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Coordinated Differential Equations

Thomas W. Judson, Austin Eide, Adam Larios

Appendix C Hints and Answers to Selected Exercises

1 A First Look at Differential Equations
1.1 Modeling with Differential Equations
1.1.9 Exercises

1.1.9.29.

Hint.
Rewriting the differential equation as \(x' + a x - q(t) = 0\) and using the fact that
\begin{equation*} x'(t) = -a Ce^{-at} - a e^{-at} \int_{t_0}^t e^{as}q(s) \, ds + b(t), \end{equation*}
we see that
\begin{align*} x'(t) + a x(t) - q(t) & = -a Ce^{-at} - a e^{-at} \int_{t_0}^t e^{as}q(s) \, ds + q(t)\\ & + \; aCe^{-at} + ae^{-at} \int_{t_0}^t e^{as}q(s) \, ds - q(t)\\ & = 0. \end{align*}

1.1.9.31.

Hint.
Think about the limit of the interaction term as the number of prey becomes very large.

1.4 Analyzing Equations Numerically
1.4.6 Exercises

1.4.6.7.

Hint.
Hints for part (2):
  • For fixed \(i\) show that
    \begin{align*} a_{i + 1} & \leq (1 + s)a_i + t\\ & \leq (1 + s)[(1 + s)a_{i - 1}+ t] + t\\ & \leq (1 + s)\{ (1 + s)[(1 + s) a_{i - 2} + t]+ t\} + t\\ & \vdots\\ & \leq (1 + s)^{i+1}a_0 + [1 + (1 + s) + (1+s)^2 + \cdots + (1 + s)^i]t. \end{align*}
  • Now use a geometric sum to show that
    \begin{equation*} a_{i + 1} \leq (1 + s)^{i + 1} a_0 + \frac{t}{s}[(1 + s)^{i + 1} - 1] = (1 + s)^{i + 1} \left( \frac{t}{s} + a_0 \right) - \frac{t}{s}. \end{equation*}
  • Apply part (1) to derive
    \begin{equation*} a_{i + 1} \leq e^{(i + 1)s} \left( \frac{t}{s} + a_0 \right) - \frac{t}{s}. \end{equation*}

1.5 First-Order Linear Equations
1.5.8 Exercises

1.5.8.16.

Hint.
\(y = \dfrac{3 x^{4} + 8 x^{3} + 6 x^{2} + 12}{6 (x + 1)^2}\)

1.5.8.17.

Hint.
\(y = (e^x + 1)/(x+1)^2\)

1.5.8.19.

Hint.
\(y = e^x \sec x\)

1.5.8.20.

Hint.
\(y = \sin x / (x + 2)\)

1.5.8.21.

Hint.
If \(x(t)\) is the amount of salt in the tank at time \(t\text{,}\) we know that \(x(0) = 10\text{.}\) The volume of the tank is \(V = 200 + 5t\text{.}\) We can model the amount of salt in the tank at time \(t\) with a differential equation,
\begin{align*} \frac{dx}{dt} & = \text{rate in} - \text{rate out}\\ & = 10(0.1) - 5 \frac{x}{V}\\ & = 1 - 5\frac{x}{200+ 5t}\\ & = 1 - \frac{x}{40 + t}. \end{align*}
The resulting equation
\begin{equation*} \frac{dx}{dt} + \frac{1}{40 + t}x = 1 \end{equation*}
is a first order linear differential equation. An integrating factor for this equation is given by
\begin{equation*} \mu(t) = \exp\left(\int \frac{1}{40 + t} \, dt\right) = 40 + t. \end{equation*}
Multiplying both sides of the differential equation by \(\mu(t)\text{,}\) we have
\begin{equation*} \frac{d}{dt} [(40 + t)x] = (40 + t) \frac{dx}{dt} + x = (40 + t)\left( \frac{dx}{dt} + \frac{1}{40 + t}x \right) = 40 + t. \end{equation*}
Integrating both sides of this equation, we obtain
\begin{equation*} (40 + t)x = 40t + \frac{t^2}{2} + C. \end{equation*}
Using the intial condition \(x(0) = 10\text{,}\) we can determine that \(C = 400\) or
\begin{equation*} x(t) = \frac{t^2 + 80t + 800}{2t + 80}. \end{equation*}
The tank is full at time \(t = 400/5 = 80\text{,}\) and the tank contains \(x(80) = 170/3 \approx 56.67\) kilograms of salt when the tank is full.

