Chapter 18

Draw the nullclines and some direction arrows and analyze the equilibria of the following competition models. $$ \begin{array}{ll} \text { a. } & x^{\prime}(t)=0.2 \times x(t) \times(1-0.2 x(t)-0.4 y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1-0.4 x(t)-0.5 y(t)) \\ \text { b. } & x^{\prime}(t)=0.2 \times x(t) \times(1-0.2 x(t)-0.8 y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1-0.4 x(t)-0.5 y(t)) \\ \text { c. } & x^{\prime}(t)=0.2 \times x(t) \times(1-0.6 x(t)-0.4 y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1-0.4 x(t)-0.5 y(t)) \\ \text { d. } & x^{\prime}(t)=0.2 \times x(t) \times(1-0.4 x(t)-0.4 y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1-0.3 x(t)-0.5 y(t)) \end{array} $$

Find the local linear approximation to the system $$ \begin{array}{l} x^{\prime}=x-x^{2}-x y \\ y^{\prime}=y-0.5 x y-2 y^{2} \end{array} $$ a. At the equilibrium point (0,0) . b. At the equilibrium point (0,0.5) . c. At the equilibrium point (1,0) . d. At the equilibrium point \((2 / 3,1 / 3)\). For each of the local linear approximations, determine whether it is stable.

Show by substitution that if \(y_{1}(t)\) and \(y_{2}(t)\) are solutions to \(y^{\prime \prime}(t)+p y^{\prime}(t)+q y(t)=0\) and each of \(C_{1}\) and \(C_{2}\) is a number then $$ y(t)=C_{1} y_{1}(t)+C_{2} y_{2}(t) $$ is a solution to \(y^{\prime \prime}(t)-p y^{\prime}(t)+q y(t)=0\).

For each of the systems, find all of the equilibrium points and determine whether the system is asymptotically stable at each equilibrium point. a. \(\quad \begin{array}{lll}x^{\prime}=-x y & \text { b. } \quad x^{\prime}=x-2 x y \\ y^{\prime}=1-x-y & \quad y^{\prime}=y+x y \\\ x^{\prime}=2-x^{2}-y^{2} & \text { d. } \quad x^{\prime}=5-x^{2}-y^{2}\end{array}\) c. \(\quad y^{\prime}=1-x y \quad\) d. \(\quad y^{\prime}=2-x y\)

Show by substitution that if \(r_{1}\) is the only root of \(r^{2}+p r+q=0\) then \(y=t e^{r_{1} t}\) is a solution to \(y^{\prime \prime}(t)+p y^{\prime}(t)+q y(t)=0\). Note: If \(r_{1}\) is the only root to \(r^{2}+p r+q=0\) then \(p^{2}-4 q=0\) and \(r_{1}=-p / 2\).

Is the equilibrium point, \((1.25,1.0),\) of predator-prey equations 18.58 , $$ x^{\prime}=0.2 x-0.2 x y $$ stable? $$ y^{\prime}=-0.1 y+0.08 x y $$ Is that equilibrium point asymptotically stable? As usual, you should explain your answers.

Show that \((1,1),(2,2),\) and (3,3) are equilibrium points of $$ \begin{aligned} x^{\prime} &=1-\frac{11}{6} x+\frac{10 x y}{11+x y} \\ y^{\prime} &=1-\frac{11}{6} y+\frac{10 x y}{11+x y} \end{aligned} $$ Determine which of these equilibrium points are stable. This example is patterned after a model Anderson and May \(^{3}\) present to suggest that antibody level may switch from a lower level stable unexposed state to an upper level stable actively immune state.

