(4)*yp_der1 + coeff(5)*yp == rx
C = solve(g, [M, K])
end

Answer 1

[Torsten]

I wonder what you expect as solution for 1 equation with 7 unset parameters.

Say you have the equation

g = x + c1 + c2 + c3 + c4 + K + M == 0

What would you like to see as output from the command

C = solve(g,[M,K])

?

One possible solution is

K = 0, M = -(x + c1 + c2 + c3 + c4)

, and that's what the symbolic toolbox offers to you.

What are the arrays x0 and y0 good for in your code ?

[Sean Rotmansky]

Hello!

x0 and y0 arrays are for the larger code this is supposed to feed into (I have to solve for yp and yh then y= yp+yh) so they are not used here. The only unknown parameters (or at least, that is how it should be) are K, M and x. I have completed code with a similar question where g = 180*M*sin(x) + 180*K*cos(x) == 90*sin(x),

and it solved for both M and K (I think one was zero and the other was 0.5). I am just super lost as to why this one doesn't work.

[Torsten]

The expression you get for g is of the form

%a*cos(y) + b*K*cos(y) + c*M*cos(y) + b*M*sin(y) = c*K*sin(y) + d*sin(y)

In order to get a solution for K and M that is independent of x, the following linear system in K and M must hold:

%(a+b*K+c*M) = 0
%(b*M-c*K-d) = 0

Now solve for K and M.

Here is a possible code:

syms x c1 c2 c3 c4 K M xx
coeff = [1,-0.02,12.5,-0.058,27.77];
rx = -0.05*exp(0.005*x)*cos(11.7*x) + 0.07*exp(0.005*x)*sin(11.7*x);
yp = 2*K*cos((117*x)/10)*exp(x/200) + 2*M*sin((117*x)/10)*exp(x/200);
yp_der1 = diff(yp);
yp_der2 = diff(yp_der1);
yp_der3 = diff(yp_der2);
yp_der4 = diff(yp_der3);
yp_der1= simplify(yp_der1);
yp_der2= simplify(yp_der2);
yp_der3= simplify(yp_der3);
yp_der4= simplify(yp_der4);
g = simplify(coeff(1)*yp_der4 + coeff(2)*yp_der3 + coeff(3)*yp_der2 + coeff(4)*yp_der1 + coeff(5)*yp == rx)

g = 

eqn1 = subs(g,x,0) 

eqn1 = 

eqn2 = subs(g,x,10/117*pi/2)

eqn2 = 

[Knum,Mnum] = solve([eqn1,eqn2],[K,M])

Knum = 

Mnum = 

simplify(subs(g,[K,M],[Knum,Mnum]))

ans = 

symtrue

[Sean Rotmansky]

last question,

do you know why g would assume that form? I derive the left side of the equation 4 times, then multiply each derivative with a coefficient. The form should look like a bunch of M*cos()*e^() and K*cos()*e^() terms, and the right side should have stayed like this:

rx = -0.05*exp(0.005*x)*cos(11.7*x) + 0.07*exp(0.005*x)*sin(11.7*x);

Is there another function or method you would recommend?

[Torsten]

You can trace how the derivatives of yp are computed by removing the semicolon behind the lines.

But it is obvious that differentiating expressions of the form

a1*sin(b*x)*exp(c*x)+a2*cos(b*x)*exp(c*x)

will give expressions of the same form, only with the values for a1 and a2 changed.

[Sean Rotmansky]

Thank you so much for your help! I finally have a number answer for M and K.