1.5.8.26.

Hint.
  1. If \(y = y_1 + 1/v\text{,}\) then \(y' = y_1' - v'/v^2\text{.}\) Substituting into our original equation, we obtain
    \begin{equation*} y' = y_1' - \frac{v'}{v^2} = p + q y_1 + r y_1^2 - \frac{v'}{v^2}. \end{equation*}
    On the other hand,
    \begin{align*} y' & = p + q \left(y_1 + \frac{1}{v} \right) + r \left(y_1 + \frac{1}{v} \right)^2\\ & = p + q y_1 + \frac{q}{v} + r y_1^2 + \frac{2r y_1}{v} + \frac{r}{v^2}\\ & = y_1' + \frac{q}{v} + \frac{2r y_1}{v} + \frac{r}{v^2}. \end{align*}
    Therefore,
    \begin{equation*} - \frac{v'}{v^2} = \frac{q}{v} + \frac{2r y_1}{v} + \frac{r}{v^2}, \end{equation*}
    which is just the first-order linear equation
    \begin{equation*} v' + [q(t) + 2 r(t) y_1(t)]v = - r(t). \end{equation*}
  2. \begin{equation*} y = t + \frac{1}{C - t} \end{equation*}
  3. \begin{equation*} y(t) = \frac{1}{C \cos t - \sin t} + \sin t \end{equation*}
  4. \begin{equation*} y(t) = 2 + \frac{1}{C e^t - 1} \end{equation*}

1.6 Existence and Uniqueness of Solutions
1.6.5 Exercises

1.6.5.2.

Hint.
(b) Make sure that the derivative of \(y(t)\) exists at \(t = t_0\text{.}\)

2 Second-Order Linear Equations
2.1 Homogeneous Linear Equations
2.1.7 Exercises

2.1.7.31.

Hint.
Pay careful attention to units.

2.1.7.32.

Hint.
  1. Observe that
    \begin{align*} a x_1'' + b_1' + cx_1 & = a \left(\frac{-b}{2a}\right)^2e^{-bt/2a} + b \left( \frac{-b}{2a} \right) e^{-bt/2a} + c e^{-bt/2a}\\ & = e^{-bt/2a} \left( \frac{b^2}{4a} - \frac{b^2}{2a} + c \right)\\ & = e^{-bt/2a} \left( \frac{-b^2 + 4ac}{4a} \right)\\ & = 0. \end{align*}
  2. If \(y = v(t) x_1(t) = v(t) e^{-bt/2a}\) is a solution to our differential equation, then
    \begin{align*} a y'' + b y' + cy & = a (v''x_1 + 2 v'x_1' + vx_1'' ) + b(v' x_1 + v x_1') + cv x_1\\ & = a v''x_1 + 2a v'x_1' + bv' x_1 + v(a x_1'' +b x_1' + c x_1)\\ & = a v'' e^{-bt/2a} + \left[2a \left( \frac{-b}{2a} \right) e^{-bt/2a} + b e^{-bt/2a} \right] v'\\ & = a v'' e^{-bt/2a}\\ & = 0. \end{align*}
    Since \(a \neq 0\text{,}\) we know that \(v'' = 0\text{.}\) Hence, \(v(t) = c_1 + c_2 t\text{.}\)

2.1.7.33.

