Solution of Ex:2.2 Differential Equations with Boundary-Value Problems (7th Edition Book)

Solution of Ex:2.2 Differential Equations. In exercise 2.2 on page 50 of the seventh edition of “Differential Equations with Boundary-Value Problems” by Dennis G. Zill, several differential equations are solved using separation of variables. Let’s take a look at a few of these solutions.

Solution of Ex:2.2 Differential Equations with Boundary-Value Problems (Book)

Differential equations are an important topic in mathematics, particularly in applied mathematics, engineering, and physics. They describe the behavior of a system over time, and can help us predict how the system will evolve given certain initial conditions. One common method for solving differential equations is separation of variables. Solution of Ex:2.2 Differential Equations

The first problem involves finding the solution to the differential equation y’ = ky, where k is a constant. We can start by separating the variables: dy/y = k dt. Integrating both sides, we get ln|y| = kt + C, where C is the constant of integration. Exponentiating both sides, we get |y| = e^(kt+C), which simplifies to y = Ae^(kt), where A is a constant.

Another problem involves finding the solution to the differential equation y’ + 2ty = 0. Again, we can separate the variables: dy/y = -2t dt. Integrating both sides, we get ln|y| = -t^2 + C. Exponentiating both sides, we get |y| = e^(-t^2+C), which simplifies to y = Ae^(-t^2), where A is a constant.

A third problem involves finding the solution to the differential equation y’ = 2y(1-y), subject to the initial condition y(0) = 1. We can separate the variables: dy/[y(1-y)] = 2 dt. We can use partial fractions to integrate the left side, which gives us ln|y/(1-y)| = 2t + C. Solving for y, we get y = e^(2t+C) / [1 + e^(2t+C)], where we can use the initial condition to solve for C and get the specific solution y = 1 / [1 + e^(-2t)].

Examples Overview of Differential Equations with Boundary-Value Problems

These examples demonstrate the power of separation of variables in solving differential equations. By separating the variables and integrating, we can often find simple solutions to complex problems. With further study and practice, anyone can become proficient at solving differential equations using separation of variables.

Leave a Comment