Solving a System of Differential Equations

Complex Eigenvalues

The Problem

x' = Ax,   x(0) = [1, 0]T
A =   3    −2 4    −1  

Find the solution to the initial value problem without using complex numbers

The Eigenvalue Method

When the eigenvalues are complex λ = α ± βi, we find a complex eigenvector and take its real and imaginary parts to obtain real solutions.

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Step 1: Finding Eigenvalues
We find the eigenvalues of matrix A by solving |λI − A| = 0

Guiding Question

Compute the characteristic polynomial:

|λI − A| = λ² − 2λ + 5 = 0

Hint: Use the quadratic formula

λ = ± i

Enter numbers: e.g. α=1, β=2

🧐 Digital Friend - Step 1

Step 1 Solution

The characteristic polynomial:
|λI − A| = (λ−3)(λ+1) − (−2)(4) = λ² − 2λ + 5
Solving the quadratic equation:
λ = (2 ± √(4−20))/2 = (2 ± √(−16))/2 = (2 ± 4i)/2

The Eigenvalues

λ = 1 ± 2i
Step 2: Finding an Eigenvector
We find an eigenvector v for λ = 1 + 2i

Guiding Question

Solve the system (λI − A)v = 0:

(−2+2i)v₁ + 2v₂ = 0

Hint: Choose v₁ = 1 and find v₂

v = [1, + i]T

Enter the second component: e.g. 1 and -1 for 1−i

🧐 Digital Friend - Step 2

Step 2 Solution

The matrix (λI − A) for λ = 1+2i:
[(−2+2i), 2; −4, (2+2i)]
From the first equation:
(−2+2i)v₁ + 2v₂ = 0 → v₂ = (1−i)v₁
Choosing v₁ = 1:
v₂ = 1 − i

The Eigenvector

v = [1, 1−i]T
Step 3: The General Solution
We decompose the complex solution into real and imaginary parts

Guiding Question

The general solution is:

x(t) = C₁et[cos2t; cos2t+sin2t] + C₂et[sin2t; sin2t−cos2t]

Remember: eαt(cosβt + i·sinβt)

y₂(t) = et( ·sin2t − ·cos2t)

Enter numbers: 1 or -1

🧐 Digital Friend - Step 3

Step 3 Solution

The complex solution:
e(1+2i)tv = ete2it[1; 1−i]
Decomposition using Euler's formula:
= et(cos2t + i·sin2t)[1; 1−i]
The real part (first solution):
x₁(t) = et[cos2t; cos2t + sin2t]
The imaginary part (second solution):
x₂(t) = et[sin2t; sin2t − cos2t]

The General Solution

x(t) = C₁et[cos2t; cos2t+sin2t] + C₂et[sin2t; sin2t−cos2t]
Step 4: Initial Conditions
We substitute the initial conditions x(0) = [1, 0]T to find C₁ and C₂

Guiding Question

Substitute t = 0 in the general solution:

[1; 0] = C₁[1; 1] + C₂[0; −1]

Hint: Solve a system of 2 equations

C₁ = C₂ =
🧐 Digital Friend - Step 4

Step 4 Solution

Substituting t = 0:
[1; 0] = C₁[cos0; cos0+sin0] + C₂[sin0; sin0−cos0]
Simplification:
[1; 0] = C₁[1; 1] + C₂[0; −1] = [C₁; C₁ − C₂]
From the first equation: C₁ = 1
From the second equation: C₁ − C₂ = 0 → C₂ = 1

The Constants

C₁ = 1, C₂ = 1