Hint.
  1. \begin{align*} x'' + px' + qx & = (v''x_1 + 2 v' x_1' + vx_1'') + p(v'x_1 + vx_1') + q(vx_1)\\ & = x_1 v'' + 2v' x_1' + p x_1 v' + v (x_1'' + p x_1' + q x_1)\\ & = x_1 v'' +(2x_1' + px_1)v' \\ & = 0. \end{align*}
  2. If \(u = v'\text{,}\) then \(x_1 u' +(2x_1' + px_1)u= 0\text{.}\)
  3. If \(x_1(t) = 1/t\text{,}\) then
    \begin{equation*} 2 t^2 x_1'' + 3t x_1' - x_1 = 2 t^2 \left(\frac{2}{t^3}\right) + 3t \left(\frac{-1}{t^2}\right) - \frac{1}{t} = 0. \end{equation*}
    If we assume that \(x = v/t\) is a second solution, then
    \begin{equation*} 2 t^2 x'' + 3t x' - x = 2tv'' - v' = 0. \end{equation*}
    If we let \(u = v'\text{,}\) then a solution of \(2tu' - u = 0\) is \(u = \sqrt{t}\) and \(v = \int \sqrt{t} \, dt = 2 t^{3/2} / 3\text{.}\) Therefore, the second solution to our equation is
    \begin{equation*} x = \frac{v}{t} = \frac{2}{3} \sqrt{t}. \end{equation*}

2.2 Forcing
2.2.7 Exercises

2.2.7.2.

Answer.
The solution to the associated homogeneous equation is \(c_1e^{2x} + c_2e^{-x}\text{,}\) and we do not see any \(e^{2x}\) or \(e^{-x}\) in the right-hand-side, so we look at Table 2.2.12 and see the solution has the form \(y(x) = Ax^2 + Bx + C\text{.}\) We then get \(y'(x) = 2Ax + B, y''(x) = 2A\text{,}\) so
\begin{equation*} y'' - y' - 2y = 2A - (2Ax + B) - 2(Ax^2 + Bx + C) = \end{equation*}
\begin{equation*} -2Ax^2 + (-2A - 2B)x + (2A - B - 2C) = 4x^2 - 2x - 9 \end{equation*}
We then have to equate the coefficients. \(-2A = 4\) gives \(A = -2\text{,}\) \(-2A - 2B = -2\) gives \(B = -A + 1 = 3\text{,}\) and \(2A - B - 2C = -9\) gives \(C = 1\text{,}\) so the answer is \(y(x) = -2x^2 + 3x + 1\)

2.2.7.25.

Hint.
Suppose that that \(f(t)\) and \(g(t)\) are linearly dependent on an interval \(I = (a, b)\text{.}\) Then one function is a multiple of the other, say \(f(t) = c g(t)\text{.}\) Thus, \(f'(t) = cg'(t)\) and
\begin{equation*} W(f, g) = f(t) g'(t) - f'(t) g(t) = c g(t) g'(t) - cg'(t) g(t) = 0. \end{equation*}
Conversely, suppose that \(W(f, g) = 0\) for all \(t\) in \((a, b)\text{.}\) If \(g = 0\text{,}\) then \(0 f = g\) and the two functions are linearly dependent. Assume that \(g(t_0) \neq 0\) for some \(t_0\) in \((a, b)\text{.}\) Since \(g\) is differentiable, it must also be continuous and there is some interval \((c, d)\) contained in \((a, b)\) such that \(t_0 \in (c, d)\) and \(g\) does not vanish on this interval. Therefore,
\begin{equation*} \frac{d}{dt} \left( \frac{f}{g} \right) = \frac{f' g - f g'}{g^2} = - \frac{W(f,g)}{g^2} = 0, \end{equation*}
and \(f/g\) is constant on the interval \((c, d)\text{.}\) Thus, \(f(t_0) = c g(t_0)\) and \(f'(t_0) = c g'(t_0)\text{.}\) Since \(f\) and \(cg\) are both solutions to the differential equation \(y'' + p y' + q y = 0\) and have the same initial condition, \(f(t) = cg(t)\) for all \(t \in (a, b)\) by the existence and uniqueness theorem. Consequently, \(f\) and \(g\) are linearly dependent.

