Step 1: Solve the Differential Equation for the Particle's Position

We are given the first-order linear differential equation:

$\frac{dx}{dt} + 2x = 3e^{-t}$

This is a linear differential equation in the standard form $\frac{dx}{dt} + p(t)x = g(t)$, where $p(t) = 2$ and $g(t) = 3e^{-t}$. To solve this equation, we will use the method of integrating factors.

Finding the Integrating Factor

The integrating factor $\mu(t)$ is given by:

$\mu(t) = e^{\int p(t) dt} = e^{\int 2 dt} = e^{2t}$

Multiply Both Sides by the Integrating Factor

Multiplying the entire equation by $e^{2t}$:

$e^{2t} \frac{dx}{dt} + 2e^{2t}x = 3e^{-t} \cdot e^{2t}$

Simplifying the right-hand side:

$e^{2t} \frac{dx}{dt} + 2e^{2t}x = 3e^t$

Write the Left Side as a Derivative

The left-hand side is now the derivative of $x(t) \cdot e^{2t}$:

$\frac{d}{dt}(x(t) \cdot e^{2t}) = 3e^t$

Integrate Both Sides

Now, integrate both sides with respect to $t$:

$x(t) \cdot e^{2t} = \int 3e^t dt = 3e^t + C$

Where $C$ is the constant of integration.

Solve for $x(t)$

Now solve for $x(t)$:

$x(t) = e^{-2t}(3e^t + C) = 3e^{-t} + Ce^{-2t}$

Step 2: Apply the Initial Condition

We are given the initial condition $x(0) = 1$. Substituting $t = 0$ into the solution:

$1 = 3e^0 + C e^0$

$1 = 3 + C$

Solving for $C$:

$C = 1 - 3 = -2$

Thus, the solution for the particle's position is:

$x(t) = 3e^{-t} - 2e^{-2t}$

Step 3: Shear Transformation

We are asked to apply a horizontal shear transformation to the position-time graph. The shear transformation mentioned is a horizontal shear by a factor of $1/2$. This means that the time coordinate $t$ will be replaced by $t - \frac{x}{2}$.

So, the new equation for the position becomes:

$x = 3e^{-(t - \frac{x}{2})} - 2e^{-2(t - \frac{x}{2})}$

This equation represents the position after applying the horizontal shear.

Step 4: Find When the Particle is at Position $x = 0$

Now, we need to find the time $t$ when $x = 0$, after applying the shear. Set $x = 0$ in the equation:

$0 = 3e^{-(t - \frac{0}{2})} - 2e^{-2(t - \frac{0}{2})}$

Simplifying:

$0 = 3e^{-t} - 2e^{-2t}$

Dividing through by $e^{-2t}$ (since $e^{-2t} \neq 0$):

$0 = 3e^{t} - 2$

Solve for $e^t$:

$3e^t = 2 \quad \Rightarrow \quad e^t = \frac{2}{3}$

Taking the natural logarithm of both sides:

$t = \ln\left(\frac{2}{3}\right)$

Thus, the time at which the particle is at position $x = 0$ after the shear is:

$t = \ln\left(\frac{2}{3}\right) \approx -0.405$

Summary of Results:

1. The position of the particle as a function of time is $x(t) = 3e^{-t} - 2e^{-2t}$.
2. After applying a shear transformation by a factor of $1/2$, the time when the particle is at position $x = 0$ is approximately $t = -0.405$.

Do you want further details or have any questions?

Related Questions:

1. How do shear transformations affect velocity-time graphs?
2. What is the significance of an integrating factor in solving differential equations?
3. Can a vertical shear be applied to this position-time graph, and how would it change the solution?
4. How does the solution of this differential equation change if the initial condition is different?
5. How would this model change if the external forcing term $3e^{-t}$ were altered?

Tip:

When solving differential equations, always check if the equation is linear or separable to determine the most efficient solving method.