(a) Explain how to solve the inhomogeneous system (d 𝐮)/(d t)=A 𝐮+𝐛 when 𝐛 is a constant vector

(a) Explain how to solve the inhomogeneous system (d 𝐮)/(d...

(a) Explain how to solve the inhomogeneous system $\frac{d \mathbf{u}}{d t}=A \mathbf{u}+\mathbf{b}$ when $\mathbf{b}$ is a constant vector belonging to img $A$. Hint: Look at $\mathbf{v}(t)=\mathbf{u}(t)-\mathbf{u}^*$ where $\mathbf{u}^*$ is an equilibrium solution. (b) Use your method to solve (i) $\frac{d u}{d t}=u-3 v+1, \frac{d v}{d t}=-u-v$, (ii) $\frac{d u}{d t}=4 v+2, \frac{d v}{d t}=-u-3$.

(a) Explain how to solve the inhomogeneous system $\frac{d \mathbf{u}}{d t}=A \mathbf{u}+\mathbf{b}$ when $\mathbf{b}$ is a constant vector belonging to img $A$. Hint: Look at $\mathbf{v}(t)=\mathbf{u}(t)-\mathbf{u}^*$ where $\mathbf{u}^*$ is an equilibrium solution. (b) Use your method to solve
(i) $\frac{d u}{d t}=u-3 v+1, \frac{d v}{d t}=-u-v$,
(ii) $\frac{d u}{d t}=4 v+2, \frac{d v}{d t}=-u-3$.

Key Concepts

Change of Variables for Homogenization

A common approach to solving inhomogeneous problems is to perform a change of variables by subtracting a particular solution, often the equilibrium solution. By defining a new variable, such as v(t) = u(t) - u*, the original inhomogeneous equation is transformed into a homogeneous one, which is simpler to solve using standard methods.

Inhomogeneous Linear Differential Equations

These are differential equations where the rate of change of the variable is determined by both a linear operator acting on the variable and an additional forcing term. The presence of this constant or time-dependent term distinguishes them from homogeneous systems and often requires specialized techniques to find the general solution.

Equilibrium (Steady-State) Solution

An equilibrium solution, also known as a steady-state solution, satisfies the condition where the derivative is zero. For linear systems, it is found by setting the time derivative to zero, leading to an algebraic equation such as Au* + b = 0. This solution represents a constant state that the system can eventually approach, under suitable conditions.

Matrix Exponential

Once the system has been converted into a homogeneous form, the matrix exponential becomes a key tool. The solution to a homogeneous linear system can be expressed as v(t) = e^(At)v(0), where e^(At) represents the matrix exponential. This technique leverages properties of matrices to solve systems where A is a square matrix.

Image of a Matrix

The concept of the image (or range) of a matrix is critical in determining the existence of solutions to equations like Au* = -b. If the forcing term b is in the image of the matrix A, then there exists at least one vector u* such that applying A to u* yields -b. This condition ensures that an equilibrium solution can be found, facilitating the transformation to an equivalent homogeneous problem.

Transcript

00:01 Okay, so in this problem, we're given this system of differential equations here.

00:04 So for part a, we need to find the equilibrium x -hat.

00:09 So to find that, we set the right -hand side of that equation equal to zero and substitute in x -hat.

00:15 So we have a x -hat plus g is equal to zero, where zero here is the zero vector.

00:23 So to solve for x, first let's move the g over to the other side.

00:27 So we have a x hat is equal to negative g.

00:33 Now to get x by itself, we're going to multiply both sides by a inverse, since a is non -singular in this example.

00:43 Okay, so we have x hat, then it's going to be equal to negative a inverse g.

00:49 And that's our answer for the equilibrium point.

00:52 Now for part b, it says that we are to substitute y is equal to x.

01:00 X hat or x minus x hat right and it asks what do the variables or what do the components of y represent well y is just x shifted by x hat so it's the components of or it's the distance really the distance uh to x hat along each component okay that's what y the components of y represent along each component.

01:39 Now we need to show that our y satisfies the equation dy d t is equal to a y like so.

01:55 Okay, so to do that we just plug in y, okay, into the original equation.

02:07 Okay, so we have a d .y d t, first off, so let's look at this, given this definition.

02:18 If we take d, y, dt on both sides, okay.

02:24 So i'm going to make that right here.

02:28 If we take this, we get that dy, dt, is just going to be equal to dx, dt.

02:39 This component here is just a constant, so the derivative of that is just zero...

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