. Solve the following initial value problems: (a) u̇1=2 u1-u2, u1(0)=0 u̇2=4 u1-3

step 1 Hello and welcome to another differential equation problem. In this one we have 3U double prime

. Solve the following initial value problems: (a) u̇1=2...

Key Concepts

Initial Value Problem

An initial value problem (IVP) is a differential equation combined with a specified value at a given initial point. It requires that the solution not only satisfies the differential equation but also passes through the point defined by the initial condition. This is essential in determining a unique solution among the many possible ones that the differential equation may admit.

Linear Ordinary Differential Equation

A linear ordinary differential equation (ODE) is one in which the unknown function and its derivatives appear to the first power and are not multiplied together. These equations have the superposition property, which allows the general solution to be expressed as a combination of the homogeneous solution and a particular solution, often making them more tractable for analysis.

Nonhomogeneous Differential Equations

Nonhomogeneous differential equations are those that include a forcing term or external input, making them different from their homogeneous counterparts. Methods such as undetermined coefficients or variation of parameters are typically used to find a particular solution that, when added to the complementary solution of the homogeneous equation, provides the complete solution.

System of Differential Equations

A system of differential equations involves multiple interrelated differential equations that must be solved simultaneously. Such systems can often be written in matrix form, and their solutions can be found using techniques such as eigenvalue–eigenvector analysis or matrix exponentials. This framework is especially useful when dealing with coupled variables.

Matrix Methods in Differential Equations

Matrix methods, including the use of matrix exponentials, are tools for solving systems of linear differential equations. By representing a system as a matrix equation, one can exploit linear algebra techniques to compute the evolution of the system over time, particularly when the system is time-invariant and can be diagonalized or otherwise simplified.

Decoupling Techniques

Decoupling techniques involve transforming a system of coupled differential equations into a set of independent equations that can be solved individually. This often involves a change of variables or the use of eigenvalue decomposition and can simplify the process of finding the overall solution by reducing the complexity of interdependencies.

Transcript

00:01 Hello and welcome to another differential equation problem.

00:02 In this one we have 3u double prime minus u prime plus 2 u equals 0.

00:07 We're also given a few initial conditions.

00:09 Now to solve this, we're going to have to take the characteristic equation of this differential equation.

00:16 That'll get us 3r squared minus r plus 2 equals 0.

00:23 Now to complete the square, i'm going to have to divide everything by 3.

00:26 This will give us r cubed minus r over three plus two thirds equals zero now we can just complete the square and that was a typo this is this is supposed to be a squared term right there if we complete the square we'll have r minus one -sixth quantity squared minus one over 36th because that's what we add in order to complete the square we'll add that to two -thirds and that is of course equal to zero.

01:05 Now i'm going to find out what this, let's get a common denominator of this two -thirds term in the 36th term.

01:15 Just by inspection, i can see that it's going to be 24 over 36.

01:20 So this will get a combined sum of 23 over 36.

01:28 Great, and then we can add that over to the right side.

01:33 So we'll have r minus 1 6th quantity squared, is equal to negative 23 over 36.

01:45 All right, so let's take the square root and add 1 .6th to both sides.

01:51 Let's do that in green.

01:53 We'll have r is equal to 1 6th plus or minus, the square root of 23 over 6 times i.

02:08 Like that.

02:09 Great.

02:10 Let's even get out that right there.

02:12 Perfect.

02:14 So if we substitute this into the solution to characteristic equations, which is just c, e to the rt, then we can solve for the, for both solutions to this differential equation.

02:34 Remember that the real solution will just be the sum of both of the branches.

02:42 So if we can conflate them, then we will have a nice and pretty solution.

02:49 Let's start off with the positive branch.

02:51 So we'll have one sixth plus square of 23 over 6i.

02:55 And just looking at this, we'll be able to take out this e to the t over 6.

03:03 And using oilis formula, we'll have this square root of 23 over 6 term.

03:09 I'm going to write this out as c1 squared of 23, oops, sorry, c1 cosine of t squared of 23 over 6, plus c2, sine of t root 23 over 6.

03:35 And i'm doing this because when we get the second branch, the negative branch, the only thing that's going to change is this sine and cosine constant of integration.

03:48 And this is because sine is an odd function and cosine is an even function.

03:53 All right, great.

03:54 So we have solved the differential equation.

03:56 And now what we have to do is plug in our initial conditions.

04:00 So since we have a value at zero, this should be fairly straightforward.

04:05 Forward let's bring us up to the top and we'll have y which is in this case two and we can set that equal to well e to the t over six when t is zero is just one times c1 cosine of cosine of zero so we'll have one times c1 times one plus what was the second one this is really just plus zero because sign of zero is zero great so that'll get a us c1 is equal to 2.

04:39 Perfect.

04:40 Now if we want to take the derivative of this, which we're going to have to get this, well to get c2 and to use our second initial condition, we're going to use the product rule.

04:53 This is going to be pretty ugly because the numbers inside here are not nice.

05:00 Anyway, let's do it.

05:02 If we write this out, we'll have y prime.

05:08 I set that equal to, whoops, my handwriting is not looking great right now.

05:15 So we write that as y prime is equal to e to the t over 6 times its derivative of this mess.

05:27 Just looking at it, it's going to be c2 times square root of 23 over 6 times cosine of cosine of this stuff, i'm just going to write that as, i'm not going to write everything because it will take so long.

05:44 And this will be minus c2 times the square of 23 over 6 times the sign of whatever.

05:56 And this will be added to the derivative of the e term, which will be just one sixth e to the t over six.

06:08 And this will just be multiplied by this thing in its entirety...

Please give Ace some feedback