Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

Solve the given differential equation subject to the given condition. Note that \(y(a)\) denotes the value of \(y a t=a\). \(\frac{d y}{d t}=0.005 y, y(10)=2\)

Step-by-Step Solution

Answer

The solution is \( y = 2e^{0.005(t-10)} \).

Step 1: Identify the Type of Differential Equation

The differential equation is a first-order linear differential equation of the form \( \frac{dy}{dt} = ky \). Here, \( k = 0.005 \).

Step 2: Write the General Solution

The general solution for the differential equation \( \frac{dy}{dt} = ky \) is \( y = Ce^{kt} \), where \( C \) is the integration constant.

Step 3: Apply the Initial Condition

Use the initial condition \( y(10) = 2 \) to determine the constant \( C \). Substituting \( t = 10 \) and \( y = 2 \) into the general solution gives:\[ 2 = Ce^{0.005 \cdot 10} \]

Step 4: Solve for the Constant C

Calculate \( e^{0.05} \) and solve for \( C \):\[ 2 = Ce^{0.05} \]\[ C = \frac{2}{e^{0.05}} \]

Step 5: Simplify and Write the Particular Solution

Substitute the value of \( C \) back into the general solution:\[ y = \frac{2}{e^{0.05}}e^{0.005t} \]Simplify to obtain the particular solution, which sometimes can be re-written as:\[ y = 2e^{0.005(t-10)} \]

Key Concepts

First-Order Linear Differential Equations

First-order linear differential equations are among the simplest types of differential equations, yet they form the backbone for understanding more complex cases. A first-order linear differential equation can generally be expressed in the form:\[ \frac{dy}{dt} = ky \]where:

• \( y \) is the function we want to solve for.
• \( t \) is the independent variable, often representing time.
• \( k \) is a constant that represents the rate of growth or decay.

This simple structure means the rate of change of \( y \) is proportional to its current value.
In practical scenarios, first-order linear differential equations frequently model exponential growth or decay processes, such as population growth, radioactive decay, or interest accumulation.

Initial Conditions

Initial conditions are crucial in differential equations as they allow us to find a specific solution from a general family of solutions. In this context, setting an initial condition involves providing a specific value for the function \( y \) at a particular point \( t \).
For example, in our exercise, we were given an initial condition \( y(10) = 2 \). This tells us that when \( t = 10 \), the value of \( y \) must be 2. Initial conditions help us determine the unique solution that satisfies both the differential equation and the specified starting condition.
This is akin to having a specific starting point in a journey, ensuring that our path fits within the larger map provided by the differential equation.

General Solution

The general solution to a differential equation provides a universal blueprint from which all particular solutions specific to initial conditions may be derived. For the form \( \frac{dy}{dt} = ky \), the general solution is:\[ y = Ce^{kt} \]where:

• \( C \) is the integration constant.

This solution demonstrates that the value of \( y \) at time \( t \) is determined by its initial value, scaled exponentially by the rate constant \( k \).
In essence, the general solution provides a model for how the system behaves over time, but doesn't yet incorporate specific conditions, making it flexible yet incomplete until those conditions are applied.

Integration Constant

In solving differential equations, the integration constant \( C \) arises from the integration process. This constant represents the family of solutions that satisfy the differential equation without initial conditions.
Determining \( C \) is where initial conditions become vital. Once we have a condition, such as \( y(10) = 2 \), we can substitute these values into the general solution and solve for \( C \). In our example, solving:\[ 2 = Ce^{0.05} \] led us to:\[ C = \frac{2}{e^{0.05}} \]
The integration constant ensures that the solution is not just a theoretical model, but one that aligns perfectly with real-world data points. Each different value of \( C \) provides a unique trajectory for \( y \) on the graph, demonstrating the system's behavior under specific circumstances.