Problem 57 - Calculus and its applications

Question

The value of a share of stock of Leslie's Designs, Inc., is modeled by $$ \frac{d V}{d t}=k(L-V) $$ where \(V\) is the value of the stock, in dollars, after \(t\) months; \(k\) is a constant; \(L=\$ 24.81,\) the limiting value of the stock; and \(V(0)=20 .\) Find the solution of the differential equation in terms of \(t\) and \(k\).

Step-by-Step Solution

Answer

\( V(t) = 24.81 - 4.81e^{-kt} \) is the solution in terms of \( t \) and \( k \).

Step 1: Recognize the Type of Differential Equation

The given differential equation \( \frac{dV}{dt} = k(L - V) \) is a first-order linear ordinary differential equation.

Step 2: Separate the Variables

Rearrange the equation to separate variables on either side: \( \frac{1}{L - V} \frac{dV}{dt} = k \). This allows easier integration with respect to \( V \) and \( t \).

Step 3: Integrate Both Sides

Integrate the left side with respect to \( V \) and the right side with respect to \( t \):\[ \int \frac{1}{L - V} \, dV = \int k \, dt \]This results in \(-\ln|L - V| = kt + C\), where \( C \) is the constant of integration.

Step 4: Solve for \( V \)

Exponentiate both sides to solve for \( V \):\[ L - V = Ce^{-kt} \] or equivalently, \( V = L - Ce^{-kt} \).

Step 5: Determine the Constant \( C \) Using Initial Conditions

Use the initial condition \( V(0) = 20 \) to find \( C \):\[ 20 = L - C \cdot e^{0} \rightarrow 20 = 24.81 - C \]Solving, \( C = 24.81 - 20 = 4.81 \).

Step 6: Write Down the Final Solution

Substitute \( C \) back into the equation from Step 4:\[ V(t) = 24.81 - 4.81e^{-kt} \].

Key Concepts

First-order Differential Equation

A first-order differential equation is one that involves the first derivative of a function but no higher derivatives. In the given problem, the equation \( \frac{dV}{dt} = k(L - V) \) is a classic example. Here, \( V \) represents the stock value, and \( t \) is time in months. An important feature of such equations is that they often model processes with rates of change, like growth or decay.

They are called "first-order" because we deal only with the first derivative here. In real-world applications, these equations help us describe dynamic systems that change over time, providing insights into how a variable evolves. The key is to recognize the form of the equation and apply the right techniques to find a solution. In this case, we use separation of variables to find \( V \) as a function of \( t \).

• Involves first derivatives only
• Used to model dynamic processes
• Solution technique: separation of variables

Separation of Variables

Separation of variables is a powerful method for solving differential equations. The essence of this method is to manipulate the equation to isolate one variable and its differential on one side and the other variable and its differential on the opposite side. In the equation \( \frac{dV}{dt} = k(L - V) \), we can rearrange it to \( \frac{1}{L - V} \frac{dV}{dt} = k \).

Now, each side depends only on a single variable: \( V \) on the left and \( t \) on the right. This setup permits easy integration:

• \( \int \frac{1}{L - V} \, dV = -\ln|L-V| \)
• \( \int k \, dt = kt + C \)

Once integrated, these give us an equation that we can solve for \( V \). This approach simplifies our solution path significantly and is often the first tool at hand for these kinds of equations.

Initial Conditions

Initial conditions are crucial for finding the specific solution of a differential equation that fits the physical situation being modeled. In this exercise, the initial condition is \( V(0) = 20 \), meaning at \( t = 0 \), the stock value is \$20.

Applying the initial condition allows us to solve for any constants introduced during the integration process, such as \( C \) in our solution \( V = L - Ce^{-kt} \). When we substitute the initial condition into the equation:

• \( 20 = 24.81 - Ce^{0} \)
• This gives \( C = 4.81 \)

These conditions ensure that the solution aligns with the specific case being modeled, unlike the general solution which represents a family of functions. This specificity is what turns a general solution into the particular solution that fits the initial criteria.