Problem 22 - Thomas Calculus

Question

Solve the following initial value problem for \(u\) as a function of \(t :\) \(\frac{d u}{d t}+\frac{k}{m} u=0 \quad\left(k \text { and } m \text { positive constants), } u(0)=u_{0}\right.\) a. as a first-order linear equation. b. as a separable equation.

Step-by-Step Solution

Answer

The solution is \( u(t) = u_0 e^{-\frac{k}{m} t} \).

Step 1: Identify the Type of Differential Equation

The given differential equation \( \frac{d u}{d t} + \frac{k}{m} u = 0 \) is a first-order linear ordinary differential equation with the initial condition \( u(0) = u_0 \).

Step 2: Solve as a First-Order Linear Equation

To solve this equation as a first-order linear equation, we use the integrating factor method. The standard form is \( \frac{d u}{d t} + P(t) u = Q(t) \), where \( P(t) = \frac{k}{m} \) and \( Q(t) = 0 \). The integrating factor \( \mu(t) \) is given by \( e^{\int P(t) \, dt} = e^{\frac{k}{m} t} \). Multiplying the equation by the integrating factor yields \( e^{\frac{k}{m} t} \frac{du}{dt} + e^{\frac{k}{m} t} \frac{k}{m} u = 0 \).

Step 3: Simplify and Integrate

The left-hand side becomes the derivative of \( e^{\frac{k}{m} t} u \), so we have \( \frac{d}{dt} \left(e^{\frac{k}{m} t} u \right) = 0 \). Integrating both sides with respect to \( t \) gives \( e^{\frac{k}{m} t} u = C \), where \( C \) is a constant. Solving for \( u \), we get \( u(t) = Ce^{-\frac{k}{m} t} \).

Step 4: Apply the Initial Condition

Using the initial condition \( u(0) = u_0 \), substitute \( t = 0 \) into \( u(t) = Ce^{-\frac{k}{m} t} \) to find \( C \): \( u_0 = Ce^0 = C \). Thus, \( C = u_0 \) and \( u(t) = u_0 e^{-\frac{k}{m} t} \).

Step 5: Solve as a Separable Equation

Rewrite the original equation \( \frac{d u}{d t} = -\frac{k}{m} u \). Separating variables, we have \( \frac{1}{u} du = -\frac{k}{m} dt \). Integrating both sides gives \( \ln |u| = -\frac{k}{m} t + C_1 \).

Step 6: Solve for \( u(t) \)

Exponentiate both sides to solve for \( u(t) \): \( |u| = e^{C_1} e^{-\frac{k}{m} t} \). Let \( C_2 = e^{C_1} \), then \( u = C_2 e^{-\frac{k}{m} t} \). Applying the initial condition \( u(0) = u_0 \), we find \( C_2 = u_0 \) leading to \( u(t) = u_0 e^{-\frac{k}{m} t} \).

Key Concepts

Initial Value Problem

An initial value problem (IVP) in differential equations involves finding a function that solves a given differential equation while satisfying an initial condition.

For instance, we are given the differential equation \( \frac{d u}{d t} + \frac{k}{m} u = 0 \) with the initial condition \( u(0) = u_0 \). This means that at time \( t = 0 \), the function \( u(t) \) should equal \( u_0 \).

Solving an initial value problem is important because it ensures the solution not only satisfies the differential equation but also starts from a specified situation. This reflects many real-world scenarios where you know the system's state at a particular starting point.

First-Order Linear Equation

A first-order linear differential equation is of the form \( \frac{dy}{dt} + P(t)y = Q(t) \). Linear refers to how the equation involves the function \( y = y(t) \) and its first derivative. Here, the term "first-order" implies that the highest derivative present is the first derivative.

In the exercise, our equation is \( \frac{d u}{d t} + \frac{k}{m} u = 0 \). This fits the first-order linear form with \( P(t) = \frac{k}{m} \) and \( Q(t) = 0 \). The lack of a separate \( Q(t) \) term simplifies the equation, making it homogeneous.

Solving such equations involves a technique called the integrating factor method, which helps simplify and solve linear ordinary differential equations efficiently.

Separable Equation

A separable equation is a type of differential equation in which variables can be separated on different sides of the equation.

In our case, the equation \( \frac{d u}{d t} + \frac{k}{m} u = 0 \) can be rearranged as \( \frac{d u}{d t} = -\frac{k}{m} u \). This allows us to separate variables, placing all \( u \) terms on one side and \( t \) terms on the other.

Formally, you rewrite it as \( \frac{1}{u} \, du = -\frac{k}{m} \, dt \). This creates two integrals that can be solved individually: \( \int \frac{1}{u} \, du \) and \( \int -\frac{k}{m} \, dt \). After integrating, we solve for \( u \) by exponentiating the result. This method is effective because it directly leverages the equation's structure to simplify solving it.

Integrating Factor Method

The integrating factor method is a strategy to solve first-order linear differential equations. This technique involves multiplying the given differential equation by a special function, called the integrating factor, which greatly simplifies it.

For the equation \( \frac{d u}{d t} + \frac{k}{m} u = 0 \), we find the integrating factor to be \( e^{\int P(t) \, dt} = e^{\frac{k}{m} t} \).

Using this integrating factor converts the left-hand side of our original equation into the derivative of a product \( \frac{d}{dt} \left(e^{\frac{k}{m} t} u \right) = 0 \).

Integrating both sides results in a simple algebraic equation for \( u \) in terms of \( t \), which, after applying the initial condition, gives the solution \( u(t) = u_0 e^{-\frac{k}{m} t} \). This method simplifies complex-looking equations and is especially powerful for linear differential equations with constant coefficients.