Help with Laplace transform method for 2nd order DE

I'm trying to solve the equation: y'' +2y' +y = g(t), with g(t) = { sin(t) if t < pi, 0 if t < pi }, and the intial conditons y(0) = 0 and y'(0) = 0.

Here's the code that I've tried:

 syms s t y(t) Y
g = heaviside(t)*(sin(t)) + heaviside(t - pi)*(-sin(t));
eqn = diff(y,2) + 2*diff(y) + 2*(y) == g;
lteqn = laplace(eqn, t, s);
neweqn = subs(lteqn, [laplace(y(t), t, s), y(0), subs(diff(y(t), t, 0)), [Y, 0, 0]]);
ytrans = simplify(solve(neweqn, Y));
Warning: Unable to find explicit solution. For options, see help. 

Invalid expression. Check for missing multiplication operator, missing or unbalanced delimiters, or other syntax error. To construct matrices, use brackets instead of parentheses.

Error in connector.internal.fevalMatlab

Error in connector.internal.fevalJSON

> In sym/solve (line 317) 
>> y = ilaplace(ytrans, s, t)
 
y =
 
Empty sym: 0-by-1

I'm not sure why ytrans is not finding a solution. Can someone help me fix this/ find a different method?

Answer 1

[Star Strider]

In the subs call, use curly braces {} instead of square brackets [].

Then, it works —

syms s t y(t) Y
g = heaviside(t)*(sin(t)) + heaviside(t - pi)*(-sin(t));
eqn = diff(y,2) + 2*diff(y) + 2*(y) == g;
lteqn = laplace(eqn, t, s);
neweqn = subs(lteqn, {laplace(y(t), t, s), y(0), subs(diff(y(t), t), t, 0)}, {Y, 0, 0})

neweqn = 

ytrans = simplify(solve(neweqn, Y));

y = ilaplace(ytrans, s, t)

y = 

figure
fplot(y, [0 10])
grid

.