Exercise \(18.2 .3 \quad\) a. Show that for any number, \(x_{0},\) the solution to $$ \begin{array}{c} x(0)=x_{0} \quad x^{\prime}=-y \\ y(0)=0 \quad y^{\prime}=25 x \\ \text { is } x(t)=x_{0} \cos 5 t, \quad y(t)=5 x_{0} \sin 5 t \end{array} $$ b. Draw the graph in the \(x, y-\) plane of the solutions for \(x_{0}=1\) and \(x_{0}=0.5\). c. Show that the origin (0,0) is a stable equilibrium point of Equations 18.28 . d. Show that the origin (0,0) not an asymptotically stable equilibrium of Equations 18.28 .

Show that if \(y_{h}(t)\) solves \(y^{\prime \prime}(t)+p y^{\prime}(t)+q y(t)=0\) and \(y_{p}(t)\) solves \(y^{\prime \prime}(t)+p y^{\prime}(t)+q y(t)=f(t)\) then for any number \(C, y_{g}(t)=y_{p}(t)+C y_{h}(t)\) solves \(y^{\prime \prime}(t)+p y^{\prime}(t)+q y(t)=f(t)\)

The predator-prey equations assume that with no predator, the prey grows exponentially. Alternatively one might assume that with no predator, the prey grow according to a logistic (Verhultz) model. Write the predator-prey equations so that without predators the prey grows according to a logistic model. Find conditions for there to be an equilibrium for which both predator and prey exist, and determine the character of that equilibrium.

For each of the systems, determine whether the origin is stable, asymptotically stable, or unstable. \(\begin{array}{ll}\text { a. } \quad x^{\prime} & =2 x-5 y \\ & y^{\prime}=x-2 y \\ \text { c. } & x^{\prime}=-6 x-2 y \\ y^{\prime} & =2 x-1 y \\ \text { e. } & x^{\prime}=3 x-2 y \\ & y^{\prime}=2 x-1 y \\ \text { g. } & x^{\prime}=-x-5 y \\ \text { i. } & y^{\prime}=2 x-3 y \\ & x^{\prime}=-3 x+1 y \\ & y^{\prime}=2 x-2 y \\ \text { k. } & x^{\prime}=x-2 y \\ & y^{\prime}=2 x+1 y\end{array}\) \(x^{\prime}=2 x-5 y\) \(y^{\prime}=2 x-4 y\) \(x^{\prime}=-9 x+4 y\) d. \(y^{\prime}=-4 x-1 y\) \(x^{\prime}=\quad y / 2\) \(\begin{aligned} y^{\prime} &=-5 x-3 y \\ x^{\prime} &=\quad-5 y \end{aligned}\) h. \(y^{\prime}=2 x+2 y\) j. \(x^{\prime}=3 x+y\) \(\begin{aligned} y^{\prime} &=2 x+2 y \\ x^{\prime} &=6 x+4 y \end{aligned}\) l. \(y^{\prime}=2 x-y\)

Show that if \(y_{p, 1}(t)\) solves \(y^{\prime \prime}(t)+p y^{\prime}(t)+q y(t)=f_{1}(t)\) and \(y_{p, 2}(t)\) solves \(y^{\prime \prime}(t)+p y^{\prime}(t)+q y(t)=f_{2}(t),\) then for any numbers \(A\) and \(B\) $$ A y_{p, 1}(t)+B y_{p, 2} \quad \text { solves } \quad y^{\prime \prime}(t)+p y^{\prime}(t)+q y(t)=A f_{1}(t)+B f_{2}(t) $$

Consider two interacting populations, \(x\) and \(y,\) that are mutually symbiotic: the presence of \(x\) enhances the growth of \(y\) and the presence of \(y\) enhances the growth of \(x\). A dynamic relation between \(x\) and \(y\) may take the form $$ \begin{aligned} x^{\prime}(t) &=r_{x} \times x(t) \times(1-a x(t)+b y(t)) \\ y^{\prime}(t) &=r_{y} \times y(t) \times(1+c x(t)-d y(t)) \end{aligned} $$ Describe the roles of the parameters \(a, b, c,\) and \(d\) in Equations \(18.63 .\)