2.2.7.26.

Hint.
  1. We can rewrite \(2 t^2 y'' + 3ty' - y = 0\) as
    \begin{equation*} y'' + \frac{3}{2t} y' - \frac{1}{2t^2} y = 0. \end{equation*}
    Since \(p(t) = 1/2t\text{,}\) Abel’s Theorem tells us that
    \begin{equation*} W(y_1, y_2) = c \exp\left( - \int \frac{3}{2t} \, dt \right) = c \exp\left( - \frac{3}{2} \ln t \right) = ct^{-3/2}. \end{equation*}
  2. Since \(y_1\) and \(y_2\) are solutions to our differential equation, we know that
    \begin{gather*} y_1'' + p(t) y_1' + q(t) y_1 = 0\\ y_2'' + p(t) y_2' + q(t) y_2 = 0. \end{gather*}
    Multiplying the first equation by \(y_2\) and the second equation by \(y_1\) and subtracting, we obtain
    \begin{equation} (y_1 y_2'' - y_1'' y_2) + p(t) (y_1 y_2' - y_1' y_2) = 0.\tag{2.2.5} \end{equation}
    If
    \begin{equation*} W(t) = W(y_1, y_2) = y_1 y_2' - y_1' y_2, \end{equation*}
    then
    \begin{equation*} W' = y_1 y_2'' - y_1'' y_2, \end{equation*}
    and equation (2.2.5) becomes
    \begin{equation*} W' + p(t) W = 0. \end{equation*}
    This equation is separable with solution
    \begin{equation*} W(t) = c \exp\left( - \int p(t) \, dt \right). \end{equation*}

2.2.7.27.

Hint.
  1. If \(y_p = u_1y_1 + u_2 y_2\text{,}\) then
    \begin{gather*} y_p' = u_1' y_1 + u_1 y_1' + u_2' y_2 + u_2 y_2' = u_1 y_1' + u_2 y_2'\\ y_p'' = u_1' y_1' + u_1 y_1'' + u_2' y_2' + u_2 y_2''. \end{gather*}
    Substituting these expressions into equation (2.2.6), we have
    \begin{align*} y_p'' + p y_p' + q y_p & = ( u_1' y_1' + u_1 y_1'' + u_2' y_2' + u_2 y_2'') + p(u_1 y_1' + u_2 y_2')\\ & + q(u_1y_1 + u_2 y_2)\\ & = u_1[y_1'' + p y_1' +q y_1] + u_2[y_2'' + p y_2' +q y_2] + u_1' y_1' + u_2' y_2'\\ & = u_1' y_1' + u_2' y_2'\\ & = f(t). \end{align*}
  2. If we solve the system
    \begin{align*} u_1'(t) y_1(t) + u_2'(t) y_2(t) & = 0\\ u_1'(t) y_1' (t)+ u_2'(t) y_2'(t) & = f(t). \end{align*}
    for \(u_1'\) and \(u_2'\text{,}\) we obtain
    \begin{align*} u_1'(t) & = \frac{- y_2(t) f(t)}{W[y_1, y_2](t)}\\ u_2'(t) & = \frac{y_1(t) f(t)}{W[y_1, y_2](t)}. \end{align*}
  3. Integrate the two equations from part (2).
  4. The general solution to the homogeneous equation \(y'' + 4y = 0\) is
    \begin{equation*} y_h = c_1 \cos 2t + c_2 \sin 2t. \end{equation*}
    To find a particular solution, assume that the solution has the form
    \begin{equation*} y_p = u_1(t) \cos 2t + u_2(t) \sin 2t. \end{equation*}
    By part (2)
    \begin{align*} u_1'(t) & = -3 \cos t\\ u_2'(t) & = \frac{3}{2} \csc t - 3 \sin t. \end{align*}
    Integrating, we obtain
    \begin{align*} u_1(t) & = -3 \sin t\\ u_2(t) & = \frac{3}{2} \ln| \csc t - \cot t| + 3 \cos t. \end{align*}
    Therefore,
    \begin{align*} y_p(t) & = u_1(t) y_1(t) + u_2(t) y_2(t)\\ & = -3 \sin t \cos 2t + \left[\frac{3}{2} \ln| \csc t - \cot t| + 3 \cos t\right] \sin 2t, \end{align*}
    and the general solution is
    \begin{align*} y & = y_h + y_p\\ & = c_1 \cos 2t + c_2 \sin 2t -3 \sin t \cos 2t + \left[\frac{3}{2} \ln| \csc t - \cot t| + 3 \cos t\right] \sin 2t. \end{align*}

