To solve for the free response of the system described by the second-order differential equation:

$\frac{d^2x(t)}{dt^2} = t$

with the initial conditions:

$x(0) = 1 \quad \text{and} \quad \frac{dx(t)}{dt} \Big|_{t=0} = 2,$

we proceed step by step.

Step 1: Integrate the given differential equation.

First, integrate the equation $\frac{d^2x(t)}{dt^2} = t$ with respect to $t$:

$\frac{dx(t)}{dt} = \int t \, dt = \frac{t^2}{2} + C_1,$

where $C_1$ is the constant of integration.

Step 2: Use the initial velocity condition.

From the initial condition $\frac{dx(t)}{dt} \Big|_{t=0} = 2$, substitute $t = 0$ into the equation for $\frac{dx(t)}{dt}$:

$2 = \frac{0^2}{2} + C_1 \quad \Rightarrow \quad C_1 = 2.$

Thus, the velocity expression becomes:

$\frac{dx(t)}{dt} = \frac{t^2}{2} + 2.$

Step 3: Integrate again to find $x(t)$.

Now integrate the expression for $\frac{dx(t)}{dt}$ to find $x(t)$:

$x(t) = \int \left(\frac{t^2}{2} + 2\right) dt = \frac{t^3}{6} + 2t + C_2,$

where $C_2$ is another constant of integration.

Step 4: Use the initial position condition.

From the initial condition $x(0) = 1$, substitute $t = 0$ into the expression for $x(t)$:

$1 = \frac{0^3}{6} + 2(0) + C_2 \quad \Rightarrow \quad C_2 = 1.$

Thus, the solution for $x(t)$ is:

$x(t) = \frac{t^3}{6} + 2t + 1.$

Conclusion:

The free response of the system is:

$x(t) = \frac{t^3}{6} + 2t + 1.$

This matches the option:

$\boxed{\frac{t^3}{6} + 2t + 1}.$

Would you like further details or clarification?

Here are 5 related questions:

1. How do initial conditions affect the solution of differential equations?
2. What is the significance of the free response in a system model?
3. How do you solve a second-order differential equation with non-constant coefficients?
4. What methods can be used to solve higher-order differential equations?
5. How do you verify the correctness of a solution to a differential equation?

Tip: Always check initial conditions after solving to confirm the constants of integration are correct.