Compute and graph the solutions to $$ y^{\prime \prime}+p y^{\prime}+y=0 \quad y(0)=1 \quad y^{\prime}(0)=0 $$ for \(p=4, p=2, p=1, p=0,\) and \(p=-2\)

Draw the nullclines and some direction arrows and analyze the equilibria of the following symbiosis models. $$ \begin{aligned} \text { a. } \quad & x^{\prime}(t)=0.2 \times x(t) \times(1-0.5 x(t)+0.4 y(t)) \\\ & y^{\prime}(t)=0.1 \times y(t) \times(1+0.4 x(t)-0.5 y(t)) \\ \text { b. } \quad & x^{\prime}(t)=0.2 \times x(t) \times(1-0.8 x(t)+0.4 y(t)) \\\ & y^{\prime}(t)=0.1 \times y(t) \times(1+0.4 x(t)-0.5 y(t)) \\ \text { c. } \quad & x^{\prime}(t)=0.2 \times x(t) \times(1-0.5 x(t)+0.4 y(t)) \\\ & y^{\prime}(t)=0.1 \times y(t) \times(1+0.4 x(t)-0.2 y(t)) \\ \text { d. } x^{\prime}(t) &=0.2 \times x(t) \times(1-5 x(t)+10 y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1+2 x(t)-5 y(t)) \\ \text { e. } \quad & x^{\prime}(t)=0.2 \times x(t) \times(1-1.1 x(t)+y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1+x(t)-y(t)) \\ \text { f. } & x^{\prime}(t)=0.2 \times x(t) \times(1-0.9 x(t)+y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1+x(t)-y(t)) \end{aligned} $$

Exercise 18.3 .6 Interpret the output of the MATLAB program. $$ \begin{array}{l} \mathrm{A}=[0.770 .1 ; 0.0680 .9] \\ \mathrm{T} 1=[\exp (\mathrm{A}) \operatorname{expm}(\mathrm{A})] \\ \mathrm{T} 2=[\log (\mathrm{A}) \operatorname{logm}(\mathrm{A})] \\ \mathrm{T} 3=[\exp (\log (\mathrm{A})) \operatorname{expm}(\operatorname{logm}(\mathrm{A}))] \\ \mathrm{T} 4=\left[\log \left(\mathrm{A}^{-2}\right) 2 * \log (\mathrm{a})\right] \end{array} $$ \(\mathrm{T} 5=\left[\operatorname{logm}\left(\mathrm{A}^{-2}\right) 2 * \operatorname{logm}\left(\mathrm{A}^{-2}\right)\right]\) \(\mathrm{T} 6=[\exp (2 * \mathrm{~A}) \exp (\mathrm{A}) * \exp (\mathrm{A})]\) \(\mathrm{T} 7=[\operatorname{expm}(2 * \mathrm{~A}) \operatorname{expm}(\mathrm{A}) * \operatorname{expm}(\mathrm{A})]\) $$ \text { Output } $$ $$ \begin{array}{rrrrr} \mathrm{T} 1= & 2.1598 & 1.1052 & 2.1674 & 0.2309 \\ & 1.0704 & 2.4596 & 0.1570 & 2.4676 \\ \mathrm{~T} 2= & -0.2614 & -2.3026 & -0.2666 & 0.1204 \\ & -2.6882 & -0.1054 & 0.0819 & -0.1100 \\ \mathrm{~T} 3= & 0.7700 & 0.1000 & 0.7700 & 0.1000 \\ & 0.0680 & 0.9000 & 0.0680 & 0.9000 \\ \mathrm{~T} 4= & -0.5113 & -1.7898 & -0.5227 & -4.6052 \\ & -2.1754 & -0.2024 & -5.3765 & -0.2107 \\ \mathrm{~T} 5= & -0.5331 & 0.2408 & -0.5331 & 0.2408 \\ & 0.1637 & -0.2201 & 0.1637 & -0.2201 \\ \mathrm{~T} 6= & 4.6646 & 1.2214 & 5.8475 & 5.1052 \\ & 1.1457 & 6.0496 & 4.9444 & 7.2326 \\ \mathrm{~T} 7= & 4.7341 & 1.0703 & 4.7341 & 1.0703 \\ & 0.7278 & 6.1254 & 0.7278 & 6.1254 \end{array} $$

Compute and graph the solutions to $$ y^{\prime \prime}+p y^{\prime}+9 y=0 \quad y(0)=0 \quad y^{\prime}(0)=1 $$ for \(p=10, p=6, p=4, p=0,\) and \(p=-6\).

