Introduction

We consider the nonhomogeneous linear system of the form:

\[ \vec{x}'(t) = \mathbf{P}(t)\vec{x} + \vec{f}(t) \]

The general solution of this system is the sum of the complementary function (the general solution of the associated homogeneous system) and one particular solution:

\[ \vec{x}(t) = \vec{x}_{c}(t) + \vec{x}_{p}(t) \]

We explore two primary techniques for finding \(\vec{x}_{p}(t)\): the Method of Undetermined Coefficients and the Method of Variation of Parameters.

(I) Method of Undetermined Coefficients

This method involves making an intelligent guess for the form of \(\vec{x}_{p}(t)\) and then determining the coefficients by substitution. Key modifications for systems include:

Example 1: Linear Term

Find a particular solution for: \(\vec{x}' = \begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix} \vec{x} + \begin{bmatrix} 3 \\ 2t \end{bmatrix}\).

Step 1: Trial Form. Since the nonhomogeneous term is linear, let \(\vec{x}_{p}(t) = \vec{a}t + \vec{b}\). Then \(\vec{x}_{p}' = \vec{a}\).

Step 2: Substitution.

\[ \vec{a} = \begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix}(\vec{a}t + \vec{b}) + \begin{bmatrix} 0 \\ 2 \end{bmatrix}t + \begin{bmatrix} 3 \\ 0 \end{bmatrix} \]

Equating coefficients of \(t\):

\[ \vec{0} = \begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix} \vec{a} + \begin{bmatrix} 0 \\ 2 \end{bmatrix} \implies \begin{cases} 3a_{1} + 2a_{2} = 0 \\ 7a_{1} + 5a_{2} + 2 = 0 \end{cases} \implies \begin{cases} a_{1} = 4 \\ a_{2} = -6 \end{cases} \]

Equating constants:

\[ \vec{a} = \begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix} \vec{b} + \begin{bmatrix} 3 \\ 0 \end{bmatrix} \implies \begin{bmatrix} 4 \\ -6 \end{bmatrix} = \begin{bmatrix} 3b_{1} + 2b_{2} + 3 \\ 7b_{1} + 5b_{2} \end{bmatrix} \implies \begin{cases} b_{1} = 17 \\ b_{2} = -25 \end{cases} \]

Particular Solution: \(\vec{x}_{p}(t) = \begin{bmatrix} 4 \\ -6 \end{bmatrix}t + \begin{bmatrix} 17 \\ -25 \end{bmatrix}\).

Example 2: Duplication Case

Find a particular solution for: \(\vec{x}' = \begin{bmatrix} 2 & -3 \\ 1 & -2 \end{bmatrix} \vec{x} + \begin{bmatrix} e^{t} \\ 3e^{t} \end{bmatrix}\).

Step 1: Complementary Solution. The eigenvalues are \(\pm 1\), giving \(\vec{x}_{c}(t) = C_{1} \begin{bmatrix} 3e^{t} \\ e^{t} \end{bmatrix} + C_{2} \begin{bmatrix} e^{-t} \\ e^{-t} \end{bmatrix}\).

Step 2: Trial Form. Since \(e^{t}\) is duplicated in \(\vec{x}_{c}\), we must use: \(\vec{x}_{p}(t) = (\vec{a}t + \vec{b})e^{t}\) (Note: Including only \(\vec{a}te^{t}\) is insufficient).

Step 3: Substitution. After plugging in and equating coefficients:

\[ \vec{a} = \begin{bmatrix} 2 & -3 \\ 1 & -2 \end{bmatrix} \vec{a} \implies a_{1} = 3a_{2} \] \[ \vec{a} + \vec{b} = \begin{bmatrix} 2 & -3 \\ 1 & -2 \end{bmatrix} \vec{b} + \begin{bmatrix} 1 \\ 3 \end{bmatrix} \implies \begin{cases} a_{1} = b_{1} - 3b_{2} + 1 \\ a_{2} = b_{1} - 3b_{2} + 3 \end{cases} \implies \begin{cases} a_{1} = -3 \\ a_{2} = -1 \end{cases} \]

Choosing \(b_{1} = 2\) gives \(b_{2} = 2\). A particular solution is: \(\vec{x}_{p}(t) = \begin{bmatrix} -3t + 2 \\ -t + 2 \end{bmatrix} e^{t}\).

Note that \(\vec{x}_{p}(t)\) is not unique.

Rule for Duplication in Systems

Unlike single equations where you multiply by the power of \(t\) needed to eliminate duplication, for a system, the trial solution must be multiplied by the required power of \(t\) and also include all lower powers of \(t\).

Example 3: Duplication Case

Find the form of a particular solution for the nonhomogeneous linear system:

\[ \vec{x}' = \begin{bmatrix} 4 & 2 \\ 3 & -1 \end{bmatrix} \vec{x} - \begin{bmatrix} 15 \\ 4 \end{bmatrix} t e^{-2t} \]

Step 1: Complementary Solution.

