Differential Equations Class Notes
Dan Wysocki
Spring 2013
Contents
1 Introduction
2
2 Classification of Differential Equations
6
2.1
Linear vs. Non-Linear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
2.2
Seperable Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
2.3
Exact Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
2.4
Integrating Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
2.5
Modeling with First Order Equations . . . . . . . . . . . . . . . . . . . . . . . . . .
21
2.6
Bernoulli’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
24
2.7
Homogenous Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
26
2.8
Second Order Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
26
3 Transformations
33
1
3.1
Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
33
3.2
Inverse Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35
3.3
Unit Step-function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
42
3.4
Gamma Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
47
4 Systems of Differential Equations
52
5 Final Examples
61
2
1
Introduction
Differential equations are equations that have one or more derivatives or differentials.
1 dy
1
=
First order
x dx
y
2x + y y 00 = 5 Second order
(1.1)
(1.2)
(1.3)
y0 =
4
dy
dx
d2 y
+ xy = sin x
dx2
(1.4)
What is a solution to a Differential Equation?
x
dy
= 2y
dx
(1.5)
Show that y(x) = x2 is a solution.
dy
= 2x
dx
x (2x) = 2(x2 )
2x2 = 2x2
For
(1 + xexy )
dy
+ 1 + yexy = 0,
dx
show that x + y + exy = 0 is a solution.
3
(1.6)
Implicit Solution
d
d
(x + y + exy ) =
(0)
dx dx
dy
dy
1+
+ exy y + x
=0
dx
dx
dy
dy
1+
+ y exy + x exy
=0
dx
dx
dy
(1 + x exy ) + 1 + y exy = 0
dx
For
dy
= 2x,
dx
show that the following are solutions
1. y(x) = x2
d
d 2
y(x) =
x
dx
dx
dy
= 2x
dx
2. y(x) = x2 + 1
d
d 2
y(x) =
x +1
dx
dx
dy
= 2x
dx
4
(1.7)
3. y(x) = x2 + C
d
d 2
y(x) =
x +C
dx
dx
dy
= 2x
dx
dp
1
= p − 450
dt
2
Show that p(t) = 900 + C et/2 is a solution.
dp
1
t
= 0 + C e /2
dt
2
C t/2
= e
2
C t
LHS = e /2
2
1
t
RHS =
900 + C e /2 − 450
2
C t/2
= e
2
What if p(0) = 850 ← (initial condition)
850 = 900 + C e /2
0
850 = 900 + C ⇒ C = −50
We now have an Initial Value Problem (IVP).



