Page 529, Question #1

Verify that the given function pair is a solution to the first-order system \[x = -e^t, y= e^t\] \[ \frac {dx} {dt} = -y, \frac {dy} {dt} = -x\]

Page 529, Answer #1

\[ \frac {dx} {dt} = \frac {d(-e^t)} {dt} \] \[ = -e^t \] \[ = -y \] Hence we have: \(-e^t = -y\)
Multiplying both side by -1, we get: \(e^t = y\)
And we have: \[ \frac {dy} {dt} = \frac {d(e^t)} {dt} \] \[ = e^t \]
And we have \[\frac {dy} {dt} = -x = e^t\]

Therefore, the given function pair (\(x = -e^t, y= e^t\)) is a solution to the first-order system.

Page 529, Question #6

Find and classify the rest points of the given autonomous system: \[ \frac {dx} {dt} = -(y -1)\] \[ \frac {dy} {dt} = x -2 \]

Page 529, Answer #6

We have:

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

\[ F(x,y) = 0 \begin{cases} -(y-1) \\ x -2 \end{cases} \]

\[ 0 = \begin{cases} -(y-1) \\ x -2 \end{cases} \]

\[ \begin{cases} y = 1 \\ x = 2 \end{cases} \]

Then the stable point is (2,1)

Next, we have:

\[ \frac {dy} {dx} = \frac {\frac {dy} {dt}} {\frac {dx} {dt} } \] \[ \frac {dy} {dx} = \frac {x -2} {-(y -1)} \] \[ \frac {dy} {dx} = \frac {x -2} {1-y} \] \[ \frac {dy} {dx} - \frac {x -2} {1-y} = 0 \] \[ (1-y) \frac {dy} {dx} - x + 2 = 0 \] \[ (1-y)dy-(x - 2)dx = 0 \] \[ \int (1-y)dy - \int (x - 2)dx = 0 \] \[ y - \frac {y^2} {2} - \frac {x^2} {2} + 2x + C = 0 \] \[ 2y - {y^2} - {x^2} + 4x + C = 0 , \qquad multiplying \ by \ 2\] \[ -2y + {y^2} + {x^2} - 4x - C = 0 , \qquad multiplying \ by \ -1\] \[ -2y + {y^2} = - {x^2} + 4x + C \] \[ y^2 -2y = - x^2 + 4x + C \qquad Equation.1\]

We know that \((y - 1)^2 = y^2 -2y +1\)   Hence: \(y^2 -2y = (y - 1)^2 - 1\)
Therefore, replacing \(y^2 -2y = (y - 1)^2 - 1 \qquad\) in Equation 1, we get: \[ y^2 -2y = - x^2 + 4x + C \qquad Equation 1\] \[ (y - 1)^2 - 1 = -x^2 + 4x + C \] \[ (y - 1) = \sqrt {-x^2 + 4x + 1+C} \] Finally, we get y in terms of x as below: \[ y = \sqrt {-x^2 + 4x +C} - 1 \qquad Equation.2\]

Now, let find C. We know that at x = 2, we get y = 1. Hence,
\[ 1 = \sqrt {-2^2 + 8 +C} - 1\] \[ 2 = \sqrt {-4 + 8 +C}\] \[ 2 = \sqrt {4 +C}\]

Clearly C= 0.

Therefore, our Final Equation is: \[ y = \sqrt {-x^2 + 4x } - 1 \qquad Equation.2\]

Let’s actually see the trajectory and plot the above function

x<- seq(0, 4, length =10000)
y = sqrt(-(x^2) + (4*x)) - 1
df<- data.frame(x,y)

library(ggplot2)
## Warning: package 'ggplot2' was built under R version 3.2.3
ggplot(data=df, aes(x=x, y=y)) +
  geom_line()

Page 546, Question 1

Apply the first and second derivative tests to the function \(f(y) = \frac{y^a}{e^{by}}\) to show that \(y = \frac{a}{b}\) is a unique critical point that yields the relative maximum f(a/b). Show also that f(y) approaches 0 as y tends to infinity.

Page 546, Answer 1

\[f(y) = \frac{y^a}{e^{by}}\] \[f(y) = {y^a}{e^{-by}}\] \[f'(y) = ay^{a-1} e^{-by} - by^a e^{-by}\]

Finding the critical point:

$$

$$

Hence: The first derivative yield to unique critical point. \[ y = a/b \]

Now let try the second derivative. We have the first is as follow:

\[f'(y) = ay^{a-1} e^{-by} - by^a e^{-by}\]

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

\[f''(y) = e^{-by} \left[a(a-1)y^{a-2} - bay^{a-1} - aby^{a-1} + b^2 y^a\right] \] \[f''(y) = e^{-by} \left[ay^{a-2} ((a-1) - by) - by^{a-1} (a - by)\right] \] \[f''(y) = e^{-by} \left[ay^{a-2} (a-1 - by) - by^{a-1} (a - by)\right] \]

