Math 308: Week-in-Review 12

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Section 7.9 — Nonhomogeneous Linear Systems

Problems

1. Form of the Particular Solution

For each of the following nonhomogeneous systems \(\mathbf{x}' = A\mathbf{x} + \mathbf{g}(t)\), do not solve for any coefficients. Instead: (i) find the eigenvalues of \(A\), (ii) decide whether a resonance occurs, and (iii) write down the correct form of a particular solution using unknown constant vectors \(\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}\).

(a) Polynomial forcing.

\[ \mathbf{x}' = \begin{pmatrix} 1 & 2 \\ 3 & -4 \end{pmatrix}\mathbf{x} + \begin{pmatrix} 2t + 1 \\ -3 \end{pmatrix} \]

(b) Exponential forcing.

\[ \mathbf{x}' = \begin{pmatrix} 1 & 2 \\ 3 & -4 \end{pmatrix}\mathbf{x} + \begin{pmatrix} 4 \\ -1 \end{pmatrix} e^{3t} \]

\[ \mathbf{x}' = \begin{pmatrix} 1 & 2 \\ 3 & -4 \end{pmatrix}\mathbf{x} + \begin{pmatrix} 2 \\ 3 \end{pmatrix} e^{2t} \]

(d) Trigonometric forcing.

\[ \mathbf{x}' = \begin{pmatrix} 0 & -4 \\ 1 & \phantom{-}0 \end{pmatrix}\mathbf{x} + \begin{pmatrix} \cos(2t) \\ \sin(2t) \end{pmatrix} \]

Hint: The matrix in parts (a), (b), (c) is the same on purpose — what changes is whether the number appearing in the forcing is an eigenvalue of \(A\). Compute the eigenvalues once, then reuse them.

SolutionSolution

Eigenvalues of the matrix used in (a), (b), (c)

For \(A=\begin{pmatrix}1&2\\3&-4\end{pmatrix}\), \(\operatorname{tr}(A)=-3\) and \(\det(A)=-4-6=-10\), so

\[\det(A-\lambda I)=\lambda^2+3\lambda-10=(\lambda-2)(\lambda+5)=0 \ \Longrightarrow\ \lambda_1=2,\ \lambda_2=-5.\]

Part (a) — Polynomial forcing

\(\mathbf{g}(t) = (2t+1,\,-3)^T\) is a polynomial of degree \(1\). The characteristic value is \(\alpha = 0\). Since \(0 \notin \{2,-5\}\), there is no resonance and \(\mathbf{x}_p\) is also a polynomial of degree \(1\):

\[\boxed{\ \mathbf{x}_p(t) = \mathbf{a}\,t + \mathbf{b}\ }\quad\text{with unknown vectors } \mathbf{a},\mathbf{b}\in\mathbb{R}^2.\]

Part (b) — Exponential forcing at \(\alpha = 3\)

Here \(\alpha = 3\) and \(3 \notin \{2, -5\}\), so no resonance:

\[\boxed{\ \mathbf{x}_p(t) = \mathbf{a}\,e^{3t}\ }\quad\text{with unknown vector } \mathbf{a}\in\mathbb{R}^2.\]

Part (c) — Exponential forcing at \(\alpha = 2\)

Here \(\alpha = 2\), and \(2 \in \{2, -5\}\) is an eigenvalue of \(A\). Resonance. The vector-valued form requires both a \(t\,e^{2t}\) term and a bare \(e^{2t}\) term:

\[\boxed{\ \mathbf{x}_p(t) = \mathbf{a}\,t\,e^{2t} + \mathbf{b}\,e^{2t}\ }\]

When substituting into the ODE, \(\mathbf{a}\) must end up being an eigenvector for \(\lambda = 2\), and \(\mathbf{b}\) is determined (up to a multiple of \(\mathbf{a}\)) by a second solvability condition. The bare \(\mathbf{b}\,e^{2t}\) term cannot be omitted.

