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Only One Word for Review
Review
Engineering Differential Equations
The Second Test
Euler the Master of Us All
Euler’s Method: Tangent Line Approximation
• For the initial value problem

y

f
(
t
,
y
),
y
(
t
)

y
,
0
0
we begin by approximating solution y = (t) at initial point t0.
• The solution passes through initial point (t0, y0) with slope
f(t0,y0). The line tangent to solution at initial point is thus



y

y

f
t
,
y
t

t
0
0
0
0
• The tangent line is a good approximation to solution curve on
an interval short enough.
• Thus if t1 is close enough to t0,
we can approximate (t1) by




y

y

f
t
,
y
t

t
1
0
0
0
1
0
Euler’s Formula
• For a point t2 close to t1, we approximate (t2) using the line
passing through (t1, y1) with slope f(t1,y1):



y

y

f
t
,
y
t

t
2
1
1
1
2
1
• Thus we create a sequence yn of approximations to (tn):
y1 y0  f0 t1 t0
y2 y1  f1t2 t1

yn1 yn  fn tn1 tn
where fn = f(tn,yn).
• For a uniform step size h = tn – tn-1, Euler’s formula becomes
y

y

f
h
,
n

0
,
1
,
2
,

n

1
n
n
Euler Approximation
• To graph an Euler approximation, we plot the points
(t0, y0), (t1, y1),…, (tn, yn), and then connect these points
with line segments.
Note (t0, y0) was the IC




y

y

f

t

t
,
whe
f

f
t
,
y
n

1
n
n
n

1
n
n
n
n
Autonomous Equations and Population
Dynamics
• In this section we examine equations of the form y' = f (y),
called autonomous equations, where the independent
variable t does not appear explicitly.
• y’(t) is the gorillaz velocity which depends on gorillaz height
y(t). Important heights are the rest points where y’ is zero;
when f(y)=0. These are Equilibrium solutions.
• Example (Exponential Growth):

y

ry
,r

0
• Solution:
rt
yy
e
0
Autonomous Equations: Equilibrium Solns
• Equilibrium solutions of a general first order autonomous
equation y' = f (y) are found by locating roots of f (y) = 0.
• These roots of f (y) are called critical points.
• For example, the critical points of the logistic equation
dy  y

r
1
 y
dt  K

• are y = 0 and y = K.
• Thus critical points are constant
functions (equilibrium solutions)
in this setting.
Autonomous Equations: Equilibrium Solns
• Equilibrium solutions of a general first order autonomous
equation y' = f (y) can be found by locating roots of f (y) = 0.
• These roots of f (y) are called critical points.
• Phase diagram is the y axis showing where the monkey
climbs (f>0), rests (f=0) and falls (f<0)
y’ = y(10 – y)
• Thus critical points are constant
functions (equilibrium solutions)
in this setting.
Population Models
• P(t)= fish pop size, b(t) = individ birth rate (births/unit
time/fish) d(t) = individ death rate( deaths/unit time/fish)
• Units for P’(t) are fish/unit time
• Balance Law
dp
 ratein  rateout  bp  dp
dt
Velocity and Acceleration
• x(t) height of an object falling in the atmosphere near sea
level; time t, velocity v(t) = x’(t), a(t) = x’’(t) accel.
• Newton’s 2nd Law: Net F = ma = m(dv/dt) net force
• Force of gravity: - mg
downward force
• Force of air resistance: -  v (opp to v)
upward force
• Then get eqn for v (F = Force Grav + Resist Force) and x
dv
 mg   v
dt
x'  v
m
Velocity and Acceleration
• We can also get one eqn for x (using F = Force Grav + Resist
Force)
• m x’’ = -mg – ϒx’ is one second order de for x which is the
same as the previous two first order DEs for x and v
dv
m
 mg   v
dt
x'  v

Homogeneous Equations, Initial Values
• Once a solution to a homogeneous equation is found, then it
is possible to solve the corresponding nonhomogeneous
equation.
• Thus consider homogeneous linear Diff equations; and in
particular, those with constant coefficients (like the previous)



a
y

b
y

cy

0
• Initial conditions typically take the form






t

y
,
y
t

y
• Thus solutiony
passes
through
(t0
0
0
0
0, y0), and slope of solution at
(t0, y0) is equal to y0'.
Characteristic Equation
• To solve the 2nd order equation with constant coefficients,



a
y

b
y

cy

0
,
we begin by assuming a solution of the form y = ert.
• Substituting this into the differential equation, we obtain
• Simplifying,
2
rt rt
rt
ar
e

bre

ce

0
rt2
e
(
ar

br

c
)

0
and hence
2
ar
c
br


0
• This last equation is called the characteristic equation of the
differential equation.
• We then solve for r by factoring or using quadratic formula.
General Solution
• Using the quadratic formula on the characteristic equation
2
ar

br

c

0
,
we obtain two solutions, r1 and r2.
• There are three possible results:
– The roots r1, r2 are real and r1  r2.
– The roots r1, r2 are real and r1 = r2.
– The roots r1, r2 are complex.
2

b
b

4
ac
r
2
a
• First assume r1, r2 are real and r1  r2.
• In this case, the general solution has the form
r
t
r
t
1
2
y
(
t
)

c
e

c
e
1
2
Theorem
• Suppose y1 and y2 are solutions to the equation



L
[
y
]

y

p
(
t
)
y

q
(
t
)
y

0
(
1
)
and that the Wronskian


W

y
y

y
y
1
2
1
2
is not zero at the point t0 where the initial conditions


y
(
t
)

y
,
y
(
t
)

y
(
2
)
0
0
0
0
are assigned. Then there is a choice of constants c1, c2 for
which y = c1y1 + c2 y2 is a solution to the differential
equation (1) and initial conditions (2).
Linear Independence and the Wronskian
• Two functions f and g are linearly dependent if there
exist constants c1 and c2, not both zero, such that
c
f
(
t
)

c
g
(
t
)