Symbiotic relationships are common and persist for long periods. It is curious that there are no known or very few symbiotic relationships between mammals. There are, however, many symbiotic relationships between mammals and other organisms, Escherichia coli, for example. Analysis of the equations for symbiosis, Equations 18.63: $$ \begin{aligned} x^{\prime}(t) &=r_{x} \times x(t) \times(1-a x(t)+b y(t)) \\ y^{\prime}(t) &=r_{y} \times y(t) \times(1+c x(t)-d y(t)) \end{aligned} $$ a. Show that the equilibrium point without zeros is $$ x_{1}=\frac{b+d}{a d-b c}, \quad y_{1}=\frac{a+c}{a d-b c} $$ $$ \text { if } a d-b c \neq 0 $$ b. \(x_{1}\) and \(y_{1}\) are positive only if \(a d-b c>0 .\) This is a surprise to us. Set \(b=c=d=1\) and examine the equilibrium point for \(a>1\) and \(a<1\). c. Assume \(a d-b c>0\) so that \(x_{1}\) and \(y_{1}\) are positive. Either work it out (no!) or accept our analysis that the Jacobian at \(\left(x_{1}, y_{1}\right)\) is $$ J=\left[\begin{array}{cc} -a \frac{b+d}{a d-b c} & b \frac{b+d}{a d-b c} \\ c \frac{a+c}{a d-b c} & -d \frac{a+c}{a d-b c} \end{array}\right]=\frac{1}{a d-b c}\left[\begin{array}{cc} -a(b+d) & b(b+d) \\ c(a+c) & -d(a+c) \end{array}\right] $$ Argue that if \(a d-b c>0,\left(x_{1}, y_{1}\right)\) is an asymptotically stable equilibrium. d. With persistence you might show that the characteristic roots are not complex. It requires showing that the discrimiant $$ \begin{array}{l} (a(b+d)+d(a+c))^{2}-4(a(b+d) d(a+c)-c(a+c) d(b+d))= \\ (a(b+d)-d(a+c))^{2}+4 c(a+c) d(b+d) \geq 0 \end{array} $$

