Laplace Transform of Given Differential Equation

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Jordan Stanley
Jordan Stanley 2022-4-25
移动：Sam Chak 2024-3-3
Hello, I have the differential equation with initial condtions: y'' + 2y' + y = 0, y(-1) = 0, y'(0) = 0.
I need to use MATLAB to find the need Laplace transforms and inverse Laplace transforms.
I'm not sure if what I have so far is correct, here is what I have...
syms s t Y;
f = 0;
F = laplace(f,t,s);
Y1 = s*Y - 0;
Y2 = s*Y1 - 0;
laplaceSol = solve(Y2 + 2*Y1 + Y - F, Y) %Laplace Transform
invlaplaceSol = ilaplace(laplaceSol,s,t) %Inverse Laplace Transform
I get the following as output.
laplaceSol = 0
invlaplaceSol = 0
I also have the following code in an m-file.
function myplot(f,interv)
% myplot(f,[a,b])
% plot f for interval [a,b]
% here f is a symbolic expression, or a string
%
% example:
% myplot('x^2',[-1,1])
% syms x; myplot(x^2,[-1,1])
f = sym(f);
tv = linspace(interv(1),interv(2),300);
T = findsym(f,1);
plot(tv,double(subs(f,T,tv)))
Thank you,
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MD.AL-AMIN
MD.AL-AMIN 2024-3-2
移动：Sam Chak 2024-3-3
y''(t) + 4y'(t) + 8y(t) = x' (t)+x(t) with x(t) = e ^ (- 4t) * u(t) y(0) = 0 and y'(0) = 0 matlab code
MD.AL-AMIN
MD.AL-AMIN 2024-3-2
移动：Sam Chak 2024-3-3
I don't know how to solve by using MATLAB?

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回答（2 个）

Sulaymon Eshkabilov
Ran in:
Laplace transform does not work at t ~0 initial conditions and thus, here dsolve() might be a better option, e.g.:
syms y(t)
Dy=diff(y,t);
D2y = diff(y,t,2);
Eqn = D2y == -2*Dy-y;
ICs = [y(-1)==0, Dy(0)==0];
S = dsolve(Eqn, ICs)
S = 
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Walter Roberson
Walter Roberson 2022-4-25
Though, considering that dsolve() using Dy(0) == 0 gives the same solution as dsolve() with y(-1)==0, then you could potentially ignore the y(-1) and go ahead with laplace.
Jordan Stanley
Jordan Stanley 2022-4-25
I'm not sure how to include the initial conditions when using the laplace() function.

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Sulaymon Eshkabilov
Ran in:
Note that if your system has "zero" ICs and not excitation force; therefore, your system solution (response) will be zero. If you set one of your ICs, non-zero varlue and then you'll see something, e.g.:
syms s Y t
y0=0;
dy0=-1; %
Y1 = s*Y - y0;
Y2 = s*Y1- dy0;
Sol = solve(Y2 + 2*Y1 + Y, Y)
Sol = 
Sol = ilaplace(Sol,s,t)
Sol = 
fplot(Sol, [-1, 1])
% Verify: alternative solution with dsolve() gives the same result
syms y(t)
Dy=diff(y,t);
D2y = diff(y,t,2);
Eqn = D2y == -2*Dy-y;
ICs = [y(0)==0, Dy(0)==-1];
S = dsolve(Eqn, ICs);
fplot(S, [-1, 1])
  1 个评论
Walter Roberson
Walter Roberson 2022-4-25
However... the posters have been clear that they have an initial condition at y(-1) not an initial condition at y(0) . Which is a problem for laplace transforms.

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