Consider a system model given by

 2 fraction numerator d squared x left parenthesis t right parenthesis over denominator d t squared end fraction plus 3 fraction numerator d x left parenthesis t right parenthesis over denominator d t end fraction plus x left parenthesis t right parenthesis equals delta left parenthesis t right parenthesis
x left parenthesis 0 to the power of minus right parenthesis equals 0
right enclose fraction numerator d x left parenthesis t right parenthesis over denominator d t end fraction end enclose subscript t equals 0 to the power of minus end subscript equals 1

What are the initial conditions at x(0+) and dx/dt (0+)?

		
x(0+)=-1 and dx/dt (0+)=0 

		
x(0+)=0 and dx/dt (0+)=1.5

		
x(0+)=1 and dx/dt (0+)=1

		
x(0+)=1 and dx/dt (0+)=0 

		
x(0+)=0 and dx/dt (0+)=1

		
x(0+)=0 and dx/dt (0+)=0



We are given a system model:

\[
2 \frac{d^2 x(t)}{dt^2} + 3 \frac{dx(t)}{dt} + x(t) = \delta(t)
\]
with the initial conditions:
\[
x(0^-) = 0, \quad \left. \frac{dx(t)}{dt} \right|_{t=0^-} = 1.
\]

### Step 1: Analyze the Differential Equation

The system is a second-order linear differential equation with an impulse input \(\delta(t)\). The impulse input suggests that there will be a discontinuity or jump in the derivative of \(x(t)\) at \(t=0\).

### Step 2: Apply Laplace Transform

To handle the discontinuity, we apply the Laplace transform to both sides of the equation. Using the fact that:

\[
\mathcal{L}\left\{ \frac{d^2x(t)}{dt^2} \right\} = s^2 X(s) - s x(0^-) - \left. \frac{dx(t)}{dt} \right|_{t=0^-},
\]
\[
\mathcal{L}\left\{ \frac{dx(t)}{dt} \right\} = s X(s) - x(0^-),
\]
\[
\mathcal{L}\left\{ x(t) \right\} = X(s),
\]
and \(\mathcal{L}\{\delta(t)\} = 1\), we transform the differential equation:

\[
2(s^2 X(s) - s x(0^-) - \left. \frac{dx(t)}{dt} \right|_{t=0^-}) + 3(s X(s) - x(0^-)) + X(s) = 1.
\]

### Step 3: Substitute Initial Conditions

We are given \(x(0^-) = 0\) and \(\left. \frac{dx(t)}{dt} \right|_{t=0^-} = 1\). Substituting these into the Laplace-transformed equation:

\[
2(s^2 X(s) - 1) + 3(s X(s)) + X(s) = 1.
\]

Simplifying:

\[
2s^2 X(s) - 2 + 3s X(s) + X(s) = 1,
\]
\[
(2s^2 + 3s + 1) X(s) = 3,
\]
\[
X(s) = \frac{3}{2s^2 + 3s + 1}.
\]

### Step 4: Inverse Laplace Transform

Now, we take the inverse Laplace transform to find \(x(t)\). The Laplace transform of the system suggests a step response for a second-order system with an impulse input. To determine the initial conditions at \(t = 0^+\), we use the fact that the system responds to the impulse input.

Since \(x(t)\) was initially at rest (\(x(0^-) = 0\)), and there was a discontinuity in \(\frac{dx(t)}{dt}\), we know:

\[
x(0^+) = x(0^-) = 0.
\]

To find \(\left. \frac{dx(t)}{dt} \right|_{t=0^+}\), we analyze the jump condition caused by the impulse input. The impulse \(\delta(t)\) forces an immediate change in the derivative. Given that \(\left. \frac{dx(t)}{dt} \right|_{t=0^-} = 1\), and considering the dynamics of the system, we expect the initial derivative to remain the same:

\[
\left. \frac{dx(t)}{dt} \right|_{t=0^+} = 1.
\]

### Final Answer:

The correct initial conditions are:

\[
x(0^+) = 0, \quad \left. \frac{dx(t)}{dt} \right|_{t=0^+} = 1.
\]

Thus, the correct choice is:
\[
\boxed{x(0^+) = 0 \text{ and } \frac{dx}{dt}(0^+) = 1.}
\]

---

Would you like further details on the steps, or do you have any questions? Here are some follow-up questions that might interest you:

1. How does the Laplace transform handle discontinuities in time-domain functions?
2. Why does an impulse input result in a jump in the system’s derivative?
3. What role do initial conditions play in solving differential equations using Laplace transforms?
4. How can we interpret second-order system behavior from its differential equation?
5. What methods can be used to find inverse Laplace transforms of rational functions?

**Tip:** Always remember that the Laplace transform is a powerful tool for solving differential equations, especially when dealing with initial conditions and discontinuous inputs like impulses.

Consider a system model given by

 2 d²x(t)/dt² + 3 dx(t)/dt + x(t) = δ(t)
with the initial conditions x(0⁻) = 0 and (dx/dt)(0⁻) = 1.
What are the initial conditions at x(0⁺) and dx/dt(0⁺)?

This article explains how to solve a second-order linear differential equation using Laplace transforms with impulse input δ(t). The problem explores system dynamics and initial conditions x(0⁺) and dx/dt(0⁺). Suitable for undergraduate students in engineering and physics.

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