Part (d) — Trigonometric forcing at \(\beta = 2\)

For \(A = \begin{pmatrix}0&-4\\1&0\end{pmatrix}\), \(\det(A - \lambda I) = \lambda^2 + 4 = 0\), so \(\lambda = \pm 2i\). The forcing frequency is \(\beta = 2\), and \(\pm i\beta = \pm 2i\) are eigenvalues of \(A\). Resonance. The correct form multiplies the trial by \(t\) and keeps an unmultiplied copy:

\[\boxed{\ \mathbf{x}_p(t) = t\bigl[\mathbf{a}\cos(2t) + \mathbf{b}\sin(2t)\bigr] + \mathbf{c}\cos(2t) + \mathbf{d}\sin(2t)\ }\]

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2. Variation of Parameters

Consider the nonhomogeneous system

\[ \mathbf{x}' = \begin{pmatrix} -5 & \phantom{-}3 \\ \phantom{-}2 & -10 \end{pmatrix}\mathbf{x} + \begin{pmatrix} e^{-t} \\ 0 \end{pmatrix}. \]

Find a fundamental matrix \(\Psi(t)\) and its inverse \(\Psi^{-1}(t)\). Then use variation of parameters to obtain a particular solution and the general solution.

SolutionSolution

Step 1 — Eigenvalues and eigenvectors

For \(A = \begin{pmatrix}-5&3\\2&-10\end{pmatrix}\), \(\operatorname{tr}(A) = -15\) and \(\det(A) = 50 - 6 = 44\), so

\[\det(A - \lambda I) = \lambda^2 + 15\lambda + 44 = (\lambda + 4)(\lambda + 11) = 0 \ \Longrightarrow\ \lambda_1 = -4,\ \lambda_2 = -11.\]

\(\lambda_1 = -4\). \(A + 4I = \begin{pmatrix}-1&3\\2&-6\end{pmatrix}\). Row 1: \(-v_1 + 3v_2 = 0\), so \(v_1 = 3v_2\). Take \(v_2 = 1\): \(\mathbf{v}_1 = (3, 1)^T\). Check: \(A\mathbf{v}_1 = (-15 + 3,\,6 - 10)^T = (-12,\,-4)^T = -4(3,1)^T.\ \checkmark\)

\(\lambda_2 = -11\). \(A + 11I = \begin{pmatrix}6&3\\2&1\end{pmatrix}\). Row 1: \(6v_1 + 3v_2 = 0\), so \(v_2 = -2v_1\). Take \(v_1 = 1\): \(\mathbf{v}_2 = (1, -2)^T\). Check: \(A\mathbf{v}_2 = (-5 - 6,\,2 + 20)^T = (-11, 22)^T = -11(1,-2)^T.\ \checkmark\)

Step 2 — Fundamental matrix and inverse

\[\Psi(t) = \begin{pmatrix} 3e^{-4t} & e^{-11t} \\ e^{-4t} & -2e^{-11t} \end{pmatrix},\qquad \det\Psi(t) = -6e^{-15t} - e^{-15t} = -7e^{-15t}.\]

By the \(2\times 2\) inverse formula,

\[\Psi^{-1}(t) = \frac{1}{-7e^{-15t}}\begin{pmatrix}-2e^{-11t} & -e^{-11t}\\ -e^{-4t} & 3e^{-4t}\end{pmatrix} = \frac{1}{7}\begin{pmatrix} 2e^{4t} & e^{4t}\\ e^{11t} & -3e^{11t}\end{pmatrix}.\]

Step 3 — Compute \(\Psi^{-1}\mathbf{g}\) and integrate

\[\Psi^{-1}(t)\,\mathbf{g}(t) = \frac{1}{7}\begin{pmatrix} 2e^{4t} & e^{4t}\\ e^{11t} & -3e^{11t}\end{pmatrix}\begin{pmatrix}e^{-t}\\ 0\end{pmatrix} = \frac{1}{7}\begin{pmatrix} 2e^{3t}\\ e^{10t}\end{pmatrix}.\]

\[\int \Psi^{-1}(t)\,\mathbf{g}(t)\,dt = \frac{1}{7}\begin{pmatrix}\tfrac{2}{3}e^{3t}\\ \tfrac{1}{10}e^{10t}\end{pmatrix} = \begin{pmatrix}\tfrac{2}{21}e^{3t}\\ \tfrac{1}{70}e^{10t}\end{pmatrix}.\]

Step 4 — Multiply by \(\Psi(t)\) to recover \(\mathbf{x}_p\)

\[\mathbf{x}_p(t) = \begin{pmatrix} 3e^{-4t} & e^{-11t} \\ e^{-4t} & -2e^{-11t} \end{pmatrix}\begin{pmatrix}\tfrac{2}{21}e^{3t}\\ \tfrac{1}{70}e^{10t}\end{pmatrix}.\]

Row 1: \(3 \cdot \tfrac{2}{21}e^{-t} + \tfrac{1}{70}e^{-t} = \tfrac{2}{7}e^{-t} + \tfrac{1}{70}e^{-t} = \tfrac{20+1}{70}e^{-t} = \tfrac{3}{10}e^{-t}\).

