Transcribed Text
Exercise 0.2.1: Check that the y given is really a solution to the equation.
Next, take the second order differential equation
=  2 y,
dx2
for some constant k > 0. The general solution for this equation is
y(x) = C1cos(kx) + C2 sin(kx).
Note
that
because
we
have
a
second
order
differential
equation,
we
have
two
constants
in
our
general
solution.
Exercise 0.2.2: Check that the y given is really a solution to the equation.
And finally, take the second order differential equation
dx² dRy = K2 y,
for some constant k > 0. The general solution for this equation is
y(x) = Ciekx + Cze
or
y(x) = D1 cosh(kx) + D2 sinh(kx).
For those that do not know, cosh and sinh are defined by
ex
+e
=
,
sinh X =
2
These
functions are sometimes easier to work with than exponentials. They have some nice familiar
properties such as cosh 0 = 1, sinh 0 = 0, and of cosh. X = sinh X (no that is not a typo) and
sinh X = cosh X.
Exercise 0.2.3: Check that both forms of the y given are really solutions to the equation.
An
interesting
note
about
cosh:
The
graph
of
cosh
is
the
exact
shape
a
hanging
chain
will
make.
This
shape
is
called
a
catenary.
Contrary
to
popular
belief
this
is
not
a
parabola.
If
you
invert
the
graph
of
cosh
it
is also the ideal arch for supporting its own weight. For example, the gateway
arch
in Saint Louis is an inverted graph of coshif it were just a parabola it might fall down. The
formula used in the design is inscribed inside the arch:
y =  127.7 ft . cosh(x/127.7 ft) + 757.7 ft.0.2.4 Exercises
Exercise 0.2.4: Show that X = eti is a solution to X"  12x" + 48x'  64x = 0.
Exercise 0.2.5: Show that X = e is not a solution to X"  12x" + 48x'   64x = 0.
Exercise 0.2.6: Is y = sint a solution to (%) 2 = 1y2?Justify. 
Exercise 0.2.7: Lety" + 2y'  8y = 0. Now try a solution of the form y = e"x for some (unknown)
constant r. Is this a solution for some r? If so, find all such r.
Exercise 0.2.8: Verify that X = Ce2i is a solution to X' = 2x. Find C to solve for the initial
condition x(0) = 100.
Exercise 0.2.9: Verify that X = Cie + C2e21 is a solution to X"  X'  2x = 0. Find C1 and C2 to
solve for the initial conditions x(0) = 10 and x' (0) = 0.
Exercise 0.2.10: Find a solution to (x')² 2 + x2 = 4 using your knowledge of derivatives of functions
that you know from basic calculus.
Exercise 0.2.11: Solve:
a)  dA dt =  10A, A(0) = 5
dH
b) = 3H, H(0) = 1
dx
change = 4y, y(0) = 0, y' (0) = 1
d) X = 9x, x(0) = 1, X' (0) = 0
d
dx²
dy²
Exercise 0.2.12: Is there a solution to y' = y, such that y(0) = y(1) ?Exercise 1.2.1: Sketch slope field for y' = exy. How do the solutions behave as x grows? Can you
guess a particular solution by looking at the slope field?
Exercise 1.2.2: Sketch slope field for y' = x2 1
Exercise 1.2.3: Sketch slope field for y' = y².
Exercise 1.2.4: Is it possible to solve the equation y' = cos for y(0) = 1? Justify.
Exercise 1.2.5: Is it possible to solve the equation y' = y v[x for y(0) = 0? Is the solution unique?
Justify.
Exercise 1.2.6: Match equations y' = 1  X, y' = x2y,  y' = x(1  y) to slope fields. Justify.
/
/
/
/

\
/
/
/
/
/
/
/
/
/
a)
b)
/
c)
/
/
Exercise 1.2.7 (challenging): Take y' = f(x,y), y(0) = 0, where f(x,y) > 1 for all x and
y.
If
the
solution exists for all X, can you say what happens to y(x) as x goes to positive infinity? Explain.
Exercise 1.2.8 (challenging): Take (y  = 0, y(0) = 0. a) Find two distinct solutions. b) Explain
why this does not violate Picard's theorem.Exercise 1.3.1: Solve y' = x/y.
