This document contains questions about differential equations relating to a poisoned duck that ate a poisonous mushroom. It provides context about the rate at which the poison is spreading through the duck and asks the reader to: 1) Sketch a slope field for the rate of change of poison dp/dt at indicated points. 2) Use Euler's Method to approximate the solution to dp/dt with an initial condition over 5 steps. 3) Find a particular solution to dp/dt given another initial condition.

1. Differential Equations:
Poisoned Bird Questions
Mushroom Eater
by Flickr user:
Just Emi

2. a) Oh no! Jamie’s pet duck ate the poisonous
mushroom! Luckily, Bench knows what type
of mushroom it is. The Fungus Differentius
has a very dangerous toxin. When eaten, it
can cause mutations. (Don’t you think the
duck has had enough torture?) The rate at
which the poison is spreading throughout
the duck is defined as dp/dt. Sketch a
slope field for at the points
indicated.

3. b) Use Euler’s Method to approximate the
solution of dp/dt with the initial condition
of with 5 steps of size 0.2.
c) Find a particular solution to dp/dt with
the initial condition .

4. Differential Equations: Mushroom Scene

5.  Ifwe take a
coordinate that’s
given on the graph
and plug the
coordinate’s x- and
y-values into the
differential
equation, we
obtain the slope at
that point

6.  Example:

7.  Bycreating a
display of lines
(the slope field),
where each line
indicates the slope
at that point, we
can see the parent
function
 That is, the solution
of a derivative via
slope fields is the
parent function

8.  Therefore, by
shortening the
distance between
the points, a
smoother line is
generated, making
the function we’ve
created further
resemble the parent
function
 Allow ∆x to be
infinitesimally
small, we have the
parent function!

9.  Thevalues on the grid below correspond to
the position of the coordinates on the graph:
 The value in cell A1 represents the coordinates
(-4, 4) on the graph

10. Differential Equations: Mushroom Scene

11.  Using Euler’s Method, we start at the
initial coordinates,
 In this case, at (3, 3.0642)
 Note that all values in the graphs above are rounded to four
decimal places for simplicity

12.  By plugging the P0 coordinates into the
differential equation, y’, we obtain y’ at
P0
 In this case, we plugged (3, 3.0642) into
to obtain dp/dt = 29.4975

13.  We know the definition of a slope as the
rise (the change in the dependent
variable) over the run (the change in the
independent variable)
 By multiplying y’ (the slope) by ∆x (the
run), we obtain ∆y (the rise)

14.  We’re given that ∆x = 0.2
 Adding ∆y to y0, we obtain y1
 We repeat this process until we reach the
number of desired steps

15.  Note that all values in the table above are given in their decimal
form, rounded to four decimal places for simplicity

16. Differential Equations: Mushroom Scene

17.  We can separate the variables, that is, in
this case, antidifferentiating t on one side
with respect to dt and p on one side with
respect to dp
 Separatingthe variables is analogous to
antidifferentiating after solving for ∆y when
given the definition of a slope

18.  Integration by parts is an
antidifferentiation technique we can use
when we have to antidifferentiate two
factors
 We’re undoing the product rule
Formula for integration
by parts:

19. LIATE is a mnemonic used to determine which of the factors should be
selected for f. LIATE tells us the order of preference for f.
L I A T E
O N L R X
G V G I
E
P
A E G O
R
R B O
S N
I E R N
T A M E
H T I E N
M R C T T
I I R I
C G I
O
A
N
C L
O
M
E
T
R
I
C

20.  Bench says: I have
discovered a rule
for differentiating
products involving
et without using
the whole process
of integration by
parts or LIATE!
MWAHAHA!

21.  Thepower rule says that the derivative of
any variable to an exponent can be found
by multiplying the term by the exponent
and decrease the exponent by 1

22. • Differentiate the algebraic
factor until we get a
constant
• Note that the signs
alternate: minus, plus,
minus, plus, etc.
• In this case: t2 – 2t + 2

23.  We’re given the initial value
 We can use this fact to determine C
* Remember, C is a constant!
Note:
 Since we’re antidifferentiating, we’d expect C’s on both
sides
 Let’s group the C’s to one side of the equation for
simplicity

24.  Puttingit all together, we now have a
general solution for p