 dp = 1 p − 450
dt
2


p(0) = 850
5
(1.8)
Solution is
p(t) = 900 − 50 e /2
t
dy
dy
= f (x, y),
is the slope.
dx
dx
Example: Consider a mouse population that reproduces at a rate proportional to the current
population, with a rate constant equal to 1/2
mice/month
assuming no owls present. When owls are
present, they eat the mice. Suppose that the owls eat on average 15 mice per day. Assuming there
are 30 days in a month, write a differential equation to describe the above.
p = mouse population, t = time in months
dp
∝p
dt
1
dp
= kp = p − 450, (450 is the mice eaten per month)
dt
2
dp
= 0. This is the equilibrium solution.
dt
dp
When p > 900,
>0
dt
dp
When p < 900,
<0
dt
When p = 900,
6
Next time: Classifications of Diff Eqns & Seperable Diff Eqns.
2
Classification of Differential Equations
Ordinary vs. Partial Differential Equation (ODE vs. PDE)
→ based on number of independent variables
(dedpendent variable) → y = f (x ← (independent variable))
• If there is one independent variable → ODE
• If there is more than one independent variable → PDE
Example 2.1
dp
1
= p − 450 → p(t) one independant variable
dt
2
(2.1)
∂ 2 v(x, t)
∂v(x, t)
=
→ PDE (Heat equation)
2
∂x
∂t
(2.2)
Example 2.2
7
Example 2.3
∂ 2 v(x, t)
∂ 2 v(x, t)
=
→ PDE (Wave equation)
∂x2
∂t2
(2.3)
Based on number of unknowns
→ system of equations
dx
= 4x + 3t
dt
dy
= 5x − 2y
dt
Order of a differential equation is the highest derivative that appears in the equation.
Equation 1
Equation 2
Equation 3
Equation
Equation
Equation
Equation
Equation
Equation
4
5
6
7
8
9
y 0 + 3y = 0
y 00 + 3y 0 − 2t = 0
d4 y d2 y
− 2 + 1 = e2t
dt4
dt
uxx + ut = 0
uxx + u uyy = sin t
uxx + sin(u)uyy = cos t
uxx + uyy + sin(u) = 0
uxx + uyy + sin(t) = 0
uxx + uyy + u = 0
1st order linear
2nd order linear
4th order
linear
2nd
2nd
2nd
2nd
2nd
2nd
linear
non-linear
non-linear
nonlinear
linear
linear
order
order
order
order
order
order
Table 1: Examples of different order and linear vs. non-linear differential equations.
2.1
Linear vs. Non-Linear
A linear ODE is of the form:
an (t)y (n) + an−1 (t)y (n−1) + . . . + a1 (t)y 0 + a0 (t)y = g(t)
(2.4)
This is similar to a polynomial, which take the form
an xn + an−1 xn−1 + . . . + a1 x + a0 = f (x)
8
(2.5)
But instead of it being the independent variables raised to decreasing powers, it’s the derivative of
the independent variables with decreasing orders of the derivatives.
2.2
Seperable Equations
Consider a first order ODE
dy
= f (x, y) . . .
dx
(2.6)
If we can express Equation 2.6 in the form of
M (x) dx + N (y) dy = 0
(2.7)
then Equation 2.6 is called a seperable equation.
Solve
Z
dy
x2
=
dx
1 − y2
(2.8)
Z
2
(1 − y ) dy = x2 dx
Z
Z
N (y) dy = M (x) dx
y−
x3
y3
=
+ C (Implicit Solution)
3
3
Solve
dy
3x2 − 4x + 2
=
dx
2(y − 1)
9
(2.9)
Z
Z
2(y − 1) dy =
3x2 − 4x + 2 dx
y 2 − 2y = x3 − 2x2 + 2x + C
Complete the square by adding 1 to both sides (note that C is still a constant, so the 1 disappears)
(y − 1)2 = x3 + 2x2 + 2x + C
√
y − 1 = ± x3 + 2x2 + 2x + C
√
y = 1 ± x3 + 2x2 + 2x + C (Explicit Solution)
If, also, y(0) = −1, solve for y
−1 = 1 ±
√
√
03 + 02 + 0 + C = 1 ± C
−
√
C
−2 = ±
C = (−2)2 = 4
√
y = 1 ± x3 + 2x2 + 2x + 4
−
√
4 = 1±2
√
y = 1 − x3 + 2x2 + 2x + 4
−1 = 1 ±
How about y(0) = 3, C = 4?
+
√
4 = 1±2
√
y = 1 + x3 + 2x2 + 2x + 4
3=1±
10
Solve



y0 =
4x−x3
4+y 3
(2.10)

 y(0) = 1
4x − x3
dy
=
dx
4 + y3
Z
Z
4 + y 3 dy = 4x − x3 dx
1
1
4y + y 4 = 2x2 − x4 + C
4
4
1 4
4(1) + 1 = 0 − 0 + C
4
1
4+ =C
4
17
C=
4
1
1
17
4y + y 4 = 2x2 − x4 +
4
4
4
2.3
Exact Differential Equations
Assume we have a solution
Ψ(x, y) = C
(2.11)
Then we will find what this solution is.
Theorem 2.4
M (x, y) + N (x, y)
∂M (x, y)
∂N (x, y)
dy
=
.
= 0 is called exact if
∂y
∂x
dx
For Equation 2.11,
∂Ψ ∂Ψ dy
+
= 0.
∂x
∂y dx
(2.12)
∂Ψ
∂Ψ
= M (x, y),
= N (x, y)
∂x
∂y
(2.13)
11
Example 2.5
2x + y 2 + 2xy
dy
=0
dx
2x + y 2 = M (x, y)
2xy = N (x, y)
∂M
∂N
= 2y,
= 2y
∂y
∂x
Z
Z
∂Ψ
=
2x + y 2 dx
∂x
= x2 + xy 2 + c1
Z
Z
∂Ψ
= 2xy dy
∂y
= xy 2 + c2
Ψ(x, y) = x2 + xy 2 + h(y)
Ψ(x, y) = xy 2 + h(x)
∂Ψ
= 2xy + h0 (y)
∂y
h0 (y) = 0
Z
Z
0
h (y)dy = 0dy
h(y) = c3
x2 + xy 2 + c3 = c4
x2 + xy 2 = C
Example 2.6
2xy + x2 + 1
dy
=0
dx
2xy = M (x, y)
x2 + 1 = N (x, y)
12
First we must see if they are exact
∂M
= 2x
∂y
∂N
= 2x
∂x
∂M
∂N
=
∂y
∂x
x2 y + y = C
∂Ψ(x, y)
= 2xy
∂x
Z
Z
∂Ψ(x, y)
dx = 2xy dx
∂x
Ψ(x, y) = x2 y + h(y)
Now take the derivative with respect to y and compare it to the known
∂Ψ
= x2 + h0 (y)
∂y
x2 + h0 (y) = x2 + 1
Z
Z
0
h (y) dy = 1 dy
h(y) = y + c1
Ψ(x, y) = x2 y + y + c1
x2 y + y = C
Example 2.7
dy
x2 + y + (x + cos(y))
=0
dx
M (x, y) = x2 + y
N (x, y) = x + cos(y)
∂M
= 2x
∂x
∂N
=
∂x
13
∂Ψ
∂y
to find h(y)
Example 2.8
dy
=0
dx
∂M
∂N
= 2x,
= 2x
∂y
∂x
2xy + (x2 + 1)
14
Example 2.9



