Question

\(y^{\prime}-y / t=-t \sin t\), with \(y(\pi)=-\pi\), on \([0,2 \pi]\).

Short answer

The solution is \( y = t \cos t \).

Step-by-step solution

Identify Equation Type

Notice that the given equation is a first-order linear differential equation of the form \( y' + P(t) y = Q(t) \). Here, \( P(t) = -1/t \) and \( Q(t) = -t \sin t \).

Determine Integrating Factor

The integrating factor \( \mu(t) \) is given by \( e^{\int P(t) \, dt} \). So, we compute \( \int -1/t \, dt = -\ln |t| \), and therefore, \( \mu(t) = e^{-\ln |t|} = \frac{1}{t} \).

Multiply Equation by Integrating Factor

Multiply the entire differential equation by the integrating factor \(\frac{1}{t}\) to obtain \( \frac{1}{t}y' - \frac{y}{t^2} = -\sin t \).

Rewrite as Total Derivative

Recognize that the left side of the equation can be expressed as a derivative: \( \frac{d}{dt}\left(\frac{y}{t}\right) = -\sin t \).

Integrate Both Sides

Integrate both sides with respect to \( t \): \( \int \frac{d}{dt}\left(\frac{y}{t}\right) \, dt = \int -\sin t \, dt \). This gives \( \frac{y}{t} = \cos t + C \), where \( C \) is the constant of integration.

Solve for y(t)

Solve the equation \( \frac{y}{t} = \cos t + C \) for \( y \), giving \( y = t(\cos t + C) \).

Apply Initial Condition

Use the initial condition \( y(\pi) = -\pi \) to find \( C \). Substitute \( t = \pi \) and \( y = -\pi \) into \( y = t(\cos t + C) \): \(-\pi = \pi (\cos \pi + C) \). Knowing \( \cos \pi = -1 \), we have \(-\pi = \pi(-1 + C) \). Solve for \( C \): \(-\pi = -\pi + \pi C \), so \( \pi C = 0 \) which implies \( C = 0 \).

Write Final Solution

Substituting \( C = 0 \) into \( y = t(\cos t + C) \), the solution is \( y = t \cos t \).

Key concepts

First-Order Linear Differential Equation

Differential equations describe various phenomena such as physics, biology, and finance. A significant category of these is the first-order linear differential equation. This type of equation appears in the form \( y' + P(t) y = Q(t) \).
In the given example, we identify the equation \( y' - \frac{y}{t} = -t \sin t \) as a first-order linear differential equation with \( P(t) = -\frac{1}{t} \) and \( Q(t) = -t \sin t \).
Recognizing the terms \( P(t) \) and \( Q(t) \) is crucial for applying the method of integrating factors to find solutions.

Integrating Factor

To solve a first-order linear differential equation, we often utilize an integrating factor. The purpose of this factor is to transform the differential equation into a form that is easier to integrate.
The integrating factor is denoted as \( \mu(t) \), given by the formula \( e^{\int P(t) \, dt} \). In our example, we compute \( \int -\frac{1}{t} \, dt = -\ln |t| \). This results in the integrating factor \( \mu(t) = e^{-\ln |t|} = \frac{1}{t} \).
Multiplying the differential equation by this integrating factor helps us to reformulate the equation into a total derivative, facilitating straightforward integration.

Initial Condition

An initial condition provides specific data to uniquely determine the solution of a differential equation from a family of possible solutions.
In our case, we have the initial condition \( y(\pi) = -\pi \). This means that when \( t = \pi \), the value of \( y(t) \) is \(-\pi\).
Initial conditions are essential because they allow us to compute the constant of integration, completing our solution. Here, we found that substituting \( t = \pi \) and \( y = -\pi \) in the solution equation provides us with \( C = 0 \).
Thus, applying the initial condition gives the particular solution applicable to the specific scenario described in the problem.

Solution to Differential Equation

Finding the solution to a differential equation involves several steps: identifying the type of equation, using an integrating factor, and applying initial conditions.
The original problem yields the differential equation \( \frac{y}{t} = \cos t + C \). Solving for \( y \) gives \( y = t(\cos t + C) \).
By using the initial condition \( y(\pi) = -\pi \), we calculated the constant \( C \) as 0.
This allows us to determine the final solution, which is \( y = t \cos t \). This solution represents the full behavior of \( y \) in the interval \([0, 2\pi]\) as described by the problem, providing an equation that respects all given conditions.