Exercise 1.3.2: Solve y' = x2y.Exercise 1.3.3: Solve dx dx (x2 1) for x(0)
=  t, = 0.
dt
Exercise 1.3.4: Solve dx dx = X sin(t), for x(0) = 1.
dt
Exercise 1.3.5: Solve dy xy y 1. Hint: Factor the right hand
= + x + + side.
dx
Exercise 1.3.6: Solve xy' = y + 2x2y, where y(1) = 1.
dy
y2
+
1
Exercise 1.3.7: Solve
=
for y(0) = 1.
dx x2 + 1
Exercise 1.3.8: Find an implicit solution for dy  , for y(0) = 1.
x2 + 1
dx
y² + 1
Exercise 1.3.9: Find an explicit solution for y' = xe>, y(0) = 1.
Exercise 1.3.10: Find an explicit solution for xy' y(1) = 1.
Exercise 1.3.11: Find an explicit solution for y' = y(0) = 1. It is alright to leave a definite
integral in your answer.
Exercise 1.3.12: Suppose a cup of coffee is at 100 degrees Celsius at time t = 0, it is at 70 degrees
at t = 10 minutes, and it is at 50 degrees at t = 20 minutes. Compute the ambient temperature.In the exercises, feel free to leave answer as a definite integral if a closed form solution cannot
be
found. If you can find a closed form solution, you should give that.
Exercise 1.4.4: Solve y' + xy=x.
Exercise 1.4.5: Solve y' + 6y = ex. .Exercise 1.4.1: Try it! Add a constant of integration to the integral in the integrating factor and
show that the solution you get in the end is the same as what we got above.
An advice: Do not try to remember the formula itself, that is way too hard. It
is
easier
to
remember the process and repeat it.
Since we cannot always evaluate the integrals in closed form, it is useful to
know
how
to
write
the solution in definite integral form. A definite integral is something that you
can
plug
into
a
computer or a calculator. Suppose we are given
y'+p(x)y=f(x), y(xo) = yo.
Look at the solution and write the integrals as definite integrals.
= plop(s)ds f(t) dt + yo )
.
(1.4)
XO
You should be careful to properly use dummy variables here. If you now plug such a formula
into
a
computer or a calculator, it will be happy to give you numerical answers.Exercise 1.4.3: Write the solution of the following problem as a definite integral, but try to simplify
as far as you can. You will not be able to find the solution in closed form.
V' + y = et?x, y(0) = 10.Exercise 1.4.6: Solve y' + 3x2y = sin(x)ex³, with y(0) = 1.
Exercise 1.4.7: Solve y' + cos(x)y = cos(x).
Exercise 1.4.8: Solve , To 1 + xy = 3, with y(0) = 0.
Exercise 1.4.9: Suppose there are two lakes located on a stream. Clean water flows into the first
lake, then the water from the first lake flows into the second lake, and then water from the second
lake flows further downstream. The in and out flow from each lake is 500 liters per hour. The first
lake contains 100 thousand liters of water and the second lake contains 200 thousand liters of water.
A truck with 500 kg of toxic substance crashes into the first lake. Assume that the water is being
continually mixed perfectly by the stream. a) Find the concentration of toxic substance as a function
of time in both lakes. b) When will the concentration in the first lake be below 0.001 kg per liter? c)
When will the concentration in the second lake be maximal?
Exercise 1.4.10: Newton's law of cooling states that dx = k(x  A) where X is the temperature,
t is time, A is the ambient temperature, and k > 0 is a constant. Suppose that A = A0 cos(wt) for
some constants A0 and W. That is, the ambient temperature oscillates (for example night and day
temperatures). a) Find the general solution. b) In the long term, will the initial conditions make
much of a difference? Why or why not?
Exercise 1.4.11: Initially 5 grams of salt are dissolved in 20 liters of water. Brine with concentration
of salt 2 grams of salt per liter is added at a rate of 3 liters a minute. The tank is mixed well and is
drained at 3 liters a minute. How long does the process have to continue until there are 20 grams of
salt in the tank?
Exercise 1.4.12: Initially a tank contains 10 liters of pure water. Brine of unknown (but constant)
concentration of salt is flowing in at 1 liter per minute. The water is mixed well and drained at 1
liter per minute. In 20 minutes there are 15 grams of salt in the tank. What is the concentration of
salt in the incoming brine?Hint: Answers need not always be in closed form.
Exercise 1.5.1: Solve y' + y(x2  1) + xy6 = 0, with y(1) = 1.
Exercise 1.5.2: Solve 2yy' + 1 = V2 + X, with y(0) = 1.
Exercise 1.5.3: Solve y' + xy = y4, with y(0) = 1.
Exercise 1.5.4: Solve yy' + X = V x2 + y2.
These solutions may offer stepbystep problemsolving explanations or good writing examples that include modern styles of formatting and construction
of bibliographies out of text citations and references. Students may use these solutions for personal skillbuilding and practice.
Unethical use is strictly forbidden.