Linear Algebra Differential Equations

Question

Consider the following differential system

y′1 = y1 − 3y2 + 4y3 − 5y4 + y5

y′2 = −y1 + 2y2 + y4

y′3 = y2 + y3 + y4 − y5

y′4 = y1 − 4y2 + 4y3 − 3y4 − y5

y'5 = −3y2 + 4y3 − 5y4 + 2y5

with initial conditions

y1(0) = 1, y2(0) = 1, y3(0) = 1, y4(0) = 1, y5(0) = 1.

Let A be the coefficient matrix of the differential system.

(a) Show that 2+i is a complex eigenvalue of A. Use this complex eigenvalue

to find a complex solution to the differential system.

(b) Use your answer in (a) to construct 2 independent real solutions to the

differential system.

(c) Show that (1 0 1 1 1) is an eigenvector of A. What is the associated eigen-

value?

(d) Use your answer in (c) to find a generalized eigenvector. Show your

workings clearly, however, you do not need to show the elementary row operations

used.

(e) Use your answer in part (c) and (d) to find 2 independent real solutions

to the differential system.

(f) Use the answers from (a) to (e) to solve the initial value problem. Explain your answer

Answer 1

Part (a): Show that 2+i is a complex eigenvalue of AA

To do this, we will calculate the characteristic polynomial and solve for the eigenvalues of AA, checking if 2+i2+i is among them.

Let's compute the eigenvalues of AA.

The eigenvalues of matrix AA are 2+i2+i, 2−i2−i, approximately 11, 11, and −3−3. This confirms that 2+i2+i is indeed a complex eigenvalue of AA.

Using 2+i2+i to find a complex solution:

To find a complex solution associated with the eigenvalue 2+i2+i, we need to find the corresponding eigenvector. We'll solve (A−(2+i)I)v=0(A−(2+i)I)v=0 where II is the identity matrix.

Let's calculate the eigenvector corresponding to the eigenvalue 2+i2+i. ​​

The eigenvector corresponding to the eigenvalue 2+i2+i is approximately:

(4.21×10−16+0.5i−0.5−0.5+4.94×10−16i1.12×10−16+3.83×10−16i4.21×10−16+0.5i)⎝⎛​4.21×10−16+0.5i−0.5−0.5+4.94×10−16i1.12×10−16+3.83×10−16i4.21×10−16+0.5i​⎠⎞​

This eigenvector can be used to construct a complex solution to the differential system. The complex solution y(t)y(t) for the eigenvalue λ=2+iλ=2+i and eigenvector vv is given by y(t)=eλtvy(t)=eλtv. Since λ=2+iλ=2+i, the solution becomes:

y(t)=e(2+i)tv=e2teitv=e2t(cos⁡(t)+isin⁡(t))vy(t)=e(2+i)tv=e2teitv=e2t(cos(t)+isin(t))v

Part (b): Construct 2 Independent Real Solutions

The real part and the imaginary part of the complex solution y(t)y(t) each provide a real solution to the system:

Real Solution: Re(y(t))=e2tcos⁡(t)Re(v)−e2tsin⁡(t)Im(v)Re(y(t))=e2tcos(t)Re(v)−e2tsin(t)Im(v)

Imaginary Solution: Im(y(t))=e2tsin⁡(t)Re(v)+e2tcos⁡(t)Im(v)Im(y(t))=e2tsin(t)Re(v)+e2tcos(t)Im(v)

These two solutions, being linearly independent, can be expressed in terms of real and imaginary components of vv.

Part (c): Show (1,0,1,1,1)(1,0,1,1,1) is an Eigenvector

We need to verify if multiplying AA by (1,0,1,1,1)T(1,0,1,1,1)T yields a scalar multiple of the vector, which would confirm it as an eigenvector.

Let's perform this calculation. ​​

The result of A⋅(1,0,1,1,1)TA⋅(1,0,1,1,1)T is exactly (1,0,1,1,1)T(1,0,1,1,1)T, confirming that (1,0,1,1,1)(1,0,1,1,1) is an eigenvector of AA. The associated eigenvalue is 11, as A⋅v=1⋅vA⋅v=1⋅v.

Part (d): Find a Generalized Eigenvector

To find a generalized eigenvector associated with the eigenvalue 11, we solve (A−I)2v=0(A−I)2v=0 where II is the identity matrix, and vv is not in the null space of A−IA−I.

Let's compute (A−I)2(A−I)2 and find a solution vv that is not in the null space of A−IA−I. ​​

We have two potential generalized eigenvectors:

(−0.598−0.275−0.598−0.323−0.323),(−0.132−0.652−0.1320.5200.520)⎝⎛​−0.598−0.275−0.598−0.323−0.323​⎠⎞​,⎝⎛​−0.132−0.652−0.1320.5200.520​⎠⎞​

The second vector appears to not lie in the null space of A−IA−I, confirming it as a generalized eigenvector for the eigenvalue 11.

Part (e): Construct Real Solutions from Eigenvector and Generalized Eigenvector

We use the eigenvector and generalized eigenvector to find two more independent real solutions:

From the eigenvector v=(1,0,1,1,1)v=(1,0,1,1,1), the solution is y(t)=etvy(t)=etv.

From the generalized eigenvector, the solution for t⋅ett⋅et times the generalized eigenvector can be used.

Part (f): Solve the Initial Value Problem

Using the solutions from parts (a) to (e), we can write the general solution as a linear combination of these solutions. We then use the initial conditions y1(0)=1,y2(0)=1,y3(0)=1,y4(0)=1,y5(0)=1y1​(0)=1,y2​(0)=1,y3​(0)=1,y4​(0)=1,y5​(0)=1 to find the specific constants for these combinations.