Laplace Transform of Given Differential Equation

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,

[Paul]

Are you sure about that initial condiion y(-1) = 0? For these types of problems, intiial conditions are typically specified at t = 0, more specifically at t = 0-, i.e., an infinitesimal to the left of 0. I'm not even sure that a condition at t = -1 can be incorporated into the Laplace trasnsform approach at all.

Having said that, the solution y(t) = 0 does satisfy the differential equation and the initial conditions.

[Jordan Stanley]

I double checked and yes y(-1) = 0 is one of the initial condtions.

[MD.AL-AMIN]

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]

I don't know how to solve by using MATLAB?

Answer 1

[Sulaymon Eshkabilov]

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 = 

[Walter Roberson]

Interestingly, the Dy(0)==0 does not add any information -- if you solve for just the first IC then you get the same result.

[Jordan Stanley]

For this task I need to specifically solve using Laplace and Inverse Laplace Transforms.

[Walter Roberson]

Then you will have to fail the assignment. Laplace transform is not valid for negative time.

[Paul]

I agree. Having said that, if both initial conditions were given at t = -1, then a solution via Laplace transform could be obtained by shifting the time variable (I think).

BTW, didn't this exact differential equation show up here no too long ago?

[Jordan Stanley]

Yes, I have an assignment using this differential equation in MATLAB.

With this being my first time using MATLAB I have been needing a lot of assistance.

[Walter Roberson]

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]

I'm not sure how to include the initial conditions when using the laplace() function.

Answer 2

[Sulaymon Eshkabilov]

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]

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.