Question

Solve the following initial-value problems with the initial condition \(y_{0}=0\) and graph the solution.\(\frac{d y}{d t}=-y-1\)

Short answer

The solution is \( y(t) = -1 + e^{-t} \). Graph it as an exponentially decaying curve approaching \( y = -1 \).

Step-by-step solution

Identify the Type of Differential Equation

The given differential equation is \( \frac{dy}{dt} = -y - 1 \). This is a first-order linear differential equation.

Separate Variables

To separate variables, rewrite the equation as \( \frac{dy}{dt} + y = -1 \). The equation is ready for an integrating factor.

Find the Integrating Factor

The integrating factor \( \mu(t) \) is calculated as \( e^{\int 1 \, dt} = e^t \). Multiply the entire differential equation by the integrating factor to facilitate integration: \( e^t \frac{dy}{dt} + e^t y = -e^t \).

Rewrite the Equation Using the Integrating Factor

The equation \( e^t \frac{dy}{dt} + e^t y = -e^t \) can be rewritten as \( \frac{d}{dt}(e^t y) = -e^t \). This simplifies the integration process.

Integrate Both Sides

Integrate both sides of the equation: \( \int \frac{d}{dt}(e^t y) \, dt = \int -e^t \, dt \). This yields \( e^t y = -e^t + C \), where \( C \) is a constant of integration.

Solve for \( y(t) \)

Divide through by \( e^t \) to solve for \( y(t) \): \( y = -1 + Ce^{-t} \).

Apply Initial Condition

Use the initial condition \( y(0) = 0 \) to solve for \( C \): \( 0 = -1 + C \cdot e^0 \). Simplifying gives \( C = 1 \).

Write the Particular Solution

Substitute \( C = 1 \) back into the general solution to get the particular solution: \( y = -1 + e^{-t} \).

Graph the Solution

The function \( y(t) = -1 + e^{-t} \) represents an exponentially decaying all-positive curve, approaching \( y = -1 \) as \( t \rightarrow \infty \). The graph is asymptotic to the line \( y = -1 \), starting at \( y(0) = 0 \).

Key concepts

First-order Linear Differential Equation

First-order linear differential equations are a specific type of differential equations characterized by having the first derivative of a function with no higher derivatives involved. The general form of a first-order linear differential equation can be written as: \(\frac{dy}{dt} + P(t) y = Q(t)\)where:

• \(y\) is the dependent variable we want to determine.
• \(t\) is the independent variable.
• \(P(t)\) and \(Q(t)\) are given functions of \(t\).

In the case of the exercise given, the equation \(\frac{dy}{dt} = -y - 1\) is a first-order linear differential equation because it can be rearranged to match this standard form, becoming \(\frac{dy}{dt} + y = -1\). This rearrangement helps identify the coefficients that help in solving the equation by known methods like finding the integrating factor.

Integrating Factor Method

The integrating factor method is a technique used to solve linear differential equations. It is particularly useful when you cannot easily separate variables. The steps for applying this method are systematic and straightforward.First, reformulate the differential equation in standard linear form \(\frac{dy}{dt} + P(t) y = Q(t)\). Here, \(P(t)\) is found directly from the reformulated equation as the coefficient of \(y\).The integrating factor \(\mu(t)\), is determined by:\[\mu(t) = e^{\int P(t) \, dt}\]This factor is then multiplied through the entire equation to rewrite it as a perfect derivative:\[\frac{d}{dt}(\mu(t) y) = \mu(t) Q(t)\]This step simplifies the equation, and allows for straightforward integration on both sides. The integrating step yields a general solution from which a particular solution can be determined once we apply any given initial conditions.

Separation of Variables

Separation of variables is another method often used for solving differential equations, although it’s not applicable directly in every case, such as with non-separable equations like our current focus. However, it remains an important technique in scenarios where it is applicable.The basic idea is to manipulate the equation such that all terms involving the dependent variable \(y\) are on one side, and all terms involving the independent variable \(t\) are on the other.This often involves rearranging and integrating both sides separately. For example, a separable differential equation would resemble: \(N(y) \, dy = M(t) \, dt\)From here, both sides can be integrated uniquely:\[\int N(y) \, dy = \int M(t) \, dt\]Once integrated, you solve for \(y\) to find the general solution, which can then be adjusted to a particular solution with initial conditions, much like in other methods. This technique shines in its simplicity when applicable, offering a direct path to the solution.