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\(\frac{d y}{d t}=-t\)

Short answer

There are two solutions: \( y(t) = -\frac{t^2}{2} + 1 \) and \( y(t) = -\frac{t^2}{2} - 1 \).

Step-by-step solution

Understand the Differential Equation

The given differential equation is \( \frac{dy}{dt} = -t \). It represents the rate of change of \( y \) with respect to \( t \) as \( -t \). This is a first-order linear ordinary differential equation.

Integrate to Find the General Solution

To find the general solution, integrate the equation \( \frac{dy}{dt} = -t \) with respect to \( t \). This gives \( y = \int -t \ dt = -\frac{t^2}{2} + C \), where \( C \) is the constant of integration.

Apply Initial Condition \(y(0) = 1\)

Using the initial condition \( y(0) = 1 \), substitute \( t = 0 \) and \( y = 1 \) into the general solution: \( 1 = -\frac{0^2}{2} + C \) which simplifies to \( C = 1 \). Thus, the specific solution for this condition is \( y(t) = -\frac{t^2}{2} + 1 \).

Apply Initial Condition \(y(0) = -1\)

Using the initial condition \( y(0) = -1 \), substitute \( t = 0 \) and \( y = -1 \) into the general solution: \( -1 = -\frac{0^2}{2} + C \) which simplifies to \( C = -1 \). Thus, the specific solution for this condition is \( y(t) = -\frac{t^2}{2} - 1 \).

Graph Both Solutions

To graph the solutions, plot \( y(t) = -\frac{t^2}{2} + 1 \) and \( y(t) = -\frac{t^2}{2} - 1 \) on the same set of axes. Both curves are parabolas opening downward, with vertices at \( t = 0 \). The first curve intersects the y-axis at \( y = 1 \) and the second at \( y = -1 \).

Key concepts

Differential Equations

Differential equations involve equations with derivatives, showing how a function changes. They are crucial in modeling real-world phenomena like motion, growth, and decay. In this exercise, we have the differential equation \( \frac{dy}{dt} = -t \). This means that the rate of change of \( y \) concerning time \( t \) is \( -t \). Here, every change in \( t \) results in a corresponding change in \( y \). Differential equations can be simple, like this one, or very complex, involving multiple variables and higher-order derivatives.
Understanding and solving differential equations help in predicting how systems change over time, crucial in fields like physics, engineering, and economics.

First-Order Linear ODE

A first-order linear ordinary differential equation (ODE) is characterized by having the derivative of the response variable of interest as the highest derivative. The given equation \( \frac{dy}{dt} = -t \) is first-order because it involves only the first derivative of \( y \). Linear refers to the fact that \( y \) and its derivative appear linearly, not raised to any power other than one. These types of equations are relatively straightforward to solve and often appear in modeling scenarios where the rate of change is directly proportional to time or space.
They serve as a vital learning foundation before moving on to more complex ODEs or even systems of ODEs, where multiple interrelated equations work together.

Integration

Integration is the process of finding a function when its derivative is known. To solve \( \frac{dy}{dt} = -t \), we use integration: \( y = \int -t \, dt = -\frac{t^2}{2} + C \). Here, \( C \) stands for the constant of integration, reflecting the indefinite nature of integrating without limits. Applying initial conditions, like given starting values for \( y(0) \), lets us find the particular solutions by determining \( C \). Integration transforms our differential equation into a practical solution that describes \( y \) at any point in time \( t \).
This step is foundational in calculus, often used to reverse differentiation and solve equations describing change.

Graphing Solutions

Graphing the solutions of differential equations provides a visual understanding. Our solutions, \( y(t) = -\frac{t^2}{2} + 1 \) and \( y(t) = -\frac{t^2}{2} - 1 \), are graphed as parabolas. These parabolas open downwards due to the negative sign with \( t^2 \). The initial conditions \( y(0) = 1 \) and \( y(0) = -1 \) determine their starting points on the y-axis. Each represents a unique solution to the original ODE based on initial conditions.
Graphing allows us to see how changing an initial condition shifts the entire curve, illustrating the dynamic nature of solutions to differential equations and how initial setups affect overall behaviors.