Anderson and May \(^{4}\) give the following model of immune effector cells (helper and cytotoxic T-cells), \(E,\) that limit viral population, \(V\), growth in a human body. $$ \begin{array}{l} d E / d t=\Lambda-\mu E+\epsilon V E \\ d V / d t=r V-\sigma V E \end{array} $$ a. \(\Lambda\) is intrinsic production rate of effector cells from bone marrow. Give similar meaning to each of the other four terms on the RHS of Equations 18.64 . b. Find the equilibrium effector cell population, \(\hat{E},\) in the absence of virus. c. Suppose an inoculum \(V_{0}\) of virus is introduced into the body with \(E=\hat{E}\). Find conditions on \(r\), \(\sigma,\) and \(\hat{E}\) in order that the viral population will increase. d. The Jacobian matrix at any \((E, V)\) is $$ J(E, V)=\left[\begin{array}{rc} -\mu+\epsilon V & \epsilon E \\ -\sigma V & r-\sigma V \end{array}\right] $$ Show that $$ J(\hat{E}, 0)=\left[\begin{array}{rr} -\mu & \sigma \frac{\Lambda}{\mu} \\ 0 & r-\sigma \frac{\Lambda}{\mu} \end{array}\right] $$ e. The characteristic values of the upper diagonal matrix \(J(\hat{E}, 0)\) are the diagonal entries, \(-\mu\) and \(r-\sigma \frac{\Lambda}{\mu} .\) What happens to a small introduction of virus into a healthy individual if both characteristic values are negative? f. In order that the viral population to expand it is necessary that \(r-\sigma \frac{\Lambda}{\mu}>0 .\) What is the role of \(r\) in the model? g. If that condition is met and the viral population increases, show that there will be an equilibrium state, $$ E^{*}=r / \sigma, \quad V^{*}=\frac{\mu r-\Lambda \sigma}{\epsilon r} $$ It is clear that in order for \(V^{*}\) to be positive, we must have (again) $$ \mu r-\Lambda \sigma>0 \quad \text { so that } \quad r>\frac{\Lambda \sigma}{\mu} $$ h. Anderson and May report this system to be asymptotically stable, at \(\left(E^{*}, V^{*}\right),\) but only weakly so, meaning that it is subject to wide oscillations. Show that $$ J\left(E^{*}, V^{*}\right)=\left[\begin{array}{cc} -\frac{\Lambda \sigma}{r} & \epsilon \frac{r}{\sigma} \\ -\sigma \frac{\mu r-\Lambda \sigma}{\epsilon r} & 0 \end{array}\right] $$ i. The characteristic equation of a \(2 \times 2\) matrix \(M\) is is \(s^{2}-\operatorname{trace}(M) s+\operatorname{det}(M)=0\). Show that the characteristic equation of \(J\left(E^{*}, V^{*}\right)\) is $$ s^{2}+\frac{\Lambda \sigma}{r} s+\mu r-\Lambda \sigma=0 $$ j. The roots of the characteristic equation 18.65 are $$ \frac{-\frac{\Lambda \sigma}{r} \pm \sqrt{\left(\frac{\Lambda \sigma}{r}\right)^{2}-4(\mu r-\Lambda \sigma)}}{2} $$ Argue that the real part is negative so that the system is stable. Note: Two cases: $$ \left(\frac{\Lambda \sigma}{r}\right)^{2}-4(\mu r-\Lambda \sigma)>0 \quad \text { and } \quad\left(\frac{\Lambda \sigma}{r}\right)^{2}-4(\mu r-\Lambda \sigma)<0 $$ k. Use \(\Lambda=1, \mu=0.5, \epsilon=0.02, r=0.25,\) and \(\sigma=0.01\) and compute \(E^{*}, V^{*},\) and the stability at \(\left(E^{*}, V^{*}\right)\) l. Let \(E_{0}=2, V_{0}=1, \Lambda=1, \mu=0.5, \epsilon=0.02, r=0.25,\) and \(\sigma=0.01 .\) Approximate the solutions to Equations 18.64 using the trapezoid rule. Observe that the rise in viral load precedes the increase in effector cells.

Compute \(y_{h}(t),\) the general solution to the homogeneous equation, and \(y_{p}(t)\) a particular solution to the nonhomogeneous equation. $$ \begin{array}{lll} \text { a. } y^{\prime \prime}+4 y^{\prime}+3 y=t & \text { b. } y^{\prime \prime}+4 y^{\prime}+3 y= & e^{t} \\ \text { c. } y^{\prime \prime}-4 y^{\prime}+3 y=e^{t} & \text { d. } y^{\prime \prime}+4 y^{\prime}+3 y=\cos t \end{array} $$

Show that the for \(m, r,\) and \(k\) positive, the characteristic roots $$ r_{1}=\frac{-r+\sqrt{r^{2}-4 m k}}{2 m} \quad \text { and } \quad r_{2}=\frac{-r-\sqrt{r^{2}-4 m k}}{2 m} $$ for \(m y^{\prime \prime}(t)+r y^{\prime}(t)+k y(t)=0\) are both negative or have negative real parts.

