Problem 1, Pg 529

Verify that the given function pair is a solution to the first-order system.

\[x = -e^t, \space y = e^t\] \[\frac{dx}{dt} = -y, \space \frac{dy}{dt} = -x\]

Taking the derivative of the pairs:

\[\frac{dx}{dt} = \frac{d(-e^t)}{dt} = -e^t\] \[\frac{dy}{dt} = \frac{d(e^t)}{dt} = e^t\]

Substituting our solution and y:

\[-e^t = -e^t\]

Substituting our solution and x:

\[e^t = e^t\]

Therefore the given functions are solutions to the differential equation.

Problem 6, Pg 529

Find and classify the rest points of the given autonomous system.

\[\frac{dx}{dt} = -(y-1), \space \frac{dy}{dt} = x-2\]

Setting \(\frac{dx}{dt} = \frac{dy}{dt} = 0\)… \[y = 1, \space x = 2\]

Therefore, the rest point of the autonomous system is at (2, 1)

In order to classify the rest point, we need to solve the differential equation.

In order to get rid of the \(t\) term, we will divide the two equations by each other:

\[\frac{\frac{dx}{dt}}{\frac{dy}{dt}} = \frac{dx}{dy} = \frac{-(y-1)}{(x-2)}\]

Separating terms:

\[\int x-2\space dx = \int -y + 1 \ \space dy\] \[\frac{x^2}{2} - 2x + C_x = \frac{-y^2}{2} + y + C_y\]

Simplify and merge constants

\[\frac{x^2}{2} - 2x + \frac{y^2}{2} - y + C_{xy} = 0\] \[x^2 - 4x + y^2 - 2y + C_0 = 0\]

Let’s solve for \(C_0\) at the rest point (2, 1)

\[2^2 - 4\times 2 + 1^2 - 2\times1 = -C_0 = -5 \rightarrow C_0 = 5\]

Let’s solve in regards to \(y\) so we can see what is going on to the left and right of the rest point.

\[y^2 - 2y = -x^2 + 4x - 5\]

\(y^2 - 2y\) can be written as \((y-1)^2 -1\), therefore:

\[(y-1)^2 = -x^2 + 4x - 4 = -(x-2)^2\]

\[y - 1 = (x-2)i\]

\[y = (x-2)i - 1\] The solution is undefined for all points outside the rest point, i.e. (2, 1).

Problem 7, Pg 536

Show that the two trajectories leading to \((m=n, a/b)\) shown in Figure 12.8 are unique.

Figure 12.8

Figure 12.8

\[\frac{dy}{dx} = \frac{(m-nx)y}{(a-by)x}\]

System (12.6) is given as follows:

\[\frac{dx}{dt} = (a-by)x \qquad \frac{dy}{dt} = (m-nx)y \qquad (12.6)\]

Dividing the second by the first gives the desired equation

\[\frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{dy}{dx} = \frac{(m-nx)y}{(a-by)x}\]

\[y^ae^{-by}=Kx^me^{-nx}\]

where K is a constant of integration

Separate and integrate \[\int\frac{a}{y} - b\space dy = \int\frac{m}{x} - n\space dx\]

\[aln(y) - by + C_y= mln(x) - nx + C_x\] Exponate

\[ln(\frac{y^a}{x^m}) = -nx + by + K\]

\[\frac{y^a}{x^m} = e^{-nx + by + K} = e^{-nx}e^{by}e^K\]

\[\therefore y^ae^{-by}=Kx^me^{-nx}\]

In order to find the maximum, we need to take the derivative and equate the function to 0.

\[\frac{df(y)}{dy} = \frac{d(\frac{y^a}{e^{by}})}{dy}\]

\[ay^{a-1}e^{-by}-by^ae^{-by} = 0\] \[e^{-by}(ay^{a-1}-by^a) = 0\] \[ay^{a-1}=by^a\] \[a/b = \frac{y^a}{y^{a-1}} = y\]

Plugging our solution back into the equation

\[\frac{\frac{a^a}{b^a}}{e^{b\frac{a}{b}}} = \frac{\frac{a^a}{b^a}}{e^a} = (\frac{a}{eb})^a = M_y\]