From Section 5.2, the eigenvalues for this matrix are \(\lambda_1 = -2\) and \(\lambda_2 = 5\). The complementary function is:

\[ \vec{x}_{c}(t) = C_1 \begin{bmatrix} 1 \\ -3 \end{bmatrix} e^{-2t} + C_2 \begin{bmatrix} 2 \\ 1 \end{bmatrix} e^{5t} \]

Step 2: Identify Duplication.

The nonhomogeneous term is \(\vec{f}(t) = \vec{k} t e^{-2t}\). Our primary trial guess would normally be \((\vec{a}t + \vec{b})e^{-2t}\). However, the term \(e^{-2t}\) is already present in the complementary function \(\vec{x}_{c}(t)\). This is the duplication case.

Step 3: Determine the Trial Form.

Because the term \((\vec{a}t + \vec{b})e^{-2t}\) duplicates a term in \(\vec{x}_{c}\), we must multiply the entire expression by \(t\) and include the lower terms:

\[ \vec{x}_{p}(t) = (\vec{a}t^2 + \vec{b}t + \vec{c}) e^{-2t} \]

(II) Method of Variation of Parameters

Given the general solution of the associated homogeneous system \(\vec{x}_{c}(t) = \Phi(t)\vec{c}\), we seek a particular solution of the form:

\[ \vec{x}_{p}(t) = \Phi(t)\vec{u}(t) \]

Differentiating and substituting into \(\vec{x}' = \mathbf{P}(t)\vec{x} + \vec{f}(t)\) leads to the relation \(\Phi(t)\vec{u}'(t) = \vec{f}(t)\).

Theorem 1: Variation of Parameters

If \(\Phi(t)\) is a fundamental matrix for \(\vec{x}' = \mathbf{P}(t)\vec{x}\), then a particular solution is:

\[ \vec{x}_{p}(t) = \Phi(t) \int \Phi^{-1}(t)\vec{f}(t) \, dt \]

For an Initial Value Problem with \(\vec{x}(a) = \vec{b}\), the solution is:

\[ \vec{x}(t) = \Phi(t)\Phi^{-1}(a)\vec{b} + \Phi(t) \int_{a}^{t} \Phi^{-1}(s)\vec{f}(s) \, ds \]

If \(\mathbf{P}(t) = \mathbf{A}\) (constant matrix) and the initial condition is \(\vec{x}(0) = \vec{x_0}\), we can use \(\Phi(t) = e^{\mathbf{A}t}\).

The unique solution is given by plugging \(\Phi(t) = e^{\mathbf{A}t}\) into the Variation of Parameters formula:

\[ \vec{x}(t) = e^{\mathbf{A}t}\big(\vec{x}_0 + \int_{0}^{t} e^{-\mathbf{A}s}\vec{f}(s) \, ds\big) \]
Example 4: Variation of Parameters IVP

Solve \(\vec{x}' = \begin{bmatrix} 2 & -3 \\ 1 & -2 \end{bmatrix} \vec{x} + \begin{bmatrix} e^{t} \\ 3e^{t} \end{bmatrix}\) with \(\vec{x}(0) = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\).

Step 1: Fundamental Matrix. \(\Phi(t) = \begin{bmatrix} 3e^{t} & e^{-t} \\ e^{t} & e^{-t} \end{bmatrix} \implies \Phi(0)^{-1} = \frac{1}{2} \begin{bmatrix} 1 & -1 \\ -1 & 3 \end{bmatrix}\).

Step 2: Matrix Exponential. \(e^{\mathbf{A}t} = \Phi(t)\Phi^{-1}(0) = \frac{1}{2} \begin{bmatrix} 3e^{t}-e^{-t} & -3e^{t}+3e^{-t} \\ e^{t}-e^{-t} & -e^{t}+3e^{-t} \end{bmatrix}\).

Step 3: Integration. Using the formula \(\vec{x}(t) = e^{\mathbf{A}t}\vec{x}_0 + e^{\mathbf{A}t} \int_{0}^{t} e^{-\mathbf{A}s}\vec{f}(s) \, ds\):

\( \int_{0}^{t} e^{-\mathbf{A}s}\vec{f}(s) \, ds\)= \[ \int_{0}^{t} \frac{1}{2} \begin{bmatrix} 3e^{-s}-e^{s} & -3e^{-s}+3e^{s} \\ e^{s}-e^{s} & -e^{-s}+3e^{s} \end{bmatrix} \begin{bmatrix} e^{s} \\ 3e^{s} \end{bmatrix} \, ds = \int_{0}^{t} \begin{bmatrix} 4e^{2s} - 3 \\ 4e^{2s} - 1 \end{bmatrix} \, ds = \begin{bmatrix} 2e^{2t} - 3t - 2 \\ 2e^{2t} - t - 2 \end{bmatrix} \]

Step 4: Final Solution. Multiplying the matrices yields:

\[\vec{x}(t)=e^{\mathbf{A}t}\begin{bmatrix} 2e^{2t} - 3t - 2 \\ 2e^{2t} - t - 2 \end{bmatrix} = \begin{bmatrix} (-3t + 2)e^{t} - 2e^{-t} \\ (-t + 2)e^{t} - 2e^{-t} \end{bmatrix} \]