a. Use a CAS to plot the slope field of the differential equation y^'=(3 x^2+4 x+2)/(2(y-1)) ove

A. Use a CAS to plot the slope field of the differential...

Key Concepts

Computer Algebra Systems (CAS)

A Computer Algebra System is a software tool that is capable of symbolically performing mathematical tasks such as integration, differentiation, equation solving, and graphing. In the context of differential equations, a CAS facilitates operations like plotting slope fields, integrating separated variables, and graphically representing both implicit general solutions and specific solution curves determined by initial conditions or arbitrary constants.

Initial Value Problems

An initial value problem (IVP) involves solving a differential equation under a given condition, typically at a specific point (the initial condition), which allows for the determination of a unique particular solution from a family of general solutions. The process involves first obtaining the general solution and then substituting the initial condition to find the specific constant that satisfies the initial setup.

Separable Differential Equations

A separable differential equation is one in which the variables can be algebraically separated on opposite sides of the equation, so that one side involves only the independent variable and the other side involves only the dependent variable. Once the variables are separated, both sides can be integrated with respect to their own variables, leading to a general solution. This is a standard technique for solving many first-order differential equations.

Slope Fields

A slope field is a graphical representation of a first-order differential equation that provides a visual insight into the behavior of solutions without having to solve the equation explicitly. It consists of small line segments or arrows drawn at a grid of points, with each segment’s slope corresponding to the value of the differential equation at that point. This method helps in understanding the general trend and tendencies of solution curves in the plane.

Implicit Solutions

An implicit solution to a differential equation is one in which the dependent variable is not isolated on one side of the equation, but rather, both the independent and dependent variables are intermingled in an equation. This form often naturally results from the process of integrating separable equations and can capture all valid solutions before applying initial conditions or further algebraic manipulation to solve explicitly.

Transcript

00:01 Okay, what we're going to do today is we're going to step through kind of a multi -step process, and we're going to use a cas system to help us plot slope fields and integrate and do also an implicit function graphing system as well.

00:26 And so there's multiple, the best known cast system is going to be that ti inspire, or there's also some online cast systems that will actually also work.

00:42 I may kind of be working in two different online ones, one for the plotting or graphing aspect and the other for the integration aspect.

00:53 So what we're going to do is we are going to start with a differential equation.

00:58 Y prime is equal to 3x squared plus 4x plus 2, all divided by 2 times y minus 1.

01:12 Okay.

01:14 And so the first thing we want to do is we want to use a cas system to plot the slope field.

01:27 And so i'm actually going to be using slope -filled generator through desmos.

01:36 And so this is desmos, and it's got a slope -filled generator.

01:43 So if you just google slope -field generator desmos, it will pop up.

01:46 I've actually already have it in here.

01:49 We want to kind of be in from negative 3.

01:57 To positive 3 and negative 3 to positive 3 as well.

02:05 So we can actually kind of, oops, went too far.

02:08 So it's kind of be in this region right here.

02:12 I probably could even do my graph settings.

02:17 Kind of like to have the bigger version though.

02:20 And so now what we want to do is so here's our slope field.

02:25 And so now what we want to do is, we want to go ahead and separate out our variables and then integrate.

02:37 So what we're going to do here is i'm going to multiply both sides by 2 times y minus 1.

02:46 And of course, you know, d .y prime is equal to d .y over dx.

02:52 And so i'm going to keep the d .y on the left side, and that will be equal to 3x squared plus 4x plus 2, and then that dx.

03:08 And then what we're going to do is we are going to integrate both sides.

03:13 And we're going to go ahead and use a cas integrator.

03:18 The integrator system i have is this integrator.

03:26 It's actually an integral calculator.

03:29 Also does derivatives through symbilabs.

03:34 And so basically what we're going to do is i already kind of got it set up in here.

03:38 Is the integral and i've done the right hand side.

03:41 I can't do both sides at the same time.

03:44 The ti inspire might be able to.

03:46 I haven't really worked with that much.

03:49 And so when i integrate that right -hand side with respect to x, i get x cubed plus 2x squared plus 2x plus some constant.

03:58 So this is my general solution.

04:01 And so if i go back here, then when i integrate, i might go ahead and raise that.

04:12 I get x cubed plus 2x squared, plus 2x plus some constant number.

04:23 And so now what i'm going to do is go ahead and go back to my integration system.

04:29 And now this time i'm going to integrate the 2y minus 1.

04:38 Whoops, y minus 1.

04:44 And we're integrating with respect to y.

04:47 I hit go.

04:49 And then it will come back and i think it's going to be y squared.

04:54 Minus 2y plus some constant number as well.

04:57 So i'm going to come back here and this is going to be y squared minus 2y plus some constant number but i can go ahead and lump all of that constant number into one on the right.

05:12 So we have y squared minus 2y equal to x q plus 2x squared plus 2x plus some constant number that includes the one that was on the left and the original one that was on the right.

05:24 Okay.

05:25 And now what we want to do is kind of a multi -step thing.

05:31 So now the next thing we want to do is we want to go ahead and use an implicit function grapher to graph.

05:42 Y squared minus 2y is equal to x -quid plus 2x squared plus 2x squared plus 2x minus c.

05:52 So the first time my constant is going to be negative c.

05:56 And then we're going to do it again for the constant value equal to a negative 4.

06:07 And then we're going to do it multiple times.

06:10 And then we're going to do it for where it's equal to a minus 2.

06:18 And then i'm going to quit writing all these down.

06:21 And then we're going to do a 0, 2, 4, and a 6...