x 2
dy
dx
= x3 − 2xy


y(1) = 3
M (x, y) = −x3 + 2xy
N (x, y) = x2
∂M
= 2x
∂y
∂N
= 2x
∂x
It is exact, so there must be a solution of the form:
Ψ(x, y) = C
∂Ψ ∂Ψ dy
+
=0
∂x
∂y dx
∂Ψ
= −x3 + 2xy
∂x
∂Ψ
= x2
∂y
Integrate:
Z
Z
∂Ψ
= x2 dy
∂y
Ψ(x, y) = x2 y + h(x)
∂Ψ
= 2xy + h0 (x)
∂x
−x3 + 2xy = 2xy + h0 (x)
h0 (x) = −x3
Z
h(x) = −x3 dx
x4
+C
4
x4
Ψ(x, y) = x2 y −
+C
4
x4
x2 y −
=C
4
=−
15
y(1) = 3
14
=C
4
1
3− =C
4
12 · 3 −
C = 11/4
x2 y −
x4 11
= /4
4
Example 2.10
(x2 + y) + (x − sin(y))
dy
=0
dx
M = (x2 + y), N = (x − sin(y))
∂M
=1
∂y
∂N
=1
∂x
It is exact
2.4
Integrating Factor
Example 2.11
x2 + y x − sin(y) dy
+
=0
x
x
dx
x2 + y
x − sin(y)
M=
,N=
x
x
∂M
= 1/x
∂y
∂N
sin(y)
=
∂x
x2
It is not exact
Both examples have the same solution, but one is exact and the other is not. Example 2.11 is
simply Example 2.10 divided by x.
16
• “Key term” (x) has been divided out!
• Key term → Integrating factor.
y 0 + p(x)y = g(x)
(2.14)
The Integrating factor is
R
p(x) dx
R
p(x) dx
R
p(x) dx
µ(x) = e
0
µ (x) = e
µ0 (x) = e
d
·
dx
Z
p(x) dx
· p(x)
µ0 (x) = µ(x) p(x)
µ(x) y 0 + µ(x) p(x) y = µ(x) g(x)
µ(x) y 0 + µ0 (x) y = µ(x) g(x)
Z
Z
d
(µ(x) y) = µ(x) g(x) dx
dx
Z
µ(x) y = µ(x) g(x) dx
Z
1
µ(x) g(x) dx
y=
µ(x)
Example 2.12
Solve y 0 + 2y = ex
p(x) = 2, g(x) = ex
R
µ(x) = e
2 dx
1
y = 2x
e
= e2x
Z
−2x
Z
=e
−2x
=e
=
e2x ex dx
e3x dx
1 3x
e +c
3
1 x
e + c e−2x
3
17
Example 2.13
dy
2y
=4−
dx
x
dy 2y
+
=4
dx
x
2
p(x) = , g(x) = 4
x
µ(x) = e
R
p(x) dx
= e2 ln x
ln(ab ) = b ln(a)
= eln x
2
= x2
Z
1
y= 2
x
=x
y(x) =
−2
4 x2
4 3
x +c
3
4
x + c x−2
3
Example 2.14



