Problem 27 - Calculus and its applications

Question

(a) find the particular solution of each differential equation as determined by the initial condition, and (b) check the solution by substituting into the differential equation. \(\frac{d B}{d t}=0.03 B,\) where \(B(0)=500\)

Step-by-Step Solution

Answer

The particular solution is \( B(t) = 500 e^{0.03t} \). Verification shows that this solution satisfies the differential equation.

Step 1: Identify the Type of Differential Equation

The given differential equation is \( \frac{d B}{d t} = 0.03 B \), which is a first-order linear differential equation with a constant coefficient. It can also be recognized as a separable differential equation.

Step 2: Separate Variables

Since the equation is separable, we can rearrange terms to separate variables: \( \frac{d B}{B} = 0.03 \, dt \).

Step 3: Integrate Both Sides

Integrate both sides of the separated equation. The integral of \( \frac{1}{B} \) is \( \ln |B| \), and the integral of \( 0.03 \) is \( 0.03t\):\[\int \frac{1}{B} \, dB = \int 0.03 \, dt \]\[\ln |B| = 0.03t + C \]

Step 4: Solve for B

Exponentiate both sides to solve for \( B \):\[B = e^{0.03t + C} = e^C e^{0.03t} \]We can rewrite \( e^C \) as a constant \( C_1 \), so\[B = C_1 e^{0.03t} \]

Step 5: Apply Initial Condition

Use the initial condition \( B(0) = 500 \) to find \( C_1 \):\[500 = C_1 e^{0.03 \times 0} = C_1 \]Thus, \( C_1 = 500 \). The particular solution is\[B(t) = 500 e^{0.03t} \]

Step 6: Verify the Solution

Substitute the particular solution back into the original differential equation to verify:\[\frac{d}{dt}(500 e^{0.03t}) = 500 \times 0.03 e^{0.03t} = 0.03 (500 e^{0.03t}) = 0.03 B \]The left side equals the right side of the given differential equation, confirming it is correct.

Key Concepts

Particular Solution

A particular solution to a differential equation is one specific solution that satisfies both the general form of the differential equation and any given initial conditions. Consider it as the exact path that fits perfectly through a specified point in the solution space.
When dealing with the problem of finding a particular solution, we first find the general solution of the differential equation. This typically involves steps like separation of variables or integrating the equation, depending on its form.
Once the general solution is found, you apply the initial conditions to identify the specific constants involved in that general solution. For example, in the problem where\[\frac{d B}{d t} = 0.03 B\]and the initial condition is \( B(0) = 500 \), our general solution after solving would be \( B(t) = C_1 e^{0.03t} \).
By applying the initial condition \( B(0) = 500 \), we substitute \( t = 0 \) and \( B = 500 \) into our general solution to find \( C_1 \). This gives the specific solution that perfectly captures the behavior of the system as described by the initial condition. Thus, the particular solution becomes \( B(t) = 500 e^{0.03t} \).
This particular solution gives us a concrete equation to describe the system or scenario represented by our differential equation, tailored to start at the initial condition's specific point.

Initial Condition

Initial conditions in differential equations are specific values assigned to the function and its derivatives at a particular point. They're essential in transforming a general solution into a particular one.
An initial condition provides the necessary information to pinpoint a specific solution among the many that satisfy the differential equation. \

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• They usually appear in the form \( y(0) = y_0 \) or \( y'(0) = y_1 \).
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• These conditions help us determine the unknown constants in a general solution.
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• They guide us in modeling real-world scenarios by narrowing down possible paths to the one that corresponds to our starting point.
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For instance, in our example, we had the initial condition \( B(0) = 500 \). This initial condition meant that at time \( t = 0 \), the value of \( B \) was 500. By substituting this into the general solution, we could determine that the constant \( C_1 = 500 \), leading us to the specific equation \( B(t) = 500 e^{0.03t} \).
Initial conditions are the bridge between a general description of how a system behaves and the specific scenario where a unique solution is necessary. They ensure solutions are not only mathematically valid but also physically or conceptually meaningful.

Separable Differential Equation

A separable differential equation is a type of differential equation where variables can be separated onto different sides of the equation. This makes them easier to manage and solve through integration.
For an expression of the form \( \frac{dB}{dt} = g(t)h(B) \), we can rearrange terms to isolate functions of \( B \) on one side and functions of \( t \) on the other. In our example, \( \frac{dB}{dt} = 0.03B \), we re-arrange it to \( \frac{dB}{B} = 0.03 \, dt \).
Here are the steps usually followed in solving a separable differential equation:\

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• Separate the variables such that each side of the equation contains only one variable along with its differential.
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• Integrate both sides. The left side with respect to \( B \) and the right side with respect to \( t \).
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• Don't forget to include the constant of integration, \( C \), on one side.
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• Solve the resulting equation for the dependent variable, if possible.
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• Apply any initial conditions to find the constant of integration and identify the particular solution.
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Using this method allows us to break down and solve differential equations step-by-step, helping to simplify potentially complex relationships between variables. It's a powerful technique, especially useful when dealing with equations modeling exponential growth, decay, and other natural phenomena.