Problem 4 - 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 \(y a t=a\). \(\frac{d y}{d t}=-0.003 y, y(-2)=3\)

Step-by-Step Solution

Answer

The solution is \( y(t) = 3e^{0.006-0.003t} \).

Step 1: Identify the Type of Differential Equation

The given equation \( \frac{d y}{d t} = -0.003 y \) is a first-order linear homogeneous differential equation with separation of variables.

Step 2: Separate the Variables

Rearrange the equation to separate the variables \( y \) and \( t \): \[ \frac{d y}{y} = -0.003 \, dt \] This sets the stage for integrating both sides.

Step 3: Integrate Both Sides

Integrate the left side with respect to \( y \) and the right side with respect to \( t \): \[ \int \frac{d y}{y} = \int -0.003 \, dt \] This yields:\[ \ln |y| = -0.003t + C \] where \( C \) is the integration constant.

Step 4: Solve for \( y \)

Exponentiate both sides to solve for \( y \): \[ |y| = e^{-0.003t + C} \] Simplify to:\[ y = C_1 e^{-0.003t} \] where \( C_1 = e^C \), adjusting the constant name.

Step 5: Apply the Initial Condition

Use the initial condition \( y(-2) = 3 \) to solve for \( C_1 \): \[ 3 = C_1 e^{0.006} \] Solving for \( C_1 \) gives:\[ C_1 = \frac{3}{e^{0.006}} \] Calculate \( C_1 \) numerically if needed.

Step 6: Write the Particular Solution

Substitute \( C_1 \) back into the general solution:\[ y(t) = \frac{3}{e^{0.006}} e^{-0.003t} \]

Key Concepts

First-order Differential Equations

First-order differential equations involve derivatives where the highest order of differentiation is one. These equations often represent a rate of change and are written in the form \( \frac{dy}{dt} = f(t, y) \). In our example, \( \frac{dy}{dt} = -0.003y \), the function on the right is simply a multiplication of \( y \) by a constant.

This equation is linear because the term \(-0.003y\) involves no powers or products of \( y \), other than the first power. Also, it is homogeneous since there is no additional term depending only on \( t \).

Understanding the classification is important because it suggests possible solution methods, and in this instance, methods like separation of variables can be effectively utilized to solve it.

Separation of Variables

Separation of Variables is a technique for solving differential equations where the variables can be rearranged, or "separated", into two single-variable expressions. This allows the equation to be integrated easily.

In the equation \( \frac{dy}{dt} = -0.003y \), separating variables means isolating the \( dy \) and the \( dt \) on opposite sides of the equation:

• \( \frac{dy}{y} = -0.003 \, dt \)

This rearrangement sets the stage for integration.

By integrating both sides separately—\( \int \frac{dy}{y} \) for the left and \( \int -0.003 \, dt \) for the right—we can determine a solution in a step-by-step, straightforward manner. After integration, we find the logarithmic expression on one side and a linear expression on the other:

• \( \ln |y| = -0.003t + C \)

The exponentiation step that follows transitions us back to functions without \( \, \ln\, \), yielding the solution form \( y = C_1 e^{-0.003t} \).

Initial Value Problems

Initial value problems introduce specific conditions that the solution of a differential equation must satisfy. In general, they provide one or more values of the unknown function at certain points.

For our equation \( y(t) = C_1 e^{-0.003t} \), the condition \( y(-2) = 3 \) specifies that the solution must pass through the point \((-2, 3)\).

To find the exact solution, we substitute this initial condition into the general solution:

• \( 3 = C_1 e^{0.006} \)

This equation allows us to solve for the constant \( C_1 \), giving us the particular solution that fits the initial condition:

• \( C_1 = \frac{3}{e^{0.006}} \)

Once \( C_1 \) is determined, it can be substituted back into the general form of the solution to give the specific curve that satisfies both the differential equation and the initial condition.