Row 2: \(\tfrac{2}{21}e^{-t} - 2 \cdot \tfrac{1}{70}e^{-t} = \tfrac{2}{21}e^{-t} - \tfrac{1}{35}e^{-t} = \tfrac{10-3}{105}e^{-t} = \tfrac{1}{15}e^{-t}\).

\[\mathbf{x}_p(t) = \begin{pmatrix}\tfrac{3}{10}\\ \tfrac{1}{15}\end{pmatrix}e^{-t} = \frac{1}{30}\begin{pmatrix}9\\ 2\end{pmatrix}e^{-t}.\]

Step 5 — General solution

\[\boxed{\ \mathbf{x}(t) = c_1 \begin{pmatrix}3\\1\end{pmatrix}e^{-4t} + c_2 \begin{pmatrix}1\\-2\end{pmatrix}e^{-11t} + \frac{1}{30}\begin{pmatrix}9\\2\end{pmatrix}e^{-t}\ }\]

Cross-check (undetermined coefficients). Since \(\alpha = -1\) is not an eigenvalue, the guess \(\mathbf{x}_p = \mathbf{a}\,e^{-t}\) should also work. Substituting gives \(-\mathbf{a} = A\mathbf{a} + (1,0)^T\), i.e. \((A + I)\mathbf{a} = -(1,0)^T\): \[\begin{pmatrix}-4&3\\2&-9\end{pmatrix}\mathbf{a} = \begin{pmatrix}-1\\0\end{pmatrix}.\] From the second row \(a_1 = \tfrac{9}{2}a_2\); substituting into the first gives \(-18a_2 + 3a_2 = -1\), so \(a_2 = \tfrac{1}{15}\) and \(a_1 = \tfrac{3}{10}\) — agreeing with \(\mathbf{x}_p\) above.

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3. Variation of Parameters (Non-Constant Coefficients)

Consider the nonhomogeneous system (for \(t > 0\))

\[ \mathbf{x}' = \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}\mathbf{x} + \begin{pmatrix} t\,e^{t} \\ e^{t} \end{pmatrix}. \]

Verify that

\[ \Psi(t) = \begin{pmatrix} e^{t} & \dfrac{t^{2}}{2}\,e^{t} \\[4pt] 0 & e^{t} \end{pmatrix} \]

is a fundamental matrix for the homogeneous system. Then use variation of parameters to find the general solution.

Observation: The coefficient matrix here is not constant in \(t\), so the method of undetermined coefficients does not apply. Variation of parameters, however, still works — as long as a fundamental matrix \(\Psi(t)\) is available.

SolutionSolution

Step 1 — Verify \(\Psi'(t) = A(t)\Psi(t)\) column by column

Column 1: \(\mathbf{x}^{(1)}(t) = (e^{t},\,0)^T\), so \((\mathbf{x}^{(1)})'(t) = (e^{t},\,0)^T\). Also

\[A(t)\,\mathbf{x}^{(1)} = \begin{pmatrix}1&t\\0&1\end{pmatrix}\begin{pmatrix}e^{t}\\ 0\end{pmatrix} = \begin{pmatrix}e^{t}\\ 0\end{pmatrix}.\ \checkmark\]

Column 2: \(\mathbf{x}^{(2)}(t) = \left(\tfrac{t^2}{2}e^{t},\ e^{t}\right)^T\). By the product rule,

\[\left(\mathbf{x}^{(2)}\right)'(t) = \begin{pmatrix} t\,e^{t} + \tfrac{t^2}{2}e^{t}\\ e^{t}\end{pmatrix} = \begin{pmatrix} (t + \tfrac{t^2}{2})e^{t}\\ e^{t}\end{pmatrix},\]

and

\[A(t)\,\mathbf{x}^{(2)} = \begin{pmatrix}1&t\\0&1\end{pmatrix}\begin{pmatrix}\tfrac{t^2}{2}e^{t}\\ e^{t}\end{pmatrix} = \begin{pmatrix} \tfrac{t^2}{2}e^{t} + t\,e^{t}\\ e^{t}\end{pmatrix}.\ \checkmark\]

Both columns match, so \(\Psi(t)\) is a fundamental matrix.