2.3 Sinusoidal Forcing
2.3.5 Exercises

2.3.5.2.

Hint.
Assume the complex solution has form \(y_c = A e^{3it}\text{.}\)

2.3.5.12.

Hint.
Assume the complex solution has form \(y_c = A e^{3it}\text{.}\)

3 The Laplace Transform
3.1 The Laplace Transform
3.1.7 Exercises

3.1.7.15.

Answer.
If the function in question were simply \(\dfrac{1}{s^4}\text{,}\) we would observe that \(\mathcal{L}(t^3)(s) = \dfrac{6}{s^4}\text{,}\) so the answer would be \(\dfrac{t^3}{6}\text{.}\) Since we subtract 2 from the argument \(s\text{,}\) we multiply by \(e^{2t}\text{,}\) so the answer is \(\dfrac{e^{2t}t^3}{6}\text{.}\)

3.1.7.16.

Answer.
We will split this into partial fractions via
\begin{equation*} \dfrac{1}{(s + 1)^2(s^2 - 4)} = \dfrac{A(s+1)+B}{(s+1)^2} + \dfrac{C}{s+2} + \dfrac{D}{s-2} \end{equation*}
Clearing denominators gives
\begin{equation*} 1 = [A(s+1) + B](s^2-4) + C(s-2)(s+1)^2 + D(s+2)(s+1)^2 \end{equation*}
Plugging in \(s = -1\) yields \(-3B = 1\text{,}\) so \(B = \frac{-1}{3}\text{.}\) Plugging in \(s = 2\) yields \(36D = 1\text{,}\) so \(D = \frac{1}{36}\text{.}\) Plugging in \(s = -2\) yields \(-4C = 1\text{,}\) so \(C = \frac{-1}{4}\text{.}\) Plugging in \(s = 0\) and using our values for \(B, C,\) and \(D\) yields \(-4A + \frac{17}{9} = 1\text{,}\) so \(A = \frac29\) So we are to find the inverse Laplace of
\begin{equation*} F(s) = \dfrac{\frac29(s+1)-\frac13}{(s+1)^2}-\dfrac{1/4}{s+2}+\dfrac{1/36}{s-2} \end{equation*}
\begin{equation*} = \dfrac{2/9}{s+1} - \dfrac{1/3}{(s+1)^2}-\dfrac{1/4}{s+2}+\dfrac{1/36}{s-2} \end{equation*}
This gives us \(f(t) = \frac29 e^{-t} - \frac13 te^{-t}-\dfrac14 e^{-2t} + \dfrac{1}{36}e^{2t}\)

3.2 Solving Initial Value Problems
3.2.6 Exercises

3.2.6.1.