 dy + 2xy = 1,
dx


y(1) = 2
p(x) = 2x, g(x) = 1
R
µ(x) = e
= ex
p(x) dx
2
1
y = x2
e
You can’t integrate ex
2
18
Z
2
ex dx
Example 2.15
x2 y 3 + x(1 + y 2 )y 0 = 0
My = 3x2 y 2 , Nx = 1 + y 2
My 6= Nx Not exact
µ(x, y) =
1
xy 3
x2 y 3 x(1 + y 2 ) 0
+
y =0
xy 3
xy 3
1
1
x+
y0 = 0
+
y3 y
My = 0, Nx = 0
My = Nx Exact!
Z
Z
x2
+C
M dx = x dx =
2
Z
Z
N dy =
y −3 + y −1 dy
=−
1
+ ln |y| + C
2y 2
1
x2
− 2 + ln |y| = C
2
2y
Theorem 2.16
If
Nx − My
= Q(y), then
M
M + N y0 = 0
R
has an integrating factor µ(y) = e
19
Q(y) dy
µM + µN y 0 = 0
e
R
Q(y) dy
R
M +e
Q(y) dy
N y0 = 0
M̃y = e
R
Q(y) dy
M
Ñx = e
R
Q(y) dy
N
M̃y = Ñx
∂ R Q(y) dy e
M
M̃y =
∂y
=e
R
Q(y) dy
=e
R
Q(y) dy
· Q(y)M + e
R
∂
Ñx =
e Q(y) dy N
∂x
R
Q(y) dy
· My
· Nx
Nx − My = M Q(y)
Nx = My + M Q(y)
Ñx = e
R
Q(y) dy
(My + M Q(y))
=e
R
Q(y) dy
· My + M Q(y)e
M̃y = Ñx , Exact!
20
R
Q(y) dy
= M̃y
Example 2.17
y + 2xy − e−2y y 0 = 0
N (x, y) = 2xy − e−2y
M (x, y) = y
Nx = 2y, My = 1
Using Theorem 2.16:
Nx − My
2y − 1
1
=
=2−
M
y
y
1
Q(y) = 2 −
y
R
µ(y) = e
2− y1 dy
= e2y−ln |y|
= e2y · e− ln |y|
1
= e2y · eln | y |
1
= e2y
y
1 2y e y + 2xy − e−2y y 0 = 0
y
1
e2y + e2y 2xy − e−2y y 0 = 0
y
1
2y
2y
y0 = 0
e + 2xe −
y
M = e2y
1
2y
N = 2xe −
y
My = 2e2y
Nx = 2e2y
My = Nx Exact!
Z
M dx = xe2y
Z
N dy = xe2y − ln |y|
xe2y − ln |y| = C
21
2.5
Modeling with First Order Equations
Example 2.18 Mixing Problem
At time t = 0, a tank contains Q0 lb of salt dissolved in 100 gal of water. Assume that water
containing 1/4 lb of salt/gal is entering the tank at a rate of r gal/min and that the well-stirred
mixture is draining from the tank at the same rate. Set up the initial value problem that describes
this flow process. Find the amount of salt Q(t) in the tank at any time, and also find the limiting
amount QL that is present after a very long time. If r = 3 and Q0 = 2QL , find the time T after
which the salt level is within 2% of QL . Also find the flow rate that is required if the value of T is
not to exceed 45 min.
Let Q(t) be the amount of salt at time t
dQ
= Rate in − Rate out
dt
Rate in = (Amount) × (Rate)
Amount = 1/4 lb/gal
Rate = r
Rate out:
gal/min



Amount :
Q [lb]
100 gal


Rate :
r [gal/min]
22
1 lb
gal
Q lb
gal
r
−
r
4 gal
min
100 gal
min
1
Q
= r−
r
4
100
dQ
Q
1
+
r= r
dt
100
4
dQ
=
dt
µ(t) = e
r
100
R
dt
rt
= e 100
Q(0) = Q0
Example 2.19 Example 3 from Page 57 in the textbook.
Let Q(t) be the amount of chemical in the pond at any time t.
dQ
= Rate in − Rate out
dt
gal
g
5
= (2 + sin(2t))
gal
year
Q g
gal
−
5
10 gal
year
Q g
g
−
= (10 + 5 sin(2t))
year
2 year
Q
= 10 + 5 sin(2t) −
2
Q(0) = 0
p(t) =
1
2
g(t) = 10 + 5 sin(2t)
µ(t) = e
R
p(t) dt
−t/2
Z
= e /2
t
e /2 · (10 + 5 sin(2t)) dt
Z
Z
t/2
t/2
−t/2
=e
10 e dt + 5 e sin(2t) dt
Q=e
t
23
Integration by parts:
Z
f (t) = e /2 sin(2t) dt



v = et/2
dv = sin(2t) dt
t


dv = 1/2 et/2 dt
v = −1/2 cos(2t)
Z
t/2
t
1
1
= − /2 e cos(2t) + /4 e /2 cos(2t) dt
Z
t
1
g(t) = /4 e /2 cos(2t) dt



v = et/2
dv = cos(2t) dt


du = 1/2 et/2 v = 1/2 sin(2t)
Z
t/2
t/2
1
1
1
= /4 /2 e sin(2t) − /4 e sin(2t) dt
Z
t
I(t) = e /2 sin(2t) dt