Step 2 — Compute \(\Psi^{-1}(t)\)

\[\det\Psi(t) = e^{t}\cdot e^{t} - \tfrac{t^2}{2}e^{t}\cdot 0 = e^{2t}.\]

\[\Psi^{-1}(t) = \frac{1}{e^{2t}}\begin{pmatrix}e^{t} & -\tfrac{t^2}{2}e^{t}\\ 0 & e^{t}\end{pmatrix} = \begin{pmatrix}e^{-t} & -\tfrac{t^2}{2}e^{-t}\\ 0 & e^{-t}\end{pmatrix}.\]

Step 3 — Compute \(\Psi^{-1}\mathbf{g}\) and integrate

\[\Psi^{-1}(t)\,\mathbf{g}(t) = \begin{pmatrix}e^{-t} & -\tfrac{t^2}{2}e^{-t}\\ 0 & e^{-t}\end{pmatrix}\begin{pmatrix}t\,e^{t}\\ e^{t}\end{pmatrix} = \begin{pmatrix} t - \tfrac{t^2}{2}\\ 1\end{pmatrix}.\]

\[\int \Psi^{-1}(t)\,\mathbf{g}(t)\,dt = \begin{pmatrix} \tfrac{t^2}{2} - \tfrac{t^3}{6}\\ t\end{pmatrix}.\]

Step 4 — Multiply by \(\Psi(t)\) to recover \(\mathbf{x}_p\)

\[\mathbf{x}_p(t) = \begin{pmatrix}e^{t} & \tfrac{t^2}{2}e^{t}\\ 0 & e^{t}\end{pmatrix}\begin{pmatrix}\tfrac{t^2}{2} - \tfrac{t^3}{6}\\ t\end{pmatrix}.\]

Row 1: \(e^{t}\!\left(\tfrac{t^2}{2} - \tfrac{t^3}{6}\right) + \tfrac{t^2}{2}e^{t}\cdot t = e^{t}\!\left(\tfrac{t^2}{2} - \tfrac{t^3}{6} + \tfrac{t^3}{2}\right) = e^{t}\!\left(\tfrac{t^2}{2} + \tfrac{t^3}{3}\right)\).

Row 2: \(0 + e^{t}\cdot t = t\,e^{t}\).

\[\mathbf{x}_p(t) = \begin{pmatrix} \left(\tfrac{t^2}{2} + \tfrac{t^3}{3}\right)e^{t}\\ t\,e^{t}\end{pmatrix}.\]

Step 5 — General solution

\[\boxed{\ \mathbf{x}(t) = c_1 \begin{pmatrix}e^{t}\\ 0\end{pmatrix} + c_2 \begin{pmatrix}\tfrac{t^2}{2}e^{t}\\ e^{t}\end{pmatrix} + \begin{pmatrix} \left(\tfrac{t^2}{2} + \tfrac{t^3}{3}\right)e^{t}\\ t\,e^{t}\end{pmatrix}\ }\]

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4. Undetermined Coefficients: Polynomial Forcing

Consider the system

\[ x' = 3x + 2y + 3, \qquad y' = 7x + 5y + 2t. \]

Rewrite in matrix–vector form and find a fundamental matrix \(\Psi(t)\) for the homogeneous part.
Use undetermined coefficients to find a particular solution \(\mathbf{x}_p(t)\), then write the general solution.

SolutionSolution

Step 1 — Matrix form

\[\mathbf{x}' = A\mathbf{x} + \mathbf{g}(t),\quad A = \begin{pmatrix}3 & 2\\ 7 & 5\end{pmatrix},\quad \mathbf{g}(t) = \begin{pmatrix}3\\ 2t\end{pmatrix} = \begin{pmatrix}0\\ 2\end{pmatrix}t + \begin{pmatrix}3\\ 0\end{pmatrix}.\]

Step 2 — Eigenvalues of \(A\)

\(\operatorname{tr}(A) = 8\) and \(\det(A) = 15 - 14 = 1\), so

\[\det(A - \lambda I) = \lambda^2 - 8\lambda + 1 = 0 \ \Longrightarrow\ \lambda = 4 \pm \sqrt{15}.\]

Denote \(\lambda_1 = 4 - \sqrt{15}\) and \(\lambda_2 = 4 + \sqrt{15}\). Both are real and nonzero (the smaller is positive: \(\sqrt{15} < 4\)), so \(0\) is not an eigenvalue — the polynomial trial form will work without resonance modification.

