To find the response \(y(t)\) for the given transfer function \(G(S) = \frac{1}{S^2+3S+2}\) with the input \(r(t)=5u(t)\), we will perform the following steps:

The first step is to express the transfer function \(G(S)\) as a partial fraction:

1. \(\frac{1}{S^2+3S+2} = \frac{1}{(S+1)(S+2)}\)

We expect it to be of the form:

1. \(\frac{A}{S+1} + \frac{B}{S+2}\)

Equating and solving for constants \(A\) and \(B\):

1. \(1 = A(S+2) + B(S+1)\)

By substituting suitable values:

• If \(S = -1\), then \(1 = A(1)\) ⇒ \(A = 1\).
• If \(S = -2\), then \(1 = -B\) ⇒ \(B = -1\).

The next step is to find the Laplace Transform of the input \(r(t) = 5u(t)\):

1. \(\mathcal{L}\{r(t)\} = \frac{5}{S}\)

Now, calculate \(Y(S) = G(S) \cdot R(S)\):

1. \(Y(S) = \left(\frac{1}{S+1} - \frac{1}{S+2}\right) \cdot \frac{5}{S}\)

Resulting in:

1. \(Y(S) = \frac{5}{S(S+1)} - \frac{5}{S(S+2)}\)

Taking the inverse Laplace transform:

1. \(y(t) = [\frac{5}{2} - 5e^{-t} + \frac{5}{2}e^{-2t}]\mu(t)\)

Thus, the response to the input \(r(t) = 5u(t)\) is:

• The correct option is: \(y(t) = \left[\frac{5}{2} - 5e^{-t} + \frac{5}{2}e^{-2t}\right]\mu(t)\)
• Other options don't satisfy the calculated expression.