Doing the same for \(g(x)\)

\[\frac{dg(x)}{dx} = \frac{d(\frac{x^m}{e^{nx}})}{dx}\]

\[mx^{m-1}e^{-nx}-nx^me^{-nx} = 0\] \[e^{-nx}(mx^{m-1}-nx^m) = 0\] \[mx^{m-1}=nx^m\] \[m/n = \frac{x^m}{x^{m-1}} = x\]

Plugging our solution back into the equation

\[\frac{\frac{m^m}{n^m}}{e^{n\frac{m}{n}}} = \frac{\frac{m^m}{n^m}}{e^m} = (\frac{m}{en})^m = M_x\] The graphs of these respective maximums are visualized in Figure 12.12

Figure 12.12

Figure 12.12

\[\lim_{\frac{y \rightarrow a/b}{x \rightarrow m/n}} \left [ \left ( \frac{y^a}{e^{by}}\right ) \left ( \frac{e^{nx}}{x^m} \right )\right ] = K\]

\[\lim_{y \rightarrow a/b}(y^ae^{-by})=K\lim_{x \rightarrow m/n}(x^me^{-nx})\]

\[\frac{\lim_{y \rightarrow a/b}(y^ae^{-by})}{\lim_{x \rightarrow m/n}(x^me^{-nx})} = K\]

\[\frac{\left ( \frac{a}{b} \right )^ae^{-a}}{\left ( \frac{m}{n} \right )^me^{-m}}=\frac{\left ( \frac{a}{eb} \right )^a}{\left ( \frac{m}{en} \right )^m}\]

This solution for K is the same as \(\frac{M_Y}{M_X}\) from above in e.

The same reasoning applies as laid out in f. Namely, choosing \(y_0 > a/b\), \(f(y_0)\) will also be less than \(M_y\) as is evident from the graphs in Figure 12.12 that shows that there is a unique defined value for \(y_0 > a/b\). In order for the following inequality to hold:

\[\frac{M_y}{M_x}(\frac{x^m}{e^{nx}}) = y^a_o/e^{by_0}< M_y \rightarrow \frac{x^m}{e^{nx}} < M_x\]

The graph for \(g(x)\) in Figure 12.12 shows that there is a unique \(x_0 < m/n\). Therefore, there is only one trajectory leading to the rest point for \(y_0 > a/b\).

Problem 1, Pg 546

Apply the first and second derivative tests to the function \(f(y) = y^a/e^{by}\) to show that \(y = a/b\) is a unique critical point that yields the relative maximum \(f(a/b)\). Show also that \(f(y)\) approaches zero as \(y\) tends to infinity.

\[\frac{df(y)}{dy} = \frac{d\left(\frac{y^a}{e^{by}} \right)}{dy} = ay^{a-1}e^{-by} - by^ae^{-by}\]

Equate to 0:

\[ay^{a-1}e^{-by} = by^ae^{-by}\] \[ay^{a-1} = by^a\]

\[y = \frac{a}{b}\]

Check second derivative to ensure concave down at \(y = \frac{a}{b}\), i.e. \(y''\) at \(y = a/b\) is less than 0.

\[\frac{df(y')}{dy} = \frac{d\left ( ay^{a-1}e^{-by} - by^ae^{-by} \right)}{dy}=a(a-1)y^{a-2}e^{-by} - aby^{a-1}e^{-by} - (aby^{a-1}e^{-by} - b^2y^ae^{-by})\] Simplify

\[e^{-by}(a(a-1)y^{a-2} - 2aby^{a-1} + b^2y^a)\]

\[y^{a-2}e^{-by}(a(a-1) - 2aby + b^2y^2)\]

for \(y = a/b\)

\[\frac{a^{a-2}}{b^{a-2}}e^{-a}(a^2-a - 2a^2 + a^2) = -\frac{a^{a-2}}{b^{a-2}}e^{-a}(a)=-\frac{a^{a-1}}{b^{a-2}}e^{-a}\] Assuming (a, b) are positive, \(y = a/b\) will be a relative maximum since the second derivative is negative at \(y = a/b\)

Intuitively, \(e^{by}\) grows much faster than \(y^a\), even considering for very large values of \(a\). So, the denominator will overtake the numerator at some finite point and grow exponentially larger than that of \(y^a\); hence \(f(y)\) will approach zero as \(y\rightarrow\infty\).