Step 3 — Eigenvectors and fundamental matrix

For either eigenvalue, \((A - \lambda I)\mathbf{v} = 0\) from the first row gives \((3 - \lambda)v_1 + 2v_2 = 0\), i.e. \(v_2 = \tfrac{\lambda - 3}{2}v_1\). Taking \(v_1 = 2\) in each case:

\[\mathbf{v}_1 = \begin{pmatrix}2\\ 1 - \sqrt{15}\end{pmatrix},\qquad \mathbf{v}_2 = \begin{pmatrix}2\\ 1 + \sqrt{15}\end{pmatrix}.\]

\[\Psi(t) = \begin{pmatrix} 2e^{(4-\sqrt{15})t} & 2e^{(4+\sqrt{15})t}\\ (1-\sqrt{15})e^{(4-\sqrt{15})t} & (1+\sqrt{15})e^{(4+\sqrt{15})t}\end{pmatrix}.\]

Step 4 — Trial form and system for the coefficients

Since \(\mathbf{g}(t)\) is a degree-\(1\) polynomial and \(0\) is not an eigenvalue, try \(\mathbf{x}_p(t) = \mathbf{a}\,t + \mathbf{b}\). Then \(\mathbf{x}_p' = \mathbf{a}\), and substituting gives

\[\mathbf{a} = A(\mathbf{a}\,t + \mathbf{b}) + \begin{pmatrix}0\\ 2\end{pmatrix}t + \begin{pmatrix}3\\ 0\end{pmatrix}.\]

Matching powers of \(t\) gives two linear systems:

\[\text{Coef. of }t:\quad A\mathbf{a} = -\begin{pmatrix}0\\ 2\end{pmatrix},\qquad \text{Coef. of }1:\quad A\mathbf{b} = \mathbf{a} - \begin{pmatrix}3\\ 0\end{pmatrix}.\]

Step 5 — Solve for \(\mathbf{a}\) and \(\mathbf{b}\)

Since \(\det A = 1\), \(A^{-1} = \begin{pmatrix}5 & -2\\ -7 & 3\end{pmatrix}\).

\[\mathbf{a} = A^{-1}\begin{pmatrix}0\\ -2\end{pmatrix} = \begin{pmatrix}5 & -2\\ -7 & 3\end{pmatrix}\begin{pmatrix}0\\ -2\end{pmatrix} = \begin{pmatrix}4\\ -6\end{pmatrix}.\]

\[\mathbf{a} - \begin{pmatrix}3\\ 0\end{pmatrix} = \begin{pmatrix}1\\ -6\end{pmatrix},\qquad \mathbf{b} = A^{-1}\begin{pmatrix}1\\ -6\end{pmatrix} = \begin{pmatrix}5 + 12\\ -7 - 18\end{pmatrix} = \begin{pmatrix}17\\ -25\end{pmatrix}.\]

Quick check. \(A\mathbf{a} = \begin{pmatrix}3\cdot 4 + 2\cdot(-6)\\ 7\cdot 4 + 5\cdot(-6)\end{pmatrix} = \begin{pmatrix}0\\ -2\end{pmatrix}\). \(\checkmark\)

Step 6 — Particular and general solution

\[\mathbf{x}_p(t) = \begin{pmatrix}4\\ -6\end{pmatrix}t + \begin{pmatrix}17\\ -25\end{pmatrix}.\]

\[\boxed{\ \mathbf{x}(t) = c_1 \begin{pmatrix}2\\ 1-\sqrt{15}\end{pmatrix}e^{(4-\sqrt{15})t} + c_2 \begin{pmatrix}2\\ 1+\sqrt{15}\end{pmatrix}e^{(4+\sqrt{15})t} + \begin{pmatrix}4\\ -6\end{pmatrix}t + \begin{pmatrix}17\\ -25\end{pmatrix}\ }\]

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5. Complex Eigenvalues with Mixed Forcing

Use the method of undetermined coefficients to find the general solution of

\[ \mathbf{x}' = \begin{pmatrix} 0 & -1 \\ 1 & \phantom{-}0 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} 3 + e^{t} \\ -2 - 2e^{t} \end{pmatrix}. \]

Observation: The forcing splits as a constant vector plus an \(e^{t}\) vector. Treat the two pieces separately (superposition) and guess \(\mathbf{x}_p\) as a sum of the corresponding forms.

