Solve the initial-value problem for x as a function of t (2 t^3-2 t^2+t-1) (d x)/(d t)=3, x(2)=0

Name: Problem 235, Chapter 3: Calculus Volume 2
Uploaded: 2020-08-07T20:05:22-08:00
Duration: 14 min 24 s
Channel: Yaw Asomani
Description: Solve the initial-value problem for x as a function of t (2 t^3-2 t^2+t-1) (d x)/(d t)=3, x(2)=0

step 1 So we have this 2T -qub minus 2T squared plus t minus 1. This is a differential equation, and it

Solve the initial-value problem for x as a function of t (2...

Key Concepts

Initial Value Problem

An initial value problem involves a differential equation along with specified values of the solution at a particular point. This provides the means to determine the unique solution by incorporating the initial condition into the integration process once the general solution is found.

Ordinary Differential Equations

The problem belongs to the study of ordinary differential equations (ODEs), which are equations involving functions of a single variable and their derivatives. Understanding ODEs is crucial as they model many phenomena in science and engineering.

Separation of Variables

Separation of variables is a technique used to solve first order ODEs by rearranging the equation so that each variable and its differentials are on opposite sides of the equation. This allows the integration of both sides with respect to their own variable.

Integration Techniques

Integration is the process of finding the antiderivative or integral of a function. In the context of solving ODEs, integration is used after separating the variables, and may involve methods such as substitution or partial fractions.

Polynomial Factorization

Polynomial factorization is a process used to rewrite a polynomial as a product of simpler polynomials. In solving differential equations, factorization can simplify the integrals, especially when the denominator of a rational function can be factored to allow for partial fraction decomposition.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down a complex rational expression into simpler fractions that are easier to integrate. This technique is especially useful when the denominator can be factored into linear or quadratic factors.

Transcript

00:01 So we have this 2t -qub minus 2t squared plus t minus 1.

00:14 This is a differential equation, and it's a separable differential equation, right? it's just like a previous tutorial.

00:24 It is an initial condition.

00:26 We have to divide both sides by this quantity right here.

00:31 So you're going to have the x d t equals 3 over 2 t, right, cube minus 2 t squared plus t minus 1, right? so then you have to find the integral in order to find the st, right? so this is going to be, i'm just going to send the t to the other way, right? because it's a separable differential equation.

01:03 So i send dt here, and then all i have to do is find an integration.

01:08 So this is what i have to integrate in order to complete my integration, right? and you have to ask yourself that what kind of technique are you going to use? we're going to use a partial fraction technique, right? so is a denominator factorable? yes, it is, right? so when you factor it, this is going to be 2t squared.

01:30 And then the parenthesis is going to have t minus one and then uh t minus one again okay so so t minus one again okay so once you have this one um then this is going to be um you know t minus one is common between the two right so i can uh i can bring it out so when i bring it out, this is going to be t minus 1 out.

02:17 What is going to be left is 2t squared plus 1, right? so this is another representation for this, right? now we're going to use the partial fraction on this guy.

02:33 So when you have this, then the first one is going to be a taken on that.

02:39 And then since this is a quadratic factor, you know how to deal with quadratic.

02:44 Factors, right, bx plus c, and then we have this one.

02:50 Okay, so now that we have this, and all we have to do is find the lcd and then equate the numerators, right? so that is what we have here.

03:08 Okay.

03:09 Now we're going to do some elimination.

03:10 We're going to let, you know, let t be 1.

03:16 So when we let t equals 1, what is happening? then we're going to have 3 equals.

03:28 Now when you let t be 1, everything is going to be 0.

03:31 So this one is effectively eliminated.

03:33 When i let t be 1 here, 1 squared is 1 times 2 is 2, and then plus 1 is 3.

03:39 So this is going to be 3a.

03:40 So when it divided both sides by 3, you can see that a is 1.

03:46 So the next elimination we can do is to let t equals 0.

03:50 So when i let t equals 0 here, you can see that this b is going to go away.

03:55 So then i'm going to have three equals.

03:58 Now, when i put zero here, i'm going to have just one times a, which is a, right? and then i'm going to have, when i put zero here and i put zero here, this one goes away.

04:10 I just have c and negative 1.

04:12 So it's going to be negative c, right? so negative c.

04:15 So i have a minus c equals 3.

04:19 Now what is a a is one right so it's just going to be three equals one minus c right and when i subtract one from either side you can have two equals negative c right therefore what is c is negative right so you have c to be negative to a to be one unfortunately for us we cannot make any elimination again so we have to foil right so we're going to foil.

04:54 This is going to be 2t squared a, right? and then this is going to be plus a.

05:01 Now i'm going to foil the second one.

05:05 This is going to be bxte.

05:08 So did i say x? this is actually going to be t, right? because the variable we're dealing with is t, not x.

05:20 So bt plus c...

Please give Ace some feedback