Solve the given non homogeneous system. x1^'=x1+x2+e^2 t, x2^'=3 x1-x2+5 e^2 t

Name: Problem 14, Chapter 9: Differential Equations and Linear Algebra
Uploaded: 2020-05-31T18:47:52-08:00
Duration: 8 min 34 s
Channel: Christian Otero
Description: Solve the given non homogeneous system. x1^'=x1+x2+e^2 t, x2^'=3 x1-x2+5 e^2 t

step 1 Okay, so we want to go ahead and solve this non -homogeneous system. And we already know that so

Solve the given non homogeneous system. x1^'=x1+x2+e^2 t, ...

Key Concepts

Systems of Linear Differential Equations

These involve multiple interlinked differential equations that determine several unknown functions. They are often written in matrix form and represent a set of relations where each derivative is expressed in terms of the functions themselves, allowing systematic analysis of the system's behavior.

Homogeneous vs Non-Homogeneous Solutions

In the context of differential equations, a homogeneous system lacks external forcing terms, while a non-homogeneous system includes an additional function or vector term that acts as a driving force. The overall solution to a non-homogeneous system is found by adding a particular solution (addressing the non-homogeneous part) to the general solution of the corresponding homogeneous system.

Method of Undetermined Coefficients

This is a technique for finding a particular solution to non-homogeneous linear differential equations when the non-homogeneous part is of a certain form like exponential, polynomial, or trigonometric functions. The method involves guessing a form for the particular solution with unknown parameters and determining these parameters by substituting back into the differential equation.

Matrix Representation and Eigenvalue Analysis

Writing the system in matrix form (X' = AX + f(t)) is a crucial step that allows the application of linear algebra methods. For the homogeneous part, the eigenvalues and eigenvectors of the coefficient matrix are computed to form the general solution. This analysis helps characterize the system's dynamics and stability.

Transcript

00:01 Okay, so we want to go ahead and solve this non -homogeneous system.

00:05 And we already know that solving our particular equation, we get that the general solution is just the sum of our, the linear combination of our homogeneous and our particular solutions.

00:17 So if we find each of those separately, we can find our final answer.

00:21 So what we'd like to do is go ahead and separate our homogeneous part and solve that system first.

00:29 And this gives us a system as follows.

00:31 D minus 1 x1h minus x2h is 0 and d plus 1 x2h is just equal to 3x1 so we'll go ahead and evaluate this to get d squared minus 1 x1 h is just d plus 1 x2 h is just d plus 1 x2 h equals 0.

01:08 And we know that this minus d plus 1x2h turns into a 3x1h.

01:14 So we get our final polynomial d squared minus 4 x1h is equal to 0.

01:22 So this gives us a characteristic equation with eigenvalues plus or minus 2 giving us the following a homogenous solution.

01:35 So we get c1e to the 2t plus c2t plus e2, to the negative 2t.

01:40 And we have to go ahead and solve the second equation.

01:45 So the first equation to get our second equation.

01:49 So we get the second x2h is just d minus 1, x1h.

01:56 And we can simplify this easily, because the exponentials of derivatives, as exponentials are very simple.

02:04 So we get derivative, brings a two out front, and then a minus one gives us c1, e to the 2t, and then derivative is a negative 2, and a minus 1 gives us a minus 3, e to the minus 2t.

02:18 So these are our homogenous solutions.

02:23 Now we want to find our particulars.

02:26 So we had to look at this carefully.

02:32 For almost any other problems, if we see our function that our non -homogeneous part is an exponential, we would say that our particular should be some function, some exponential function with the same power.

02:51 But the first thing we'll notice is that our homogenous solutions already include, already have e to 2t in them because 2 is an eigenvalue of our problem.

03:07 So then we'll have to think, okay, so the next step forward is for any other equation where we had multiplicity.

03:14 We would say that we have our, bring our linear term, our t, into in front of our solution...

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