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 \( y = -1 + e^{t} \).

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 and can be solved using methods for linear differential equations.

Rearrange the Equation

Rewrite the given equation in standard linear form: \( \frac{dy}{dt} - y = 1 \). The standard form is \( \frac{dy}{dt} + P(t)y = Q(t) \). Here, \( P(t) = -1 \) and \( Q(t) = 1 \).

Find the Integrating Factor

The integrating factor \( \mu(t) \) for a linear differential equation is \( e^{\int P(t) \, dt} \). For this problem, \( \int P(t) \, dt = \int -1 \, dt = -t \). Therefore, the integrating factor \( \mu(t) = e^{-t} \).

Multiply Through by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( e^{-t} \): \( e^{-t} \frac{dy}{dt} - e^{-t}y = e^{-t} \).

Solve the Equation

The left side is now the derivative of \( e^{-t} y \). So, \( \frac{d}{dt} (e^{-t} y) = e^{-t} \). Integrate both sides with respect to \( t \):\[ e^{-t} y = \int e^{-t} \, dt = -e^{-t} + C \].

Solve for y

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

Apply Initial Condition

Use the initial condition \( y(0) = 0 \) to solve for \( C \):\[ y(0) = -1 + Ce^{0} = 0 \Rightarrow -1 + C = 0 \Rightarrow C = 1 \].

Final Solution

The solution to the differential equation with the given initial condition is \( y = -1 + e^{t} \).

Graph the Solution

To graph the solution \( y = -1 + e^{t} \), plot \( y \) against \( t \). The curve will be above the line \( y = 0 \) due to the exponential term \( e^{t} \) minus 1, starting from \( y(0) = 0 \). This indicates exponential growth.

Key concepts

First-Order Linear Differential Equations

A first-order linear differential equation is a type of differential equation that involves the function and its first derivative. These equations are characterized by having one variable with respect to which differentiation is performed. In general, a first-order linear differential equation can be expressed in the form:\[ \frac{dy}{dt} + P(t)y = Q(t) \]In this equation, \( P(t) \) and \( Q(t) \) are functions of \( t \). The main idea behind these equations is that the unknown function, here denoted as \( y(t) \), and its derivative show up linearly in the equation. Some key points about first-order linear differential equations are:

• The derivative \( \frac{dy}{dt} \) is always of first degree.
• There are no powers or products of the function \( y(t) \) and \( \frac{dy}{dt} \).
• These equations can usually be solved using standard techniques like the integrating factor method.

By understanding this type of equation, it becomes easier to identify and solve them in various mathematical problems.

Integrating Factor Method

The integrating factor method is a powerful technique used to solve first-order linear differential equations. It involves simplifying the differential equation to make it easier to solve. With this method, an integrating factor is calculated and used to rewrite the equation in a form that makes finding the solution straightforward.To solve an equation like \( \frac{dy}{dt} + P(t)y = Q(t) \) using the integrating factor method, follow these steps:1. Find the Integrating Factor: Calculate the integrating factor \( \mu(t) = e^{\int P(t) \, dt} \). This transforms the differential into a more manageable form.2. Multiply the Equation: Apply the integrating factor to every term in the differential equation. This step ensures that the left side becomes the derivative of the product of the integrating factor and \( y(t) \).3. Solve the Resulting Equation: Once the equation is transformed, you can integrate both sides to find the solution for \( y(t) \).This method works effectively because it leverages the properties of exponential functions to simplify the differential equation into a form that is easy to integrate.

Initial-Value Problem

An initial-value problem in the context of differential equations is when you're given a differential equation along with a specific value of the unknown function at a certain point. This given condition helps to determine the constant that arises when integrating.Solving an initial-value problem usually involves:

• Finding a general solution to the differential equation.
• Using the given initial condition to solve for the constant in the general solution.

As seen in our example, the initial condition \( y(0) = 0 \) was used to find the constant, making the solution specific to the problem. Without using the initial condition, you'd only find a family of solutions, not the specific one fitting the problem's requirements.