Getting empty solution for second order differential equation
显示 更早的评论
initial conditions are, x(0)=0 and dx/dt(0)=0
I want equation of x in terms of t. But I am not getting.
5 个评论
Jan
2022-4-7
What ist your question? Do you consider, that most functions do not have a closed form integral? That Matlab cannot find one, could mean, that there is none.
Torsten
2022-4-7
Maybe you mean sin(t) instead of sin(x) ?
Gopalji Kalojiya
2022-4-7
Gopalji Kalojiya
2022-4-7
Star Strider
2022-4-7
Note that 
implies 
.
implies . MATLAB makes the appropriate assumption about it, and defines x as 
.
. 采纳的回答
It apparently does not have an analytic solution.
Integrate it numerically —
syms x(t) t Y
ode = diff(x,2) + 2*diff(x) + 5*sin(x) == 5
Dx = diff(x);
% cond = [x(0) == pi/2, Dx(0) == 0]
[VF,Subs] = odeToVectorField(ode)
ode_fcn = matlabFunction(VF, 'Vars',{t,Y})
tspan = [0 10];
ic = [pi/2, 0];
[t,y] = ode45(ode_fcn, tspan, ic);
figure
plot(t,y)
grid
legend(string(Subs))
.
更多回答（1 个）
Sam Chak
2022-4-8
1 个投票
I believe that the question never ask you to find the analytical solution of the damped pendulum system, because it doesn't exist. The analytical solution only exists for undamped pendulum system:
and it is in the form of Jacobi elliptic function and the inverse sine function. For more info, please check the suggested article:
Ochs, K. (2011). A comprehensive analytical solution of the nonlinear pendulum. European Journal of Physics, 32(2), 479–490. doi:10.1088/0143-0807/32/2/019.
Script for the forced responses of the damped pendulum system:
% Damped Pendulum ODE
f = @(t, x) [x(2); ...
5 - 5*sin(x(1)) - 2*x(2)];
tStart = 0;
tEnd = 10;
step = 0.01;
numel = length(tStart : step : tEnd);
tspan = linspace(tStart, tEnd, numel)';
init = [pi/2 0]; % initial condition
% Runge-Kutta Dormand-Prince 4/5 solver
[t, x] = ode45(f, tspan, init);
plot(t, x, 'linewidth', 1.5)
grid on
xlabel('Time, t [sec]')
ylabel({'x(t), and x''(t)'})
title('Forced responses of the damped pendulum system')
legend('x(t)', 'x''(t)')
Result:
If you can continuously supply a torque that generates the angular acceleration of 5 rad/s², then the bob of a nearly 2-m length pendulum will look as if it is suspended at 90°. When the pendulum is in steady-state, 
, and thus,
, and thus,
.Since the initial value for 
is already in equilibrium, it will continue to stay so until the constant torque is removed or changed.
is already in equilibrium, it will continue to stay so until the constant torque is removed or changed.Script for the free responses of the damped pendulum system:
% Damped Pendulum ODE
f = @(t, x) [x(2); ...
0 - 5*sin(x(1)) - 2*x(2)];
tStart = 0;
tEnd = 10;
step = 0.01;
numel = length(tStart : step : tEnd);
tspan = linspace(tStart, tEnd, numel)';
init = [pi/2 0]; % initial condition
% Runge-Kutta Dormand-Prince 4/5 solver
[t, x] = ode45(f, tspan, init);
plot(t, x, 'linewidth', 1.5)
grid on
xlabel('Time, t [sec]')
ylabel({'x(t), and x''(t)'})
title('Free responses of the damped pendulum system')
legend('x(t)', 'x''(t)')
Result:
If the torque is removed, the bob takes approximately 5 seconds to converge to 0°. You may think that 5-sec is rather slow. After all, the pendulum length is nearly 2 meters long 
.
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