# Laplace Transform of Given Differential Equation

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Jordan Stanley on 25 Apr 2022
Commented: Walter Roberson on 25 Apr 2022
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,
##### 2 CommentsShowHide 1 older comment
Jordan Stanley on 25 Apr 2022
I double checked and yes y(-1) = 0 is one of the initial condtions.

Sulaymon Eshkabilov on 25 Apr 2022
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 =
Jordan Stanley on 25 Apr 2022
I'm not sure how to include the initial conditions when using the laplace() function.

Sulaymon Eshkabilov on 25 Apr 2022
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])
Walter Roberson on 25 Apr 2022
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.

R2022a

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