Answer.
Taking the Laplace transformation of both sides yields
\begin{equation*} s^2Y(s) - s - 2sY(s) + 2 - 3Y(s) = \frac{24}{s+3} \end{equation*}
so
\begin{equation*} Y(s) = \frac{24}{(s-3)(s+1)(s+3)} + \frac{s-2}{(s-3)(s+1)} \end{equation*}
A partial fractions decomposition reveals
\begin{equation*} Y(s) = \frac{1}{s-3} + \frac{-3}{s+1} + \frac{2}{s+3} + \frac{1/4}{s-3} + \frac{3/4}{s+1} \end{equation*}
So the answer is
\begin{equation*} y(t) = e^{3t} - 3e^{-t} + 2e^{-3t} + \frac{1}{4}e^{3t} + \frac34 e^{-t} \end{equation*}
\begin{equation*} y(t) = \frac54 e^{3t} - \frac{9}{4}e^{-t} + 2e^{-3t} \end{equation*}

3.4 Convolution
3.4.5 Exercises

3.4.5.13.

Answer.
Taking the Laplace transformation of both sides yields
\begin{equation*} 2s^2Y(s) + 5sY(s) + 2Y(s) = G(s) \end{equation*}
Therefore, \(Y(s) = \frac{G(s)}{2s^2+5s+2}\text{.}\) We now want to factor \(2s^2+5s+2\text{.}\) We multiply the \(s^2\) and constant coefficients to get 4, and search for two numbers multiplying to 4 and adding to the \(s\) coefficient (\(5\)). We see that \(4\) and \(1\) work, and we can write \(2s^2 + 5s + 2 = 2s^2 + 4s + s + 2 = 2s(s + 2) + (s + 2) =\)\((2s + 1)(s + 2)\text{,}\) so we get
\begin{equation*} y(t) = \mathcal{L}^{-1}\lbrace\frac{G(s)}{(2s+1)(s+2)}\rbrace = g(t)*\mathcal{L}^{-1}\lbrace\frac{1}{(2s+1)(s+2)}\rbrace \end{equation*}
We could use partial fractions or another convolution to calculate \(\mathcal{L}^{-1}\lbrace\frac{1}{(2s+1)(s+2)}\rbrace\text{.}\) If we choose convolution, we see \(\mathcal{L}^{-1}\lbrace\frac{1}{(2s+1)(s+2)}\rbrace\) \(= \frac12e^{-t/2}*e^{-2t}\) \(= \frac12\int_0^te^{-\tau /2}e^{-2(t-\tau)}d\tau\) \(= \frac13(e^{-t/2}-e^{-2t})\text{,}\) so the final answer is
\begin{equation*} \frac13 g(t)*(e^{-t/2}-e^{-2t}) = \frac13\int_0^tg(\tau)(e^{\frac{\tau-t}{2}}-e^{2(\tau-t)})d\tau \end{equation*}

5 Linear Systems and Linearization
5.1 Linear Algebra in a Nutshell
5.1.6 Exercises

5.1.6.4.

Answer.
\(det(A) = 1(3) - 4(2) = -5\)

5.1.6.6.

Answer.
\(det(A) = 2(4) - (-6)(-2) = -4\)

5.2 Planar Systems
5.2.6 Exercises

5.2.6.2.

Answer.
This is equivalent to \(x' = Ax\text{,}\) where \(A = \begin{pmatrix}-1 & 2 \\ 2 & 2\end{pmatrix}\text{.}\) The characteristic polynomial is \((-1-\lambda)(2-\lambda) - 2(2)\) \(= \lambda^2 - \lambda - 6\text{,}\) with zeros \(\lambda = -2, 3\text{.}\) We then get
\begin{equation*} A - (-2)I = \begin{pmatrix}-1 + 2 & 2 \\ 2 & 2 + 2\end{pmatrix} = \begin{pmatrix}1 & 2 \\ 2 & 4\end{pmatrix} \end{equation*}
So \(\lambda = -2\) has corresponding eigenvector \(\begin{pmatrix}-2 \\ 1\end{pmatrix}\text{.}\) Analogously,
\begin{equation*} A - 3I = \begin{pmatrix}-1 - 3 & 2 \\ 2 & 2 - 3\end{pmatrix} = \begin{pmatrix}-4 & 2 \\ 2 & -1\end{pmatrix} \end{equation*}
So \(\lambda = 3\) has corresponding eigenvector \(\begin{pmatrix}1 \\ 2\end{pmatrix}\text{.}\) Therefore the set of solutions is
\begin{equation*} c_1e^{-2t}\begin{pmatrix}-2 \\ 1\end{pmatrix} + c_2e^{3t}\begin{pmatrix}1 \\ 2\end{pmatrix} \end{equation*}

5.2.6.6.

