# Nonhomogeneous Linear Systems

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 9.4 – Differential Operators and the Elimination Method for Systems
• Section 9.4 -- Linear Systems in Nominal Form
• Section 9.5 -- Homogeneous Linear Systems with Constant Coefficients
• Section 9.6 -- Complex Eigenvalues
• Section 9.5 -- Nonhomogeneous Linear Systems with Constant Coefficients

## Expected Educational Results

• Objective 18–1: I can non-homogeneous linear systems using eigenvalues and eigenvectors.

## Nonhomogeneous Linear Systems

### Normal Form of a System of $n$ Linear DEs

$\pmb{x}^{\,\prime}(t)=\pmb{A}(t)\pmb{x}(t)+\pmb{f}(t)$

where $\displaystyle \pmb{x}(t)=\left[ \begin{array}{c} x_1(t) \\ x_2(t) \\ \vdots \\ x_n(t) \end{array} \right]$,

$\displaystyle \pmb{f}(t)=\left[ \begin{array}{c} f_1(t) \\ f_2(t) \\ \vdots \\ f_n(t) \end{array} \right]$, and

$\displaystyle \pmb{A}(t)=\left[a_{ij}\right]$ is an $n\times n$ matrix.

NOTE: Lowercase variables in boldface are vectors; uppercase variables in boldface are matrices.

### Normal Form of a Nonhomogenous System of $n$ Linear DEs with Constant Coefficients

where $a_{ij}$ are all constants, $\pmb{f}(t)\ne\pmb{0}$, and $f_i(t)$ are polynomials, exponential functions, sines and cosines, sums and products of these functions.

#### Undetermined Coefficients

##### Particular Solution

Example 01:

Solve the following system of DEs.

$\begin{array}{ll} x^{\,\prime}_1 &= x_1-5x_2 \\ x^{\,\prime}_2 &= -5x_1+x_2+e^{2t} \end{array}$

Solution:

1. Rewrite system as: $\pmb{x}^{\,\prime}=\left[\begin{array}{cc} 1 & -5 \\ -5 & 1 \end{array}\right]\pmb{x}+e^{2t}\left[\begin{array}{c} 0 \\ 1 \end{array}\right]$

2. Using technology, find the homogeneous solution to the system of DEs is\newline $\displaystyle \pmb{x}_h(t)=c_1e^{6t}\left[\begin{array}{c} -1 \\ 1 \end{array}\right]+c_2e^{-4t}\left[\begin{array}{c} 1 \\ 1 \end{array}\right]$

3. Using Method of Undetermined Coefficients, identify the linearly independent particular solution to the system of DEs: $\displaystyle \pmb{x}_p(t)=e^{2t}\left[\begin{array}{c} a_1 \\ a_2 \end{array}\right]$

4. Find the values for $a_1$ and $a_2$ by substituting $\displaystyle \pmb{x}_p(t)$ into the system of DEs:

1. $\pmb{x}_p^{\,\prime}=\left[\begin{array}{cc} 1 & -5 \\ -5 & 1 \end{array}\right]\pmb{x}_p+e^{2t}\left[\begin{array}{c} 0 \\ 1 \end{array}\right] \Rightarrow 2e^{2t}\left[\begin{array}{c} a_1 \\ a_2 \end{array}\right]=\left[\begin{array}{cc} 1 & -5 \\ -5 & 1 \end{array}\right]e^{2t}\left[\begin{array}{c} a_1 \\ a_2 \end{array}\right]+e^{2t}\left[\begin{array}{c} 0 \\ 1 \end{array}\right]$
2. Rewrite as: $2e^{2t}\left[\begin{array}{c} a_1 \\ a_2 \end{array}\right]=e^{2t}\left(\left[\begin{array}{cc} 1 & -5 \\ -5 & 1 \end{array}\right]\left[\begin{array}{c} a_1 \\ a_2 \end{array}\right]+\left[\begin{array}{c} 0 \\ 1 \end{array}\right]\right) \Rightarrow 2\left[\begin{array}{c} a_1 \\ a_2 \end{array}\right]=\left[\begin{array}{c} a_1 - 5a_2\\ -5a_1 +a_2 +1 \end{array}\right]$
3. Simplify to:\newline $a_1=-5a_2$ and $a_2=-5a_1+1$ and solve for unknown coefficients: $a_1=\frac{5}{24}$ and $a_2=-\frac{1}{24}$.
4. Thus, $\displaystyle \pmb{x}_p(t)=e^{2t}\left[\begin{array}{c} \frac{5}{24} \\ -\frac{1}{24} \end{array}\right]$
5. The general solution is $\displaystyle \pmb{x}=\pmb{x}_h(t)+\pmb{x}_p(t)$.

NOTE: See CPT_12_Method_Undetermined_Coefficients.pdf for a refresher on the Method of Undetermined Coefficients. Keep in mind the coefficients needed for $\displaystyle \pmb{x}_p(t)$ are vectors.

#### Investigation 01

Solve the following systems of DEs:

1. $\displaystyle \begin{array}{ll} x_1^{\,\prime} &= x_2 + 3t\\ x_2^{\,\prime} &= -3x_2-2x_1 \end{array}$
2. $\displaystyle \pmb{x}^{\,\prime}=\left[ \begin{array}{ll} 5 & 4 \\ -1 & 0 \end{array} \right]\pmb{x}+\left[ \begin{array}{l} 0 \\ \sin{(2t)} \end{array} \right]$
3. $\displaystyle \pmb{x}^{\,\prime}=\left[ \begin{array}{cccc} 2 & 1 & 1 & -1 \\ 0 & -1 & 0 & 1 \\ 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 7 \end{array} \right]\pmb{x}+\left[ \begin{array}{l} \sin{(5t)} \\ -2t \\ e^{-t} \\ 3 \end{array} \right]$

#### Variation of Parameters

NOTE: This section is not part of the syllabus and you will not be tested on this method.

See 2652_INV_13_Variation_Of_Parameters.pdf for a refresher on the method of Variation of Parameters.

##### Method
1. Find the fundamental solution set of linearly independent solutions, $\pmb{x}_h(t)=c_1\pmb{x}_1(t)+c_2\pmb{x}_2(t)+\cdots$, where $\pmb{x}_i$ are column vectors.
2. Let $\pmb{\Phi}(t)=\left[\begin{array}{cccc} \pmb{x}_1(t) & \pmb{x}_2(t) & \cdots & \pmb{x}_n(t) \end{array}\right]$
3. Evaluate $\displaystyle \pmb{x}_p(t)=\pmb{\Phi}(t)\left(\int{\pmb{\Phi}^{-1}(t)\pmb{f}(t)\,dt}+\pmb{C}\right)$

#### Investigtion 02

Use the Method of Variation of Parameters to solve the following systems of DEs:

1. $\displaystyle \begin{array}{ll} x_1^{\,\prime} &= 6x_1 + 5x_2 + e^t\\ x_2^{\,\prime} &= x_1 + x_2 + e^{5t} \end{array}$

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Last Modified: Monday, 19 October 2020 8:23 EDT