ode(t) =
Solving ODE using laplace
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This is the question I'm struggling on
Using the Laplace transform find the solution for the following ODE:
d^2/dt(y(t)) + 16y(t) = 16[1(t-3) -1(t)]
initial conditions:
y(0) = 0
dy(t)/dt = 0
I have to solve the ODE with laplace and with inverse laplace
Save the inverse laplace in y_sol.
This is what I wrote but it gives me the wrong answer:
syms t s y(t) Y X
y0 = 0;
dot_y0 = 0;
dot_y = diff(y,t);
ddot_y = diff(dot_y,t);
ode = ddot_y + 16*y == 16*(1*(t-3)-1*(t))
Y1 = laplace(ode,t,s)
ysol1 = subs(Y1,laplace(y,t,s),X)
ysol2 = subs(ysol1,y(0),y0)
ysol3 = subs(ysol2, subs(diff(y(t), t), t, 0), dot_y0)
ysol = solve(ysol3, X)
Y = simplify(expand(ysol))
y_sol = ilaplace(Y)
7 个评论
Torsten
2024-4-21
I guess 1(t-3) and 1(t) are dirac(t-3) and dirac(t), not 1*(t-3) and 1*t.
Star Strider
2024-4-21
That may be correct. I certainly didn’t interpret the ‘1*’ notation as the delta function.
Not '1*', just '1'.
I think the notation is quite common for the indicator function:
But I'm not sure how it translates to a known Matlab function.
Paul
2024-4-21
1(t), as in the question, most likely means the unit step function. I’d be surprised if it means 1*t.
Considering that this appears to be a homework question at a typical undergraduate level, and given that the initial values start from zero, I am inclined to agree with @Paul's interpretation of 
as the unit step function, a.k.a. the Heaviside function. It seems that the original poster may not be proficient in properly expressing mathematical notations, which naturally leads to the interpretation of the missing asterisk as a multiplication operator between two values.
as the unit step function, a.k.a. the Heaviside function. It seems that the original poster may not be proficient in properly expressing mathematical notations, which naturally leads to the interpretation of the missing asterisk as a multiplication operator between two values.Additionally, it appears that the undamped mass-spring system has an analytical solution, though I'm unsure how to get that from Inverse Laplace Transform.
%% Input signal
sympref('HeavisideAtOrigin', 1);
t = 0:0.001:6;
u = 16*(heaviside(t-3) - heaviside(t));
plot(t, u, 'linewidth', 1.5), grid on, ylim([-25 25]), xlabel('t'), title('u(t)')
I didn't simplify the analytical solution from dsolve, but it seems to yield the similar plot as WolframAlpha.
sympref('HeavisideAtOrigin', 1);
syms y(t) t s
dy = diff( y,t);
ddy = diff(dy,t);
massSpring = ddy + 16*y == 16*(heaviside(t-3) - heaviside(t))
sol = dsolve(massSpring, y(0) == 0, dy(0) == 0)
fplot(sol, [0 13]), grid on, xlabel('t'), title('y(t)')
syms t s y(t) Y X
y0 = 0;
dot_y0 = 0;
dot_y = diff(y,t);
ddot_y = diff(dot_y,t);
ode = ddot_y + 16*y == 16*(heaviside(t-3)-heaviside(t));
Y1 = laplace(ode,t,s);
ysol1 = subs(Y1,laplace(y,t,s),X);
ysol2 = subs(ysol1,y(0),y0);
ysol3 = subs(ysol2, subs(diff(y(t), t), t, 0), dot_y0);
ysol = solve(ysol3, X);
Y = simplify(expand(ysol));
y_sol = ilaplace(Y)
Check_Laplace_Solution = dsolve(ode, y(0) == 0, dot_y(0) == 0)
hold on
fplot(y_sol,[0 13])
fplot(Check_Laplace_Solution,[0 13])
hold off
grid on
回答（1 个）
Your code looks correct to me, and when I checked the result with dsolve, its solution agreees with yours —
syms t s y(t) Y X
y0 = 0;
dot_y0 = 0;
dot_y = diff(y,t);
ddot_y = diff(dot_y,t);
ode = ddot_y + 16*y == 16*(1*(t-3)-1*(t))
Y1 = laplace(ode,t,s)
ysol1 = subs(Y1,laplace(y,t,s),X)
ysol2 = subs(ysol1,y(0),y0)
ysol3 = subs(ysol2, subs(diff(y(t), t), t, 0), dot_y0)
ysol = solve(ysol3, X)
Y = simplify(expand(ysol))
y_sol = ilaplace(Y)
Check_Laplace_Solution = dsolve(ode, y(0) == 0, dot_y(0) == 0)
.
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