Question

How can I make these equations into 6 first order equations to input them in MATLAB as :

function yp = ivpsys_fun_oscillator(t, y)
% IVPSYS_FUN evaluates the right-hand-side of the ODE's.
% Inputs are current time and current values of y. Outputs are values of y'.
% Call format: yp = ivpsys_fun_oscillator(t, y)
global m k l
%% Define ODE for IVP
% Reduce high order derivative to first order
% y1 -> x1, y2 -> x2, y3 -> x3
% y4 -> z1, y5 -> z2, y6 -> z3
% y7 -> u1, y8 -> u2, y9 -> u3
% y10 -> w1, y11 -> w2, y12 -> w3

yp(1) = y(7);
yp(2) = y(8);
yp(3) = y(9);
yp(4) = y(10);
yp(5) = y(11);
yp(6) = y(12);
yp(7) = (k/m) * ( (y(2)-y(1)) + (y(3)-y(1)) );
yp(8) = (k/m) * ( (y(1)-y(2)) + (y(3)-y(2)) - 3*l/2);
yp(9) = (k/m) * ( (y(1)-y(3)) + (y(2)-y(3)) + 3*l/2);
yp(10) = (k/m) * ( (y(5)-y(4)) + (y(6)-y(4)) + sqrt(3)*l );
yp(11) = (k/m) * ( (y(4)-y(5)) + (y(6)-y(5)) - sqrt(3)*l/2);
yp(12) = (k/m) * ( (y(4)-y(6)) + (y(5)-y(6)) - sqrt(3)*l/2);

end

This is probably wrong but it is as far as I got, thank you

k [2-)(3 т k ( 2)(x3- 2)- 31/2], т k 2 - s)31/2] т k (2-y) (s- )3, т k )(s-2)-31/2, y2 т k [(yy) (y2-y3)- 31/2], т where and y for i [1, 3 are the position of the three masses (numbered from 1 to 3) in the x and y directions, respectively. At time t 0, the masses are located at the following position (in m): 1, (t = 0) = [1,1 3/2], [ar2, y2](t= 0) = [1/4,1 - 3/4], 3, Ya (t= 0) = [7/4,1 - 3/4], and they have the following initial velocities (in m/s): u1, 1|(t= 0) = [-0.1, 0.1], u2, U2(t 0)(0.2,0.1 (13)] us, v3|(t= 0) [0.2, 0.1 (1 - V3)

Answer 1

, and a, X;'. So, Let - a, and a2 i. e A2 (1) (22-a)(3-a A2 and u t 0) = -0.1 . here fore, Also, a2 (t (t = 0) = u, (t0)=-0., D) - 1 and 0s (t 0) Thus the initial conditions in terms of 'a' and 'a' are and a2(t 0) = -0.1 (t-0) = 1 Again, let b 2band - 3l - m = [(a,-b) (-b)- m

(2) .. b - (0-b,+ (z3-b-2J m u2 (t 0) = 0. 2 , thenfore, X2 (t 0) = 1/ 4 and Since, and b, (t 0 ) = 0.2 bi (t 0) 1 4 So, And, let and k =3= (*,-3)2-2) m C C2 i.e (3) 3l [(a-C(b-G) - m and u t 0) = 0.2, the refoye, (t0) 4 since and C2 (t 0) = 0.2 C (t0) 4 Further rewrite system of First order equations (1) and (2) by substituting values oF X, 72 as C; , b, espec tive a,a2 - (b-aG-a,) 02 / = b2 and, 2 (a) b2 q-1(q- 10) ]

Hence For the second order differntial equations in x , 22, and the system OF First order diffe rential equations are a, 3 а, а, к [(Ь, -aj) + ( Gg - a,) m as t 0)= 1 , a, (t =0) = -0.1 b2 (0-b) (C,- b)- 30 m b, (t 0) , b2 (t 0) = 0.2 4 k m +- - ] C, (t0) = C2 (t 0) = 0. 2 4- - Likewise, For, y, = Þ, and y,'= P, . Therefore k [9,-P)(y-,) + JE A m Pt0) = 0.1 Note, let

NOw and y,= q, . Therefoye, ,'= q,'a=q, and [ (P-,) - q, ) - v31/2] m ai (t 0)=, q, (t =0) = 0.1 1+ V3) 4 And, and 2= y. Therefore 3A2 ,= (9 -4 J3) (t0) 1 - 13 4 Thus FoT the second order differenhal equations in y, y, , and y the system OF First order differential equations are m [rE4-) (-0) P, (t 0) 1 3/2 , p2(t=0) = 0.1 2 [-,) -4, ) - 3 A/2] . m (t0)- 1/4 42(t-0) = 0.1 1+ V3)

- 2 (t 0)= 0.1 (1- V3) (t 0) = - /3/4 Hence in terms of a,', a2', b;, b2, ;', c', ,, b2', i,2,and Y,', r,you can input them in Matlab. Ard define yvector in matlab such that y(2)= a2,y3) = b, y (4) = b2 y(6)= 2,y (7) = P, , y(8) = (5) C y (10)2 y(9) Likewise For the derivatives, the yp vector such that yb7)= ,', ypl8) = yp(5)= Cy (6 ) = ', 1 Thus, the system of equations a re りわ(1) = 9(2) [y13)-y 5) - y (1) yb(2) m yP(3) y(4) k[yy3+ { y5)-y( 3)}- 3! ] yP(4) m

yp(5) y (6) [y-y5 yp (6) y3) - yl5)} + 3] And y (8) k [y(9)-y y1)-y(7) 1] yb18) - m pC9) y(10) y)y)y(11) - y (9)} - B ] k yp (10) m y2 yP11) ypu2) = y-y11y)-)} - m