Now, lets y = a/b, we get:

\[f''(y) = e^{-by} \left[ay^{a-2} (a-1 - b a/b) - by^{a-1} (a - ba/b)\right] \] \[f''(y) = e^{-by} \left[ay^{a-2} (a-1 - a) - by^{a-1} (a - a)\right] \] \[f''(y) = e^{-by} \left[ay^{a-2} (-1 ) - by^{a-1} (0)\right] \] \[f''(y) = e^{-a} \left[ay^{a-2} (-1 ) \right] \] \[f''(y) = -e^{-a} \left[ay^{a-2} \right] \] \[f''(y) = -e^{-a} \left[a(a/b)^{a-2} \right] \] \[f''(y) = -e^{-a} \left[a.a^{a-2} b^{2-a} \right] \] \[f''(y) = -e^{-a} \left[a^{a-1} b^{2-a} \right] \] \[f''(y) = -e^{-a}a^{a-1}b^{2-a} \qquad Equation.4 \]

let us see if we can figure out the value by converting to polar coordinates…

\[ a^{a-1} = (e ^{ln(a)})^{a-1}\] \[ b^{2-a} = (e ^{ln(b)})^{2-a}\]

lets replace in Equation.4: \[f''(y) = -e^{-a}a^{a-1}b^{2-a} \qquad Equation.4 \] \[f''(y) = -e^{-a}(e ^{ln(a)})^{a-1}(e ^{ln(b)})^{2-a} \]

Show also that f(y) approaches 0 as y tends to infinity.

\[ \lim_{x\to \infty} f(y) = \frac{y^a}{e^{by}}\] \[ \lim_{x\to \infty} f(y) = \frac {(e ^{ln(y)})^{a}} {e^{by}}\]

As y tends to infinity constant a and b become negligible

\[ \lim_{x\to \infty} f(y) = \frac {(e ^{ln(y)})} {e^{y}}\]

\[ \lim_{x\to \infty} f(y) = \frac {y} {e^{y}}\]

Note that both y and e^y approach infinity as y approaches infinity, so we can use l’Hôpital’s Rule. Also, the derivative of y is 1, and the derivative of e^y is (still) e^y. Hence our limit becomes:

\[ \lim_{x\to \infty} f(y) = \frac {1} {e^{y}}\] \[ \lim_{x\to \infty} f(y) = 0\]

Page 566: Question #1

Use Euler method to solve the first order system subject to the specified initial condition.

\[dx/dt = 2x + 3y\] \[dy/dt = 3x + 3y\]

x(0)=1, y(0) = 1, \(\Delta t = 1/5\)

\[ x(t)= 1/2 e^{-t} + 1/2 e^{5t}, y(t)= -1/2 e^{-t}+ 1/2 e{^5t}\]

Page 566: Answer #1

First:   \[x_1 = x_0 +f(t_0,x_0,y_0)\Delta t\] \[ = x_0 + (2x_0 + 3y_0)\Delta t\] \[ = 1 + (2 + 3)(.25)\] \[ = 2.25\] \[y_1 = y_0 +f(t_0,x_0,y_0)\Delta t\] \[ = y_0 + (3x_0 + 2y_0)\Delta t\] \[ = 1 + (3 + 2)(.25) \] \[ = 2.25 \]

Second:
\[x_2 = x_1 +f(t_1,x_1,y_1)\Delta t\] \[ = x_1 + (2x_1 + 3y_1)\Delta t\] \[ = x_1 + (2x_1 + 3y_1)\Delta t\] \[ = 2.25 + (2x2.25 + 3x2.25)(.25)\] \[x_2 = 5.0625\] \[y_2 = y_1 +f(t_1,x_1,y_1)\Delta t\] \[ = y_1 + (3x_1 + 2y_1)\Delta t\] \[ = x_1 + (3x_1 + 2y_1)\Delta t\] \[ = 2.25 + (3x2.25 + 2x2.25)(.25)\] \[y_2 = 5.0625\]

Third:
\[x_3 = x_2 +f(t_2,x_2,y_2)\Delta t\] \[ = x_2 + (2x_2 + 3y_2)\Delta t\] \[ = x_2 + (2x_2 + 3y_2)\Delta t\] \[ = 5.0625 + (3x5.0625 + 2x5.0625)(.25)\] \[x_2 = 11.390625\] \[y_3 = y_2 +f(t_2,x_2,y_2)\Delta t\] \[ = y_2 + (3x_2 + 2y_2)\Delta t\] \[ = x_2 + (3x_2 + 2y_2)\Delta t\] \[ = 5.0625 + (3x5.0625 + 2x5.0625)(.25)\] \[x_3 = 11.390625\]