Question

Solve the following initial-value problems starting from \(y(t=0)=1\) and \(y(t=0)=-1 .\) Draw both solutions on the same graph\(\frac{d y}{d t}=2\)

Short answer

The solutions are \(y = 2t + 1\) and \(y = 2t - 1\). Plot these lines on the same graph.

Step-by-step solution

Identify the given differential equation

The problem provides a differential equation:\[ \frac{dy}{dt} = 2 \].This is a simple first-order differential equation.

Integrate the differential equation

To find the solution, integrate both sides of the equation with respect to time \(t\):\[ y = \int 2 \, dt = 2t + C \],where \(C\) is the constant of integration.

Apply the initial condition \(y(0) = 1\)

For the initial condition \(y(0) = 1\), substitute \(t = 0\) and \(y = 1\) into the equation:\[ 1 = 2(0) + C \].Solving for \(C\) gives \(C = 1\).Thus, the first solution is:\[ y = 2t + 1 \].

Apply the initial condition \(y(0) = -1\)

For the initial condition \(y(0) = -1\), substitute \(t = 0\) and \(y = -1\) into the equation:\[ -1 = 2(0) + C \].Solving for \(C\) gives \(C = -1\).Thus, the second solution is:\[ y = 2t - 1 \].

Draw the solutions on the same graph

Now we have the two solutions:1. \( y = 2t + 1 \), and2. \( y = 2t - 1 \).To draw them, plot both lines on the same set of axes with \(t\) on the horizontal axis and \(y\) on the vertical axis.- For \(y = 2t + 1\), the line passes through \((0, 1)\) with a slope of 2.- For \(y = 2t - 1\), the line passes through \((0, -1)\) with a slope of 2. Both lines are parallel.

Key concepts

differential equation

A differential equation is a mathematical equation that relates some function with its derivatives. In simpler terms, it describes how a certain quantity changes as other quantities change. This can be crucial for modeling real-world phenomenona. In our given problem, the differential equation is \[ \frac{dy}{dt} = 2 \], a very basic form, which suggests that the rate of change of the function \(y\) over time \(t\) is constantly \(2\).
This indicates a steady rate at which \(y\) increases or decreases, depending on the context. Differential equations come in many forms and complexities, but fortunately, this first-order linear type is among the simplest. Understanding the type of differential equation you're working with can greatly aid in finding solutions.

integration

Integration is a fundamental concept in calculus used to find the area under curves, solve differential equations, and much more. When we talk about solving a differential equation like \[ \frac{dy}{dt} = 2 \],integration is used to determine the function \(y\) from its derivative. Think of integration as the reverse process of differentiation.
In the problem, integrating the constant \(2\) with respect to \(t\) gives us:\[ y = \int 2 \, dt = 2t + C \],where \(C\) is the essential constant of integration. This process gives us the general solution — a family of functions that satisfy the differential equation. To determine a specific solution, known as a particular solution, initial conditions are applied to find the value of \(C\).

constant of integration

The constant of integration, often denoted by \(C\), arises when we perform integration. Since differentiation of a constant term is zero, any constant could have been part of the function prior to differentiation. Therefore, when integrating, we must include \(C\) to account for this unknown constant.
In the initial-value problem, once we integrate to find \[ y = 2t + C \], applying initial conditions helps us find the exact value of \(C\) for the situation given.

• With \(y(0) = 1\), substituting into the equation gives \(C = 1\), so \(y = 2t + 1\).
• With \(y(0) = -1\), substituting gives \(C = -1\), so \(y = 2t - 1\).

This illustrates how the constant of integration is pivotal in finding specific solutions to differential equations with given initial conditions.