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MA 2930, Jan 26, 2011
Worksheet 1
1.
Write down in your own words what a differential equation is.
What infor-mation does it give you? Can you think of at least two
different ways ofspecifying a differential equation?
A differential equation is a relation among a function and its
derivatives.It tells you how the values of a function and its
derivatives are related atany value of the independent variable.
Two different ways of specifying adifferential equation are via an
algebraic equation and a direction field.
2.
Suppose someone gives you the direction field of a differential
equation. Howwould you find its solutions?
Start at any point (corresponding to an initial value) and
follow the ar-rows from that point on; youll get the solution (aka
integral curve) passingthrough that point.
3.
(a) Is the function y(t) = t a solution of y + 4y + 3y = t?To
check if it is a solution plug it into the two sides of the
equation and
see if they come out equal: if they do, it is a solution,
otherwise its not. Inthe given problem the left hand side gives (t)
+ 4(t) + 3t = 0 + 0 + 3t = 3twhereas the right had side is t, so
the y = t is not a solution.(b) Which value of the constant r would
make y(x) = erx a solution ofy + 2y = 0?
If y = erx is a solution it should satisfy the equation,
therefore (erx) +2erx = 0, i.e., (r + 2)erx = 0. Since erx is never
zero, r = 2.
4.
Look at the direction fields in figures 1.1.5 and 1.1.6 in the
text (page 9).(1) What happens to the values of y as time passes if
you start at (a)
y = 1, (b) y = 2, (c) y = 3?
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8/4/2019 MATH 2930 - Worksheet 1 Solutions
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In both graphs y = 2 is the equilibrium point; ify(0) = 2, y(t)
= 2 for all
times t.If y(0) = 1, with time y(t) approaches the value 2 in
1.1.5 whereas itmoves away from 2 towards in 1.1.5
If y(0) = 3, with time y(t) approaches the value 2 in 1.1.5
whereas itmoves away from 2 towards in 1.1.6
(2) Sketch some solutions (integral curves.) What do the shapes
of thesolutions seem to be?
If you sketch the solutions (by following the arrows from a
starting point),they seem to be exponential in shape, decaying in
1.1.5 and growing in 1.1.6.
(3) Match the direction fields with the differential equations
from amongthe choices given on page 8 (a j). Then solve the
equations. Do the graphs
of functions you find match the shapes of the integral
curves?(4) Apply the initial conditions of part (1) to your
solutions and verify
your answers in part (1).For 1.1.5, note that y = 2 is an
equilibrium point, so in its differential
equation y(2) 0. That leaves us with (c) and (j). Now note that
wheny < 2 (say at y = 1) the slopes in the direction field are
positive, so in thedifferential equation y > 0 when y < 2.
This is true of the choice (j), so thematching diff. eq. is y = 2
y.
To solve it note that its separable: dy2y
= dt, so when we integrate, we
get ln |2 y| = t + c, so |2 y| = cet, since the sign of (2 y)
can be
absorbed in the sign of the unknown constant c, we have 2 y =
cet
, i.e.,y(t) = 2 cet.Since y(0) = 2 c, we can say y(t) = 2 (2
y(0))et.Its clear that (a) ify(0) = 1, y(t) = 2et which approaches
2 from below
as t , (b) ify(0) = 2, y(t) = 2 for all t, and (c) ify(0) = 3,
y(t) = 2 + et
which approaches 2 from above as t , all of which agrees with
theconclusions in part (1).
Similar analysis applies to 1.1.6 which matches with the
equation y =y 2.
5.
A raindrop evaporates at a rate proportional to its surface
area. Supposethe proportionality factor k is 2ml1/3/s. How long
would it take raindrop tocompletely disappear if it had 1 ml at the
beginning of its career?
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