Answer.
From exercise 2, we see the general solution to the system \(x' = -x + 2y, y' = 2x + 2y\) is
\begin{equation*} c_1e^{-2t}\begin{pmatrix}-2 \\ 1\end{pmatrix} + c_2e^{3t}\begin{pmatrix}1 \\ 2\end{pmatrix} \end{equation*}
Plugging in \(t = 0\) and setting the top to \(19\text{,}\) the bottom to \(8\text{,}\) we end up at the system
\begin{gather*} -2c_1 + c_2 = 19 \\ c_1 + 2c_2 = 8 \end{gather*}
Solving this yields \(c_1 = -6, c_2 = 7\text{,}\) and we thus have the solution
\begin{equation*} -6e^{-2t}\begin{pmatrix}-2 \\ 1\end{pmatrix} + 7e^{3t}\begin{pmatrix}1 \\ 2\end{pmatrix} \end{equation*}
which can also be written as
\begin{align*} x(t) = & 12e^{-2t} + 7e^{3t}\\ y(t) = & -6e^{-2t} + 14e^{3t} \end{align*}

5.2.6.10.

Hint.
Assume that your solution must be of the form
\begin{equation*} {\mathbf x}_p = \begin{pmatrix} a_2 t^2 + a_1 t + a_0 \\ b_2 t^2 + b_1 t + b_0. \end{pmatrix} \end{equation*}
This is called the method of undetermined coefficients.

5.3 Phase Plane Analysis of Linear Systems
5.3.6 Exercises

5.3.6.2.

Answer.
(a) The characteristic polynomial is \((4-\lambda)(-3-\lambda) + 6\text{,}\) with zeros (and thus eigenvalues) \(\lambda = -2, 3\)
(b) If the eigenvalues have the same sign, the one with a bigger absolute value is dominant. Since these eigenvalues have different signs, there is no dominant eigenvalue.
(c) For \(\lambda = -2\) we subtract \(-2\) from the main diagonal of the matrix to obtain \(A + 2I = \begin{pmatrix}6 \amp -2 \\ 3 \amp -1\end{pmatrix}\text{,}\) so \(v = \begin{pmatrix}1 \\ 3\end{pmatrix}\) is an eigenvector.
For \(\lambda = 3\text{,}\) we subtract \(3\) from the main diagonal to obtain \(A - 3I = \begin{pmatrix}1 \amp -2 \\ 3 \amp -6\end{pmatrix}\text{,}\) so \(v = \begin{pmatrix}2 \\ 1\end{pmatrix}\)
(d) There are two straight-line solutions, one corresponding to each eigenvalue. For \(\lambda = -2\) we got \(\begin{pmatrix}1 \\ 3\end{pmatrix}\text{,}\) so the solution set here is the set of multiples of \(\begin{pmatrix}1 \\ 3\end{pmatrix}\text{,}\) or equivalently, the line \(1y = 3x\text{.}\)
For \(\lambda = 3\) we got the eigenvalue \(\begin{pmatrix}2 \\ 1\end{pmatrix}\text{,}\) so the straight-line solution is the line passing through \(\begin{pmatrix}2 \\ 1\end{pmatrix}\) and the origin, or the line \(2y = 1x\text{.}\)
(e) Since there is one positive and one negative eigenvalue, the equilibrium solution is a saddle.
(f)
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