SolutionSolution

Step 1 — Eigenvalues and homogeneous solution

For \(A = \begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}\), \(\det(A - \lambda I) = \lambda^2 + 1 = 0\), so \(\lambda = \pm i\) (purely imaginary).

For \(\lambda_1 = i\): \((A - iI)\mathbf{v} = 0\) gives \(-i\,v_1 - v_2 = 0\), i.e. \(v_2 = -i\,v_1\). Take \(v_1 = 1\): \(\mathbf{v} = (1,\,-i)^T\).

Applying Euler's formula to \(e^{it}(1,-i)^T\) gives the two real solutions

\[\mathbf{w}_1(t) = \begin{pmatrix}\cos t\\ \sin t\end{pmatrix},\qquad \mathbf{w}_2(t) = \begin{pmatrix}\sin t\\ -\cos t\end{pmatrix}.\]

So \(\mathbf{x}_h(t) = c_1 \begin{pmatrix}\cos t\\ \sin t\end{pmatrix} + c_2 \begin{pmatrix}\sin t\\ -\cos t\end{pmatrix}\).

Step 2 — Split the forcing and check for resonance

Write \(\mathbf{g}(t) = \mathbf{g}_1 + \mathbf{g}_2(t)\) with

\[\mathbf{g}_1 = \begin{pmatrix}3\\ -2\end{pmatrix}\ (\text{constant}),\qquad \mathbf{g}_2(t) = \begin{pmatrix}1\\ -2\end{pmatrix}e^{t}.\]

Constant piece: characteristic value \(\alpha = 0\), which is not in \(\{\pm i\}\) — no resonance. Trial form: \(\mathbf{x}_{p,1} = \mathbf{a}\).

Exponential piece: \(\alpha = 1\), which is not in \(\{\pm i\}\) either — no resonance. Trial form: \(\mathbf{x}_{p,2} = \mathbf{c}\,e^{t}\).

By superposition, \(\mathbf{x}_p(t) = \mathbf{a} + \mathbf{c}\,e^{t}\).

Step 3 — Solve for the constant vector \(\mathbf{a}\)

Substituting \(\mathbf{x}_{p,1} = \mathbf{a}\) into \(\mathbf{x}' = A\mathbf{x} + \mathbf{g}_1\) gives \(\mathbf{0} = A\mathbf{a} + \mathbf{g}_1\), i.e. \(A\mathbf{a} = -\mathbf{g}_1 = (-3, 2)^T\):

\[\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}\begin{pmatrix}a_1\\ a_2\end{pmatrix} = \begin{pmatrix}-3\\ 2\end{pmatrix} \ \Longrightarrow\ -a_2 = -3,\ a_1 = 2 \ \Longrightarrow\ \mathbf{a} = \begin{pmatrix}2\\ 3\end{pmatrix}.\]

Step 4 — Solve for the exponential vector \(\mathbf{c}\)

Substituting \(\mathbf{x}_{p,2} = \mathbf{c}\,e^{t}\) into \(\mathbf{x}' = A\mathbf{x} + \mathbf{g}_2\) and cancelling \(e^{t}\) gives \(\mathbf{c} = A\mathbf{c} + (1,-2)^T\), i.e. \((I - A)\mathbf{c} = (1,-2)^T\):

\[(I - A) = \begin{pmatrix}1 & 1\\ -1 & 1\end{pmatrix},\quad \det(I - A) = 2,\quad (I-A)^{-1} = \frac{1}{2}\begin{pmatrix}1 & -1\\ 1 & 1\end{pmatrix}.\]

\[\mathbf{c} = \frac{1}{2}\begin{pmatrix}1 & -1\\ 1 & 1\end{pmatrix}\begin{pmatrix}1\\ -2\end{pmatrix} = \frac{1}{2}\begin{pmatrix}3\\ -1\end{pmatrix}.\]