v = et/2
dv = sin(2t) dt


du = 1/2 et/2
v = −1/2 cos(2t)
Z
t/2
t
1
1
= − /2 e cos(2t) + /4 e /2 cos(2t) dt
= −1/2 e /2 cos(2t) + 1/8 e /2 sin(2t) −
t
t
I
16
17
t
t
I = −1/2 e− /2 cos(2t) + 1/8 e /2 sin(2t) + e
16
Now we can substitute I(t) back into g(t) and then f (t) to get our final solution.
Example 2.20 Page 59, Problem 3
A tank originally contains 100 gal of fresh water. Then water containing 1/2 lb of salt per gallon is
poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at the same rate.
After 10 min the process is stopped, and fresh water is poured into the tank at a rate of 2 gal/min,
with the mixture again leaving at the same rate. Find the amount of salt in the tank at the end of
an additional 10 min.
24
dQ̃
= Rate in − Rate out
dt
Q̃(0) = 0
lb
Q lb
dQ̃
=1
−
dt
min 50 min
Q
dQ̃
=1−
dt
50
dQ̃
Q
1=
+
dt
50
µ(t) = e
R
Q(t) = e
−t/50
1/50 dt
Z
= e /50
t
e /50 · 1 dt
t
= e− /50 50 e /50 + C
t
t
= 50 + Ce− /50
t
Q(0) = 0 = 50 + C
C = −50
Q(t) = 50 − 50e− /50
t
Q(10) = 50 − 50e−
10/50
≈ 9.06 lb
2.6
Bernoulli’s Equation
Bernoulli’s Equation is of the form
dy
+ p(x)y = g(x)y n
dx
(2.15)
This is a more general case of the form we’ve used for integrating factors, where n was equal to 0
1
dy
0
7
+ p(x)y = g(x)
y
dx
25
n
1
Let y = v 1−n ; y n = v 1−n
1
n
dy
v 1−n −1 dv
v 1−n dv
=
=
dx
1 − n dx
1 − n dx
n
1
n
v 1−n dv
+ p(x)v 1−n = g(x)v 1−n
1 − n dx
1
n
1−n
1
n
− 1−n
v 1−n
Now divide everything by
1−n
= v 1−n = v 1 = v
n
1 − n v 1−n = v
dv
+ (1 − n)p(x)v = g(x)(1 − n)
dx
dy
+ p(x)y = g(x)y n
dx
Example 2.21
Solve
dy
+ 3y = ex y 2
dx
This is a Bernoulli with n = 2
1
y = v 1−2 = v −1 ⇒ y =
1
v
dv
+ (1 − 2)p(x)v = g(x)(1 − 2)
dx
dv
− p(x)v = −g(x)
dx
dv
− 3v = −ex
dx
µ(x) = e
R
−3 dx
Z
= e−3x
e−3x · (−1)ex dx
Z
3x
= −e
e−3x ex dx
Z
3x
= −e
e−2x dx
−2x
e
3x
= −e
−
+ c1
2
ex
=
− c1 e3x
2
ex
=
+ Ce3x
2
v(x) = e
3x
26
1
= v −1
v
x
−1
e
3x
so y(x) =
+ Ce
2
but y =
y = 0 is a solution as well
2.7
Homogenous Equations
wrote it in my notebook
2.8
Second Order Linear Equations
of the form
d2 y
= f (x, y, y 0 ) or p(x)y 00 + q(x)y 0 + r(x)y = g(x)
dx2
Example 2.22 Solve y 00 − y = 0 or y 00 = y.
y1 (x) = ex ,
y2 (x) = e−x
y(x) = c1 y1 (x) + c2 y2 (x)
= c1 ex + c2 e−x
y 0 (x) = c1 ex − c2 e−x
y 00 (x) = c1 ex + c2 e−x = y(x)
So y(x) = c1 ex + c2 e−x is the general solution.
27
If we have ay 00 +by 0 +cy = 0 where a, b, and c are constants, then y = emx is a solution if y 0 = memx ,
y 00 = m2 emx
am2 emx + bmemx + cemx = 0
emx am2 + bm + c = 0
So am2 + bm + c = 0 (characteristic equation)
There are three cases for characteristic equations
1. Two distinct roots r1 6= r2
y(x) = c1 er1 x + c2 er2 x
2. Complex roots m = α ± βi
y(x) = c1 eαx cos(βx) + c2 eαx sin(βx)
3. Repeated roots r1 = r2 = r
y(x) = c1 erx + c2 xerx
A second order linear equation
y 00 + p(x)y 0 + q(x)y = g(x)
is called homogenous if g(x) = 0 and inhomogenous if g(x) 6= 0.
If r1 and r2 are two distinct roots of a characteristic equation, then the general solution is given by
y(x) = c1 er1 x + c2 er2 x
28
Example 2.23 Solve y 00 + 3y 0 − 10y = 0
emx m2 + 3m − 10 = 0
m2 + 3m − 10 = 0
(m + 5)(m − 2) = 0
m = −5, m = 2
y(x) = c1 e−5x + c2 e2x
For an IVP you would need y(x0 ) = y0 and y 0 (x0 ) = ỹ0
Example 2.24
I.V.P.



