Homework 1 Oracle

Chapter 1.1

Problem 7


Problem 10


Problem 15 [FOR GRADE]

We can see that the direction field results in constant rate of change of zero at level where \(y=0\) and \(y=3\). Then we can also see that the direction of the graph decreases when \(y<0\) and \(y>3\), while also increasing in section of \(y>0\) and \(y<3\). That means, we are looking for a separate \(y\) to get constant change at \(0\) and \((3-y)\) term, so that the constraints above are satisfied. \(\therefore y' = y(3-y)\). Answer is h.

Chapter 1.2

Problem 7 [FOR GRADE]

Our differential equation is as given

\begin{equation*} \frac{dp}{dt} = \frac{p}{2} - 450 \end{equation*}

a. Find the time at which the population becomes extinct if \(p(0)=850\)

We find that \(\mu(t) = e^{\int -\frac{1}{2} dt} = e^{-\frac{1}{2}t}\). Then

\begin{align*} \frac{d}{dt}(\mu(t)p) = -450 \mu(t) \\ \implies p(t) & = \frac{\int -450\mu(t) dt}{\mu(t)} \\ & = \frac{\int -450 e^{-\frac{1}{2}t}dt}{e^{-\frac{1}{2}t}} \\ & = \frac{-450 \times (-2e^{-\frac{1}{2}t}) + C}{e^{-\frac{1}{2}t}} \\ & = 900 + C e^{\frac{1}{2}t} \end{align*}

Then by using \(p(0)=850\), we find the constant \(C\) to be \(-50\). We need to find the time \(t_e\), at which \(p(t_e)=0\). We simply solve the equation \(0 = 900 - 50 e^{\frac{1}{2}t_e}\), where we find \(t_e = \ln(324) \approx 5.78\).

b. Find the time of extinction if \(p(0)=p_0\), where \(0

Using the above, we find that \(t_e = 2 \ln(\frac{900}{900-p_0})\)

c. Find the initial population \(p_0\) if the population is to become extinct in 1 year.

Recall from Example 1 that \(t\) is measured in months. So we simply solve

\begin{equation*} 0 = 900 - (900-p_0)e^{6} \implies p_0 = 900 - \frac{900}{e^6} \approx 897.77 \end{equation*}

Problem 9

a. If the limiting velocity is 49m/s (the same as in Example 2), show that the equation of motion can be written as

\begin{equation*} \frac{dv}{dt} = \frac{1}{245}(49^2-v^2) \end{equation*}

Recall the base equation from Example 2, by using the given, we have

\begin{equation*} 0 = 9.8 - C (49^2) \implies C = \frac{9.8}{49^2} = \frac{1}{245}\\ \implies \frac{dv}{dt} = \frac{1}{245}(49^2-v^2) \end{equation*}

b. If \(v(0) = 0\), find an expression for \(v(t)\) at any time.

We can view this problem as a first-order separable equation and rewrite it as

\begin{equation*} \int{\frac{dv}{49^2-v^2}} = \int{\frac{dt}{245}} \end{equation*}

After some computations, the expression for \(v(t)\) is \(49\tanh(\frac{t}{5})\)

c. Plot your solution from part b and the solution (26) from Example 2 on the same axes.


d. Based on your plots in part c, compare the effect of a quadratic drag force with that of a linear drag force.

The quadratic drag damps the speed faster and has a bigger effect on the speed than the linear dependance.

e. Find the distance \(x(t)\) that the object falls in time \(t\).

We know that \(\frac{dx}{dt} = v(t) = 49\tanh(\frac{t}{5})\) so then \(x(t) = 245\ln(\cosh(\frac{t}{5})) + C\). If \(x(0) = 0\), then \(C = 0\).

f. Find the time \(T\) it takes the object to fall 300m.

Simply solve \(300 = 245 \ln(\cosh(\frac{t}{5}))\), it will be something like \(t = 5\cosh^{-1}(e^{\frac{60}{49}}) \approx 9.477\)

Chapter 1.3

Problem 1

2nd order and linear.

Problem 3

4th order and linear.

Problem 5

Plug them both in. They are solutions.

Problem 9

Plug them both in. They are solutions.

Problem 11 [FOR GRADE]

\(y'+2y=0\) form yields solutions for \(r=-2\). You can solve a characteristic polynomial of this equation to get it. I guess you guys aren't ready for that yet. But your kids are gonna love it

Problem 16

\(u_{xx} + u_{yy} + u_{zz} = 0\) is linear and second order.

Problem 20

Plug them both in. They are solutions.

Date: 44; 12021 H.E.

Author: Sandy Urazayev

Email: University of Kansas (ctu@ku.edu)

Created: 2021-02-28 Sun 16:57