Key Concepts and Formulas

• First-Order Separable Differential Equations: A differential equation of the form $\frac{dy}{dx} = f(x)g(y)$ can be separated and integrated as $\int \frac{dy}{g(y)} = \int f(x) dx$.
• Integration: The process of finding the antiderivative of a function. Remember to include the constant of integration, "+ C".
• Initial Value Problem: Using a given initial condition (e.g., $V(0) = I$) to find a particular solution to a differential equation.

Step-by-Step Solution

Step 1: Understanding the Problem and Separating Variables

We are given the differential equation $\frac{dV(t)}{dt} = -k(T - t)$, where $V(t)$ is the value of the equipment after $t$ years, $k > 0$ is a constant, and $T$ is the total life of the equipment. We want to find the scrap value, $V(T)$, given that the initial purchase value is $V(0) = I$. The first step is to separate the variables.

Given: $\frac{dV(t)}{dt} = -k(T - t)$

Multiply both sides by $dt$: $dV(t) = -k(T - t) dt$

Step 2: Integrating Both Sides

Now, we integrate both sides of the equation to find the general solution for $V(t)$.

$\int dV(t) = \int -k(T - t) dt$

The left side integrates to: $\int dV(t) = V(t)$

For the right side, we integrate: $\int -k(T - t) dt = -k \int (T - t) dt = -k \left(Tt - \frac{t^2}{2}\right) + C = -k Tt + k\frac{t^2}{2} + C$ Alternatively, we can use substitution. Let $u = T - t$, so $du = -dt$. Then $dt = -du$. $\int -k(T - t) dt = \int -k u (-du) = k \int u du = k \frac{u^2}{2} + C = k \frac{(T - t)^2}{2} + C$ Expanding this gives: $k \frac{(T^2 - 2Tt + t^2)}{2} + C = k \frac{T^2}{2} - k Tt + k\frac{t^2}{2} + C$ Both methods are correct, so we have: $V(t) = -k Tt + k\frac{t^2}{2} + C$ Or, equivalently: $V(t) = k \frac{(T - t)^2}{2} + C$

Step 3: Applying the Initial Condition

We are given that $V(0) = I$. We use this to find the value of $C$. Using $V(t) = -k Tt + k\frac{t^2}{2} + C$: $V(0) = -k T(0) + k\frac{(0)^2}{2} + C = I$ $0 + 0 + C = I$ $C = I$ So, the particular solution is: $V(t) = -k Tt + k\frac{t^2}{2} + I$

Alternatively, using $V(t) = k \frac{(T - t)^2}{2} + C$: $V(0) = k \frac{(T - 0)^2}{2} + C = I$ $k \frac{T^2}{2} + C = I$ $C = I - k \frac{T^2}{2}$ So, the particular solution is: $V(t) = k \frac{(T - t)^2}{2} + I - k \frac{T^2}{2}$

Step 4: Calculating the Scrap Value, V(T)

We want to find $V(T)$, the value of the equipment at the end of its life. Using the first form of the particular solution: $V(T) = -k T(T) + k\frac{(T)^2}{2} + I = -kT^2 + k\frac{T^2}{2} + I = I - k\frac{T^2}{2}$ Using the second form of the particular solution: $V(T) = k \frac{(T - T)^2}{2} + I - k \frac{T^2}{2} = k \frac{0}{2} + I - k \frac{T^2}{2} = I - k \frac{T^2}{2}$

Thus, the scrap value is $I - \frac{kT^2}{2}$.

Common Mistakes & Tips

• Forgetting the Constant of Integration: Always include "+ C" when performing indefinite integration.
• Incorrectly Applying the Initial Condition: Ensure you substitute the correct values for $t$ and $V(t)$ when solving for $C$.
• Algebra Errors: Be careful with signs and exponents during integration and simplification.

Summary

We solved the first-order separable differential equation $\frac{dV(t)}{dt} = -k(T - t)$ with the initial condition $V(0) = I$. By separating variables, integrating, and applying the initial condition, we found the particular solution and then evaluated it at $t = T$ to determine the scrap value. The scrap value is $I - \frac{kT^2}{2}$. This corresponds to option (A).

Final Answer

The final answer is $\boxed{I - {{k{T^2}} \over 2}}$, which corresponds to option (A).