It may be that recovered individuals do not have life time immunity; they become susceptible after a period \(p,\) and one may write $$ \begin{array}{l} s^{\prime}(t)=\alpha r(t-p)-\beta \times s(t) \times i(t) \\ i^{\prime}(t)=\beta \times s(t) \times i(t)-\gamma i(t) \\ r^{\prime}(t)=\gamma i(t)-\alpha r(t-p) \end{array} $$ This system is considerably more complex than Equations \(18.61,\) and is simplified by letting \(p=0 .\) $$ \begin{aligned} s^{\prime}(t) &=\alpha r(t)-\beta \times s(t) \times i(t) \\ i^{\prime}(t) &=\beta \times s(t) \times i(t)-\gamma i(t) \\ r^{\prime}(t) &=\gamma i(t)-\alpha r(t) \end{aligned} $$ This system involves three functions and three equations and is beyond our exposition. However, you may be able to analyze it. a. Show that \(\left(s_{0}, 0,0\right)\) is an equilibrium point, for any \(s_{0}\). b. Guess or compute the Jacobian matrix at \(\left(s_{0}, 0,0\right)\). c. The characteristic values of the Jacobian matrix at \(\left(s_{0}, 0,0\right)\) are \(\beta s_{0}-\gamma\) and \(-\alpha .\) What is the criterion for an epidemic (the number of infected will increase when a small number, \(\epsilon,\) of infected individuals enter a population of \(s_{0}\) susceptible). d. Solve the equations $$ \begin{array}{ll} x^{\prime}(t)=-\beta s_{0} y(t)+\alpha z(t) & x(0)=s_{0} \\ y^{\prime}(t)=\left(\beta s_{0}-\gamma\right) y(t) & y(0)=\epsilon \\ z^{\prime}(t)=\gamma y(t)-\alpha z(t) & z(0)=0 \end{array} $$ Solve first for \(y(t)\), then for \(z(t)\) and then for \(x(t)\).

Consider a special case of a spring-mass system in which there is no resistance and with a a harmonic forcing function, \(f(t)=\cos \omega t\). Thus examine $$ m y^{\prime \prime}(t)+k y(t)=\cos \omega t $$ Let \(\omega_{0}=\sqrt{\mathbf{k} / \mathbf{m}}\). It is routine to show that if \(\omega \neq \omega_{0},\) the general solution to \(m y^{\prime \prime}(t)+k y(t)=\cos \omega t\) is $$ y(t)=\frac{1}{m\left(\omega_{0}^{2}-\omega^{2}\right)} \cos \omega t+C_{1} \cos \omega_{0} t+C_{2} \sin \omega_{0} t . $$ a. Suppose that the mass is initially at rest so that \(y(0)=0,\) and \(y^{\prime}(0)=0\) and \(\omega \neq \omega_{0} .\) Show that the motion of the mass is approximated by $$ \begin{aligned} y(t) &=\frac{1}{m\left(\omega_{0}^{2}-\omega^{2}\right)} \cos \omega t-\frac{1}{m\left(\omega_{0}^{2}-\omega^{2}\right)} \cos \omega_{0} t \\ &=\frac{1}{m\left(\omega_{0}^{2}-\omega^{2}\right)}\left(\cos \omega t-\cos \omega_{0} t\right) \end{aligned} $$ b. Sketch the graph of \(y(t)\) in Equation 18.16 for the case \(m=1, \omega=1\) and \(k=0.01\) (weak spring) \(\left(\omega_{0}=0.1\right) .\) The impressed force \(\cos t\) appears as the rapid oscillations, and the inherent system frequency appears as the overall gradual wave due to the term \(\cos \omega_{0} t=\cos 0.1 t\) c. Sketch the graph of \(y(t)\) in Equation 18.16 for the case \(m=1, \omega=1\) and \(k=0.81\) (stiff spring) \(\left(\omega_{0}=0.9\right)\) From the identity, \(\cos x-\cos y=2 \sin \frac{x+y}{2} \sin \frac{x-y}{2},\) $$ y(t)=\frac{1}{m\left(\omega_{0}^{2}-\omega^{2}\right)} 2 \sin \left(\frac{\omega_{0}+\omega}{2} t\right) \sin \left(\frac{\omega_{0}-\omega}{2} t\right) $$ The amplitude of the rapid vibrations, \(\sin \left(\frac{\omega_{0}+\omega}{2} t\right)\) is $$ \frac{1}{m\left(\omega_{0}^{2}-\omega^{2}\right)} 2 \sin \left(\frac{\omega_{0}-\omega}{2} t\right) $$ and results in a beat of frequency \(4 \pi /\left(\omega_{0}-\omega\right)\) (that may be heard in mechanical systems.