y 00 + 4y 0 + 3y = 0


y(0) = 1,
y 0 (0) = 1
m2 + 4m + 3 = 0
(x + 1)(x + 3) = 0
y(x) = c1 e−x + c2 e−3x
y 0 (x) = −c1 e−x − 3c2 e−3x
y(0) = 1 = c1 + c2
y 0 (0) = 1 = −c1 − 3c2

 

1 1 1 0 2 
1

∼

−1 −3 1
0 1 −1
y(x) = 2e−x − e−3x
29
Example 2.25 Solve x2 + 2x + 2 = 0
x=
−2 ±
p
22 − 4(1)(2)
2(1)
= −1 ± i
Example 2.26 Solve y 00 + 2y 0 + 2y = 0
m2 + 2m + 2 = 0
m = −1 ± i
r1 = −1 + i
r2 = −1 − i
y(x) = c1 e(−1+i)x + c2 e(−1−i)x
= c1 e−x+xi + c2 e−x−xi
= c1 e−x [cos(x) + i sin(x)] + c2 e−x [cos(−x) + i sin(−x)]



− sin(θ) = sin(θ)


cos(θ) = cos(−θ)
= c1 e−x cos(x) + ic1 e−x sin(x) + c2 e−x cos(x) − ic2 e−x sin(x)
= (c1 + c2 )e−x cos(x) + (c1 − c2 )ie−x sin(x)
A = c1 + c2
B = i(c1 − c2 )
y(x) = Ae−x cos(x) + Be−x sin(x)
So, if α ± iβ are the roots, the general solution is
y(x) = Aeαx cos(βx) + Beαx sin(βx)
Example 2.27 The roots are: m = −2 ± 7i
30
Then the general solution must be
y(x) = Ae−2x cos(7x) + Be−2x sin(7x) = e−2x [A cos(7x) + B sin(7x)]
Example 2.28



y 00 + 14y 0 + 149y = 0
y 0 (0) = 8


y(0) = −10,
Example 2.29



y 00 + 4y 0 + 13y = 0
y 0 (0) = 7


y(0) = 0,
For repeated roots, r1 = r2 , so y(x) = er1 x = er2 x
Example 2.30




y 00 + 4y 0 + 4y = 0




2
m + 4m + 4 = 0






(m + 2)2 = 0,
m = −2
y(x) = v(x)e−2x
y 0 (x) = v 0 (x)e−2x − 2v(x)e−2x
= e−2x [v 0 (x) − 2v(x)]
y 00 (x) = v 00 (x)e−2x − 2v 0 (x)e−2x + 4v(x)e−2x − 2v 0 (x)e−2x
= e−2x [v 00 (x) − 4v 0 (x) + 4v(x)]
e−2x [v 00 (x) − 4v 0 (x) + 4v(x)] + 4e−2x [v 0 (x) − 2v(x)] + 4e−2x v(x) = 0
e−2x [v 00 (x) − 4v 0 (x) + 4v(x) + 4v 0 (x) − 8v(x) + 4v(x)] = 0
31
e−2x v 00 (x) = 0
v 0 = c1
Z
v = c1 dx = c1 x + c2
v(x) = c1 x + c2
y(x) = (c1 x + c2 ) e−2x
y(x) = c2 e−2x + c1 xe−2x
y 0 (x) = −2c2 e−2x + c1 e−2x − 2c1 xe−2x

−2x
xe


e−2x − 2xe−2x
 
e
 c1 
 
−2e−2x
c2
−2x
If r1 = r2 = r, the general solution is y(x) = c1 erx + c2 xerx
Example 2.31



16y 00 − 40y 0 + 25y = 0
y 0 (0) = −9/4


y(0) = 3,
16m2 − 40m + 25 = 0
(4m − 5)2 = 0
m = 5/4
32
y(x) = c1 e5x/4 + c2 xe5x/4
y 0 (x) = 5/4 c1 e5x/4 + c2 e5x/4 + 5/4 c2 xe5x/4
y(0) = 3 = c1 + 0
y 0 (0) = −9/4 = 5/4 c1 + c2 + 0
c1 = 3
c2 = −9/4 − 5·3/4 = −6
y(x) = 3e5x/4 − 6xe5x/4
= 3e5x/4 (1 − 2x)
Example 2.32



y 00 + 14y 0 + 149y = 0
y 0 (0) = 8


y(0) = −10,
m2 + 14m + 149 = 0
m = −7 ± 10i
y(x) = c1 e−7x cos(10x) + c2 e−7x sin(10x)
y 0 (x) = −7c1 e−7x cos(10x) − 10c1 e−7x sin(10x) − 7c2 e−7x sin(10x) + 10c2 e−7x cos(10x)
= (−7c1 + 10c2 )e−7x cos(10x) + (−10c1 − 7c2 )e−7x sin(10x)
Example 2.33



y 00 + 4y 0 + 13y = 0
y 0 (0) = 7


y(0) = 0,
33
m2 + 4m + 13 = 0
m = −2 ± 3i
y(x) = c1 e−2x cos(3x) + c2 e−2x sin(3x)
y 0 (x) = −2c1 e−2x cos(3x) − 3c1 e−2x sin(3x) − 2c2 e−2x sin(3x) + 3c2 e−2x cos(3x)
y(0) = 0 = c1
y 0 (0) = 7 = −2c1 + 3c2

