# Homework 1 Oracle

## Chapter 1.1

### 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.

Created: 2021-02-28 Sun 16:57