Problem 1 - Applied Calculus

Question

Find solutions to the differential equations, subject to the given initial condition. $$ \frac{d w}{d r}=3 w, \quad w=30 \text { when } r=0 $$

Step-by-Step Solution

Answer

The solution is \( w = 30 e^{3r} \).

Step 1: Identify the Type of Differential Equation

The given differential equation is \( \frac{d w}{d r} = 3w \). This is a first-order linear differential equation with a separable variable form.

Step 2: Separate Variables

Separate the variables by rewriting the equation such that all \(w\) terms are on one side and all \(r\) terms are on the other side. This gives us \( \frac{1}{w} dw = 3 dr \).

Step 3: Integrate Both Sides

Integrate both sides of the equation: \( \int \frac{1}{w} dw = \int 3 dr \). The left side integrates to \( \ln|w| \) and the right side integrates to \( 3r + C \), where \( C \) is the constant of integration.

Step 4: Solve for w

Exponentiate both sides to solve for \( w \): \( |w| = e^{3r + C} \). Rewriting, we have \( w = Ce^{3r} \), where \( C = e^C \) is a constant.

Step 5: Apply Initial Condition

Substitute the initial condition \( w = 30 \) when \( r = 0 \) into the equation \( w = Ce^{3r} \). This gives \( 30 = Ce^{0} \), therefore \( C = 30 \).

Step 6: Write the Solution

The particular solution to the differential equation with the initial condition is \( w = 30 e^{3r} \).

Key Concepts

First-Order Linear Differential Equation

First-order linear differential equations are a type of differential equation commonly encountered in various fields of science and engineering. They are called "first-order" because they involve the first derivative of the unknown function. In general, these equations take the form:

• \( \frac{dy}{dx} + P(x)y = Q(x) \)

For the equation given in this exercise, \( \frac{d w}{d r} = 3w \), the function \( w \) depends on \( r \), which plays the role of the independent variable.
These equations can often be solved by methods that simplify their properties, including converting them into separable forms.

Separable Variables

The method of separable variables is a powerful tool for solving differential equations. It works by rearranging the equation so that all terms containing the dependent variable are on one side, and all terms involving the independent variable are on the other side.
By doing this, you simplify the problem into two separate integrals. For example, in the equation \( \frac{dw}{dr} = 3w \), we separate variables to obtain \( \frac{1}{w} dw = 3 dr \).
This allows us to independently integrate both sides and solve for the unknown function. This method is particularly helpful for solving first-order linear differential equations like the one in this exercise.

Initial Condition

An initial condition is a piece of information that allows us to find a specific solution to a differential equation. It tells us the value of the function at a particular point.
In this exercise, the initial condition given is \( w = 30 \) when \( r = 0 \).
This information is essential to determine the constant of integration when solving the separated and integrated equation, ensuring that the solution is not just general but applicable to the specific scenario depicted in the problem.

Integration

Integration is a fundamental mathematical operation used to solve differential equations after the variables have been separated. When you integrate both sides of an equation like \( \frac{1}{w} dw = 3 dr \), you essentially find the antiderivative of each side.

• The left side integrates to \( \ln|w| \).
• The right side results in \( 3r + C \), where \( C \) is the constant of integration.

Integration brings together all the pieces, allowing us to rebuild the original function \( w(r) \). Solving the integral equation gives us the general form of the solution.

Exponential Function

The exponential function is key when solving differential equations of the form presented in this exercise. Once integration is performed, the result \( \ln|w| = 3r + C \) can be solved as an exponential function.

• By rewriting, it becomes: \( |w| = e^{3r + C} \).
• This further simplifies to \( w = Ce^{3r} \), where \( C = e^C \).

The exponential function characterizes solutions to such equations by describing growth or decay processes. Applying initial conditions often helps to determine the exact value of the constant, leading to a particular solution like \( w = 30 e^{3r} \).