0
1
2
for all t in I. Note that this reduces to determining
whether f and g are multiples of each other.
• If the only solution to this equation is c1 = c2 = 0, then f
and g are linearly independent.
Theorem
• If f and g are differentiable functions on an open interval I
and if W(f, g)(t0)  0 for some point t0 in I, then f and g
are linearly independent on I. Moreover, if f and g are
linearly dependent on I, then W(f, g)(t) = 0 for all t in I.
Repeated Roots
• Recall our 2nd order linear homogeneous ODE



a
y

b
y

cy

0
• where a, b and c are constants.
• Assuming an exponential soln leads to characteristic
equation:
rt 2
y
(
t
)

e

ar

br

c

0
• Quadratic formula (or factoring) yields two solutions, r1 & r2:
2

b
b

4
ac
r
2
a
• When b2 – 4ac = 0, r1 = r2 = -b/2a, since method only gives

b
t
/
2
a
one solution: y
(
t
)

ce
1
General Solution
When roots are the same, then two linearly independent
solutions are e^{rt} and te^{rt}
• Thus the general solution for repeated roots is

bt
/
2
a 
bt
/
2
a
y
(
t
)

c
e

c
te
1
2
Complex Roots of Characteristic Equation
• Recall our discussion of the equation



a
y

b
y

cy

0
where a, b and c are constants.
• Assuming an exponential soln leads to characteristic
equation:
rt 2
y
(
t
)

e

ar

br

c

0
• Quadratic formula (or factoring) yields two solutions, r1 & r2:
2

b
b

4
ac
r
2
a
• If b2 – 4ac < 0, then complex roots: r1 =  + i, r2 =  - i


i
t


i
t
• Thus
y
(
t
)

e
,
y
(
t
)

e
1
2




Euler’s Formula; Complex Valued Solutions
• Substituting it into Taylor series for et, we obtain Euler’s
formula:
n
n
2
n
n

1
2
n

1


it
(
it
) (

1
)
t (

1
)
t
e



i

cos
t

i
sin
t






n
!
2
n
!n
2
n

1
!
n

0
n

0

1
• Generalizing Euler’s formula, we obtain







i

t
e

cos
t

i
sin
t





• Then


i

t
t
i
t
t
t
t


e

e
e

e
cos
t

i
sin
t

e
co
t

ie
s
t
• Therefore



i


t

t

t
y
(
t
)

e

e
cos
t

ie
sin
t
1
 



 

y
(
t
)

e

e
cos

t

ie
sin

t

it
2
t
t
Real Valued Solutions: The Wronskian

• Thus we have the following real-valued functions:

t
t
y
(
t
)

e
cos
t
,
y
(
t
)

e
sin
t
3
4
• Checking the Wronskian, we obtain


 


t

t
e
cos
t
e
sin
t
W


t

t
cos
e
sin

e
t

sin
t
t

cos
t
2

t

e

0
• Thus y3 and y4 form a fundamental solution set for our ODE,
and the general solution can be expressed as



t
t
y
(
t
)

c
e
cos
t

c
e
sin
t
1
2
Real Valued Solutions
• Our two solutions thus far are complex-valued functions:

t

t
y
(
t
)

e
cos

t

ie
sin

t
1

t

t
y
(
t
)

e
cos

t

ie
sin

t
2
• We would prefer to have real-valued solutions, since our
differential equation has real coefficients.
• To achieve this, recall that linear combinations of solutions
are themselves solutions:

t
y
(
t
)

y
(
t
)

2
e
cos

t
1
2

t
y
(
t
)

y
(
t
)

2
ie
sin

t
1
2
• Ignoring constants, we obtain the two solutions


t
t
y
(
t
)

e
cos
t
,
y
(
t
)

e
sin
t
3
4
Theorem (Nonhomogenous Des)
• If Y1, Y2 are solutions of nonhomogeneous equation



y

p
(
t
)
y

q
(
t
)
y

g
(
t
)
then Y1 - Y2 is a solution of the homogeneous equation



y

p
(
t
)
y

q
(
t
)
y

0
• If y1, y2 form a fundamental solution set of homogeneous
equation, then there exists constants c1, c2 such that
Y
(
t
)

Y
(
t
)

c
y
(
t
)

c
y
(
t
)
1
2
1
1
2
2
Theorem (General Solution)
• The general solution of nonhomogeneous equation



y

p
(
t
)
y

q
(
t
)
y

g
(
t
)
can be written in the form
y
(
t
)

c
y
(
t
)

c
y
(
t
)

Y
(
t
)
1
1
2
2
where y1, y2 form a fundamental solution set of
homogeneous equation, c1, c2 are arbitrary constants and Y
is a specific solution to the nonhomogeneous equation.
Thanks
• Leonhard Euler for his insights and beautiful mathematics.
You are number –e^{(pi)i} in my book.
• Ry Cooder’s music I think it’s going to work (as long as you
do)
• Good luck on Test 2
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