 

 1 0 1 1 0 1

∼

−2 3 7
0 1 3
y(x) = e−2x cos(3x) + 3e−2x sin(3x)
MISSING NOTES GO HERE
3
3.1
Transformations
Laplace Transform
The Laplace transform of f (t) is denoted by
Z
L {f (t)} =
∞
f (t)e−st dt = F (s)
0
34
(3.1)
Example 3.1 Compute L {eat }
Z
∞
F (s) =
eat e−st dt
Z0 ∞
e(a−s)t dt
∞
1 (a−s)t
=
e
a−s
0
1 −∞
=
e
− e0 , s > a
a−s
1
1
=−
=
, s>a
a−s
s−a
=
0
Example 3.2 Compute L {sin(at)}
Z
∞
sin(at)e−st dt
F (s) =
 0


u = e−st ,
du = −se−st
IBP


v = − cos(at) , dv = sin(at) dt
a
Z
cos(at)e−st s ∞
cos(at)e−st dt
F (s) = −
−
a
a 0



u = e−st ,
du = −se−st dt


v =
sin(at)
,
a
dv = cos(at) dt
Z
cos(at)e−st s sin(at)e−st s ∞
−st
−
+
sin(at)e dt
F (s) = −
a
a
a
a 0
cos(at)e−st s sin(at)e−st s
=−
−
+ F (s)
a
a
a
a
−st
2
cos(at)e
s
s
=−
− 2 sin(at)e−st − 2 F (s)
a
a
a
s2
e−st s
F (s) 1 + 2 = −
cos(at) + sin(at)
a
a
a
−st a2 ∞
e
s
F (s) = −
cos(at) + sin(at) 2
a
a
a + s2 0
1
a2
, s>0
= 0 − − (1 + 0)
a
a2 + s 2
a
= 2
, s>0
a + s2
35
Properties
1. L{f (t) + g(t)} = L{f (t)} + L{g(t)}
2. L{c f (t)} = c L{f (t)}
Example 3.3 Compute L{3 sin(2t)}
L{3 sin(2t)} = 3 L{sin(2t)}
=3
=
s2
s2
2
, s>0
+4
6
, s>0
+4
Example 3.4 Compute L{et − e2t }
L{et − e2t } = L{et } − L{e2t }
1
1
−
, s>2
s−1 s−2
1
=−
, s>2
(s − 2)(s − 1)
=
3.2
Inverse Laplace Transform
L{f (t)} = F (s)
L−1 {L{f (t)}} = L−1 {F (s)}
Properties
1. L−1 {F (s) + G(s)} = L−1 {F (s)} + L−1 {G(s)}
2. L−1 {c F (s)} = c L−1 {F (s)}
36
(3.2)
−1
Example 3.5 Compute L
3
s
L
−1
1
3
−1
= 3L
s
s
= 3(1) = 3
Example 3.6
L
−1
4
s+1
= 4L
−1
1
s − (−1)
= 4e−t
Example 3.7
L
−1
2
1
−1
= 2L
s2 + 3s + 2
(s + 1)(s + 2)
B
2
A
+
=
s+1 s+2
(s + 1)(s + 2)
A(s + 2) + B(s + 1) = 2
Set s = −2
B = −2, A = 2
f (s) = 2 L
−1
1
s − (−1)
= 2e−t − 2e−2t
Do for homework:
37
−1
+ (−2) L
1
s − (−2)
Example 3.8
f (t) = L
−1
2s + 4
2
s + 2s + 5
s2 + 2s + 5 = s2 + 2s + 1 + 4
= (s + 1)2 + 4
= (s + 1)2 + 22
2s + 4
−1
f (t) = L
(s + 1)2 + 22
From the table:
b
−1
at
e sin(bt) = L
, s>a
(s − a)2 + b2
s−a
−1
at
e cos(bt) = L
, s>a
(s − a)2 + b2
2(s + 1) + 2
−1
f (t) = L
(s + 1)2 + 22
s+1
−1
= 2L
(a = −1, b = 2)
(s + 1)2 + 22
2
−1
+ L
(a = −1, b = 2)
(s + 1)2 + 22
= 2e−t cos(2t) + e−t sin(2t)
Theorem 3.9
L {f 0 (t)} = s L {f (t)} − f (0)
38
Proof
0
Z
∞
L {f (t)} =
f 0 (t)e−st dt
 0


