Find an explicit solution of the given initial-value problem. (d y)/(d t)+2 y=1, y(0)=(5)/(2)

Name: Problem 26, Chapter 2: A First Course in Differential Equations with Modeling Applications
Uploaded: 2021-07-26T17:24:52-08:00
Duration: 6 min 31 s
Channel: Joseph Calomo
Description: Find an explicit solution of the given initial-value problem. (d y)/(d t)+2 y=1, y(0)=(5)/(2)

Find an explicit solution of the given initial-value...

Key Concepts

First Order Linear Differential Equations

These are differential equations of the form dy/dt + p(t)y = q(t), where p(t) and q(t) are functions of t. They are a fundamental class of equations in differential equations, characterized by their linear dependence on the unknown function and its first derivative. The structure of these equations allows for systematic solution methods such as the integrating factor technique.

Integrating Factor Method

This is a standard method used to solve first order linear differential equations. The idea is to multiply the entire equation by a function—known as the integrating factor—that simplifies the left-hand side into the derivative of a product, allowing direct integration. The integrating factor is typically given by e^(?p(t) dt), making it a powerful tool for finding explicit solutions.

Initial Value Problem

An initial value problem (IVP) specifies the value of the unknown function at a certain point, which is used to determine any arbitrary constants that appear in the solution of the differential equation. The presence of an initial condition ensures that the solution is not only general, but uniquely determined by the given data.

Explicit Solution

This refers to expressing the solution of a differential equation in a form where the dependent variable is isolated as a function of the independent variable. An explicit solution clearly shows how the variables relate without requiring further manipulation, providing an immediate functional description of the solution’s behavior.

Transcript

00:01 So in this problem, we are given a differential equation that i've written on the left side right over here and an initial condition.

00:08 And we're asked to solve this using separation of variables.

00:13 So first step of this, we're going to try to get the ys and t's on either side of the equation.

00:19 That's the beginning.

00:20 And since there are no t's, we're just going to start by adding or starting subtracting this 2y from both sides.

00:27 So we're going to end up with dy over dt equals 1 minus.

00:34 2y.

00:35 Now, to get this d .y over to the other side, we're going to have to flip the right side of the equation.

00:42 So this is really going to end up looking something like this.

00:44 We're going to end up with a dt equals 1 over 1 minus 2y, dy, and you can kind of see how that happens, right? you know, if we were to, let's get this in red, if we were to multiply this dt over here, then we'd have to divide this to get it back over here, and then you end up with the equation to see right here.

01:13 I just skipped that step, you know, to make it easier.

01:16 From here, we're just gonna integrate.

01:17 We're gonna integrate both sides.

01:18 You can excuse my integration symbol.

01:20 They're basically big asses to me.

01:23 The integral of dt is simply t.

01:26 So we ended up with, that's really easy.

01:28 The other one we're gonna have to actually work out here.

01:31 We're gonna use u substitution for this.

01:33 So we're gonna say that, let's see what, let's move this around here for a second.

01:38 Let's just work with the left side.

01:40 I'm going to put the t down here.