Step 5 — General solution

\[\mathbf{x}_p(t) = \begin{pmatrix}2\\ 3\end{pmatrix} + \frac{1}{2}\begin{pmatrix}3\\ -1\end{pmatrix}e^{t}.\]

\[\boxed{\ \mathbf{x}(t) = c_1 \begin{pmatrix}\cos t\\ \sin t\end{pmatrix} + c_2 \begin{pmatrix}\sin t\\ -\cos t\end{pmatrix} + \begin{pmatrix}2\\ 3\end{pmatrix} + \frac{1}{2}\begin{pmatrix}3\\ -1\end{pmatrix}e^{t}\ }\]

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6. Undetermined Coefficients with Resonance

Use the method of undetermined coefficients to determine the general solution of

\[ \mathbf{x}' = \begin{pmatrix} -3 & \phantom{-}4 \\ -2 & \phantom{-}3 \end{pmatrix}\mathbf{x} + \begin{pmatrix} -3 \\ -1 \end{pmatrix} e^{t}. \]

Hint: Before guessing the form of \(\mathbf{x}_p\), compute the eigenvalues of the coefficient matrix. You will find that \(1\) is one of them — so the natural guess \(\mathbf{a}\,e^{t}\) fails. Modify the guess by including a \(t\,e^{t}\) term.

SolutionSolution

Step 1 — Eigenvalues and eigenvectors

For \(A = \begin{pmatrix}-3 & 4\\ -2 & 3\end{pmatrix}\), \(\operatorname{tr}(A) = 0\) and \(\det(A) = -9 - (-8) = -1\), so

\[\det(A - \lambda I) = \lambda^2 - 1 = (\lambda - 1)(\lambda + 1) = 0 \ \Longrightarrow\ \lambda_1 = 1,\ \lambda_2 = -1.\]

\(\lambda_1 = 1\). \(A - I = \begin{pmatrix}-4 & 4\\ -2 & 2\end{pmatrix}\). Row 1: \(-4v_1 + 4v_2 = 0\), so \(v_1 = v_2\). Take \(v_2 = 1\): \(\mathbf{v}_1 = (1, 1)^T\).

\(\lambda_2 = -1\). \(A + I = \begin{pmatrix}-2 & 4\\ -2 & 4\end{pmatrix}\). Row 1: \(-2v_1 + 4v_2 = 0\), so \(v_1 = 2v_2\). Take \(v_2 = 1\): \(\mathbf{v}_2 = (2, 1)^T\).

Step 2 — Resonance and trial form

The forcing exponent \(\alpha = 1\) matches the eigenvalue \(\lambda_1 = 1\), so the guess \(\mathbf{a}\,e^{t}\) is insufficient. Use

\[\mathbf{x}_p(t) = \mathbf{a}\,t\,e^{t} + \mathbf{b}\,e^{t}.\]

Substituting into \(\mathbf{x}' = A\mathbf{x} + \mathbf{g}_0 e^{t}\) with \(\mathbf{g}_0 = (-3, -1)^T\) and collecting the coefficients of \(t\,e^{t}\) and \(e^{t}\) separately yields the two master equations

\[(A - I)\mathbf{a} = \mathbf{0},\qquad (A - I)\mathbf{b} = \mathbf{a} - \mathbf{g}_0.\]

Step 3 — Solve for \(\mathbf{a}\) using the solvability condition

The first equation forces \(\mathbf{a}\) to be an eigenvector for \(\lambda = 1\), so \(\mathbf{a} = k\,(1,1)^T\) for some scalar \(k\).

For the second equation to be solvable, \(\mathbf{a} - \mathbf{g}_0\) must be orthogonal to every left null vector of \(A - I\). A left null vector \(\mathbf{w}\) satisfies \(\mathbf{w}^T(A - I) = \mathbf{0}\): with \(A - I = \begin{pmatrix}-4 & 4\\ -2 & 2\end{pmatrix}\), this gives \(-4w_1 - 2w_2 = 0\), i.e. \(w_2 = -2w_1\). Take \(\mathbf{w} = (1, -2)^T\).

