I am fairly new to MATLAB and i don't really know how to solve this equation. y''+(exp(-10t)+sen(10t))y'+y=0 If anyone could help me it would me much appreciated
MATLAB: How do i solve a second order ODE with variable coefficients
differential equationsequationssecond order
Related Solutions
The easiest way is to ask for a second output from your odeToVectorField call. It tells you the substitutions the function made to define each element in the returned column:
[yode,Sbs] = odeToVectorField(eqns);here, the substitutions are:
Sbs = y Dy x Dx z DzThe columns of ‘Yodefcn’ exactly reproduces those returned by odeToVectorField, and ode45 returns them in that order.
That can be used with some other assignments to automatically produce a legend that labels every line in the plot.
The Code —
syms x(t) y(t) z(t) p(t) Yp=sin((0.5+3.5*t)*t);D1y = diff(y,t);D2y = diff(y,t,2);D1x = diff(x,t);D2x = diff(x,t,2);D1z = diff(z,t);D2z = diff(z,t,2);eqns=[D2x+2*x-1*y==p,D2y+(0.5*10*exp(-10*t))*D1y-x+2*y-z==0,D2z-y+2*z==0];[yode,Sbs] = odeToVectorField(eqns);Yodefcn = matlabFunction(yode, 'Vars',[t Y]);tspan = (0:0.001:10);Y0 = [0 0,0 0,0 0];[T,Y] = ode45(Yodefcn, tspan, Y0);plot(T, Y)Lgndc = sym2cell(Sbs);Lgnds = regexp(sprintf('%s\n', Lgndc{:}), '\n', 'split');legend(Lgnds(1:end-1), 'Location','N')The Plot —
The symbolic derivation is similar to that in your previous Question. I list it here (commented-out) for reference.
Symbolic Derivation —
% syms p t y(t) Y
% p = sym('p%d', [1 2]);
% D1y = diff(y,t);
% D2y = diff(y,t,2);
% Eqn = D2y + p(1)*(1-exp(-10*t)) + p(2)*(1-cos(10*t))*D1y + y == 0;
% yode = odeToVectorField(Eqn);
% Yodefcn = matlabFunction(yode, 'Vars',[t Y p])
The code with the varying parameters is therefore straightforward with the results of the symbolic derivation.
The Code —
Yodefcn = @(t,Y,p1,p2) [Y(2);p1.*(exp(t.*-1.0e1)-1.0)-Y(1)+p2.*(cos(t.*1.0e1)-1.0).*Y(2)];Pmat = [-0.11674 0.05351 1.8989; -0.18115 0.05157 0.6279]; % Parameter & Initial Condition Matrix
tspan = linspace(0, 15, 100);for k1 = 1:size(Pmat,1) Y0 = [0 Pmat(k1,3)]; [T{k1},Y{k1}] = ode45(@(t,Y) Yodefcn(t,Y,Pmat(k1,1),Pmat(k1,2)), tspan, Y0);endfigure(1)plot(T{1}, Y{1})hold onplot(T{2}, Y{2}, '--')hold offgridxlabel('\itt\rm')ylabel('\ity\rm(\itt\rm)')The ‘p’ vector in the symbolic derivation are the relevant parameters. These are then entered into the ‘Pmat’ matrix for convenience, the first two columns corresponding to the first two parameters in each differential equation, and the third column the initial condition for the derivative. These are then called in the corresponding iterations of the for loop, and the results saved in the corresponding cell arrays, ‘T’ and ‘Y’. Then figure(1) plots them, the first set with a solid line, and the second set with a dashed line. (I would have added a legend if I knew what to call the plotted values.)
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