Exercise 18.1 .13 Show that $$ \text { Solution: } y(t)=-\frac{50}{\sqrt{0.99}} e^{-0.1 t} \sin \sqrt{0.99} t+50 \sin t $$ is the solution to $$ y(0)=0, \quad y^{\prime}(0)=0, \quad y^{\prime \prime}(t)+0.2 y^{\prime}(t)+y(t)=10 \cos t $$ Graph the solution. Note that the magnitude of the vibrations are five times those of the forcing function, \(10 \cos t\)

Consider now the case that there is resistance \(r>0\) in the spring-mass equation with a harmonic forcing function, $$ m y^{\prime \prime}(t)+r y^{\prime}(t)+k y(t)=\cos \omega t $$ a. Show that in the real root case, \(\mu_{1}\) and \(\mu_{2}\) are negative. b. Show that in the repeated root case, the value of \(\mu\) is negative. c. Show that in the and complex roots case, the value of \(\mu\) is negative. d. Show that in all of the cases, $$ \lim _{t \rightarrow \infty} y_{h}(t)=0 $$ The next few exercises examine the importance of \(y_{p},\) the particular solution of Equation \(18.20 .\) We will need $$ \begin{array}{ll} y_{p}=A \cos 0.3 t+B \sin 0.3 t & \text { and one of } \\ y_{h}=C_{1} e^{\mu_{1} t}+C_{2} e^{\mu_{2} t}, & \text { real roots, } \\ y_{h}=C_{1} e^{\mu t}+C_{2} t e^{\mu t}, & \text { repeated root, and } \\ y_{h}=e^{\mu t}\left(C_{1} \cos (\omega t)+C_{2} \sin (\omega t)\right) & \text { complex roots. } \end{array} $$

Exercise 18.1 .16 Find the particular solution, \(y_{p}=A \cos 0.3 t+B \sin 0.3 t,\) and \(y_{h},\) the solution to the homogeneous equation, for the systems below. a. $$ y^{\prime \prime}+3 y^{\prime}+2 y=\cos 0.3 t \quad \text { b. } \quad y^{\prime \prime}+0.1 y^{\prime}+0.12 y=\cos 0.3 t $$ $$ \text { c. } y^{\prime \prime}+0.05 y^{\prime}+0.0004 y=\cos 0.3 t $$ d. \(y^{\prime \prime}+0.13 y^{\prime}+0.0036 y=\cos 0.3 t\) The algebra to solve these problems is extensive and you will find the MATLAB code below helpful at least to read and at best to run. It is written for the case of distinct real roots, and the system is initially at rest: \(y(0)=0, y^{\prime}(0)=0 .\) NOTE: \(\omega \neq 0.3\) in the code. Solutions to \(y^{\prime \prime}+0.1 y^{\prime}+0.12 y=\cos \omega t, y(0)=0, y^{\prime}(0)=0,\) for \(\omega=0.1\) and \(0.2 .\) The blue curves are the solution; the green curves are the solutions to the homogeneous equation, \(y^{\prime \prime}+0.1 y^{\prime}+0.125 y=0,\) and the red curves are the particular solutions. The solution curves (blue) are asymptotic to the particular solutions (red).