Imposing \(\mathbf{w}^T(\mathbf{a} - \mathbf{g}_0) = 0\):

\[(1)(k + 3) + (-2)(k + 1) = (k + 3) - 2(k + 1) = -k + 1 = 0 \ \Longrightarrow\ k = 1.\]

So \(\mathbf{a} = (1, 1)^T\).

Step 4 — Solve for \(\mathbf{b}\)

\[(A - I)\mathbf{b} = \mathbf{a} - \mathbf{g}_0 = \begin{pmatrix}1 + 3\\ 1 + 1\end{pmatrix} = \begin{pmatrix}4\\ 2\end{pmatrix}.\]

Both rows of \((A - I)\mathbf{b} = (4,2)^T\) reduce to \(-b_1 + b_2 = 1\). Choose \(b_1 = 0\), so \(b_2 = 1\): \(\mathbf{b} = (0, 1)^T\). (Any other choice differs by a multiple of \(\mathbf{v}_1 = (1,1)^T\) and absorbs into the \(c_1\) term of \(\mathbf{x}_h\).)

Step 5 — General solution

\[\mathbf{x}_p(t) = \begin{pmatrix}1\\ 1\end{pmatrix}t\,e^{t} + \begin{pmatrix}0\\ 1\end{pmatrix}e^{t}.\]

\[\boxed{\ \mathbf{x}(t) = c_1 \begin{pmatrix}1\\ 1\end{pmatrix}e^{t} + c_2 \begin{pmatrix}2\\ 1\end{pmatrix}e^{-t} + \begin{pmatrix}1\\ 1\end{pmatrix}t\,e^{t} + \begin{pmatrix}0\\ 1\end{pmatrix}e^{t}\ }\]

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7. Trigonometric Resonance (Form Only)

Without solving for the coefficients, determine the correct form of the particular solution \(\mathbf{x}_p(t)\) for

\[ \mathbf{x}' = \begin{pmatrix} 2 & -5 \\ 1 & -2 \end{pmatrix}\mathbf{x} + \begin{pmatrix} -2\cos t + 4\sin t \\ 2\sin t \end{pmatrix}. \]

Hint: First find the eigenvalues of the coefficient matrix and compare them to \(\pm i\), the characteristic values associated with \(\cos t\) and \(\sin t\). Adjust the guess for \(\mathbf{x}_p\) if a resonance occurs.

SolutionSolution

Step 1 — Eigenvalues of the coefficient matrix

For \(A = \begin{pmatrix}2 & -5\\ 1 & -2\end{pmatrix}\), \(\operatorname{tr}(A) = 0\) and \(\det(A) = -4 - (-5) = 1\). Hence

\[\det(A - \lambda I) = \lambda^2 + 1 = 0 \ \Longrightarrow\ \lambda = \pm i.\]

Step 2 — Resonance check

The forcing is a combination of \(\cos t\) and \(\sin t\), so its angular frequency is \(\beta = 1\). The critical test values for trigonometric forcing are \(\pm i\beta = \pm i\), which are precisely the eigenvalues of \(A\). Resonance occurs.

Step 3 — Correct form of \(\mathbf{x}_p\)

With resonance at \(\pm i\beta = \pm i\), the trial form must include a \(t\)-multiplied piece and an unmultiplied piece:

\[\boxed{\ \mathbf{x}_p(t) = t\bigl[\mathbf{a}\cos t + \mathbf{b}\sin t\bigr] + \mathbf{c}\cos t + \mathbf{d}\sin t\ }\]

with four unknown constant vectors \(\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d} \in \mathbb{R}^2\). Omitting the \(\mathbf{c}\cos t + \mathbf{d}\sin t\) piece would give an underdetermined system when you substitute, just as in the exponential-resonance case.

Remark. If you were to solve for the coefficients, substituting \(\mathbf{x}_p\) into the ODE and matching \(t\cos t\) and \(t\sin t\) terms would first force \(\mathbf{a}\) and \(\mathbf{b}\) to be the real and imaginary parts of an eigenvector for \(\lambda = i\) (up to scale). The leftover \(\cos t\) and \(\sin t\) terms then give two \(2\times 2\) linear systems for \(\mathbf{c}\) and \(\mathbf{d}\), together with a solvability condition that pins down the scale of \(\mathbf{a}, \mathbf{b}\) — exactly analogous to Question 6.

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