u = e−st , du = −se−st dt


v = f (t), dv = f 0 (t) dt
Z ∞
−st
∞
= e f (t) 0 −
(−s)e−st f (t) dt
0
Z ∞
0
:
−s·∞
−s·0
= f (t)e−st dt
e f (∞) − e f (0) + s
0
= 0 − f (0) + 5 L {f (t)}
Example 3.10
0
L {f 00 (t)} = L [f 0 (t)]
= s L {f 0 (t)} − f 0 (0)
= s (s L {f (t)} − f (0)) − f 0 (0)
= s2 L {f (t)} − sf (0) − f 0 (0)
Example 3.11
0
L{f 000 (t)} = L{[f 00 (t)] }
= s L{f 00 (t)} − f 00 (0)
= s s2 L {f (t)} − sf (0) − f 0 (0) − f 00 (0)
= s3 L{f (t)} − s2 f (0) − sf 0 (0) − f 00 (0)
1. L{f 0 (t)} = s L{f (t)} − f (0)
2. L{f 00 (t)} = s2 L{f (t)} − sf (0) − f 0 (0)
3. L{f 000 (t)} = s3 L{f (t)} − s2 f (0) − sf 0 (0) − f 00 (0)
39
4. L{f (n) (t)} = sn L{f (t)} − sn−1 f (0) − sn−2 f 0 (0) − . . . − f (n−1) (0)
= sn L{f (t) +
Pn
i=1
−sn−i f i−1 (0)
Example 3.12
Solve



y 0 = y



 dy = y
dx
or


y(0) = 1
Z


y(0) = 1
1
dy =
y
Z
dx
ln |y| = (x + c1 )
|y| = ex+c1
= ec1 ex
y = ±ec1 ex
= Cex
y = Cex
but y(0) = 1
1 = Ce0
C=1
y = ex
40
Now let’s do it again using the Laplace transform
y 0 (x) = y(x)
L{y 0 (x)} = L{y(x)}
s L{y(x)} − y(0) = L{y(x)}
s L{y(x)} − L{y(x)} = y(0)
L{y(x)}(s − 1) = 1
1
s − 1
L{y(x)} =
−1
L {L{y(x)}} = L
−1
1
s−1
, a=1
y(x) = ex
Example 3.13
Solve



y 00 + 2y 0 + 5y = 0


y(0) = 2, y 0 (0) = 0
L{y 00 } + 2 L{y 0 } + 5 L{y} = L{0}
s2 L{y} − sy(0) − y 0 (0) + 2 (s L{y} − y(0))5 L{y} = 0
L{y} s2 + 2s + 5 = 2s + 4
2s + 4
+2s + 5
2s + 4
−1
−1
L {L{y}} = L
s2 + 2s + 5
L{y} =
s2
y(x) = 2e−x cos(2x) + e−x sin(2x)
Example 3.14
Solve: y 00 + y = sin(2t), y(0) = 2, y 0 (0) = 1
41
L {y 00 } − L {y} = L {sin(2t)}
2
, s>0
+4
2
s2 L {y} − 2s − 1 + L {y} = 2
s +4
1
L {y} (s2 + 1) − 2s − 1 = 2
s +4
1
L {y} (s2 + 1) = 2
+ 2s + 1
s +4
2 + 2s3 + 8s + s2 + 4
=
s2 + 4
2s3 + s2 + 8s + 6
L {y} =
(s2 + 4)(s2 + 1)
3
2s + s2 + 8s + 6
−1
y=L
(s2 + 4)(s2 + 1)
2s3 + s2 + 8s + 6
as + b cs + d
= 2
+
2
2
(s + 4)(s + 1)
s + 1 s2 + 4
s2 L {y} − sy(0) − y 0 (0) + L {y} =
s2
2s3 + s2 + 8s + 6 = (as + b)(s2 + 4) + (cs + d)(s2 + 1)
= as3 + bs2 + 4as + 4b + cs3 + ds2 + cs + d
= s3 (a + c) + s2 (b + d) + s(4a + c) + (4b + d)
a+c=2
b+d=1
4a + c = 8

1

0


4


0
0 1
1 0
0 1
4 0
4b + d = 6
 
0 2 1
 

1 1
 0
∼

0 8
 0
 
1 6
0

2 

1 0 0 5/3 


0 1 0 0 


0 0 1 −2/3
0 0 0
2s3 + s2 + 8s + 6 = 2s3 + s2 (5/3 − 2/3) + 8s + 20/3 − 2/3
42
3.3
Unit Step-function
Definition The function u(t) is called the unit step-function.
u(t) =



0, t < 0
(3.3)


1, t > 0
uc (t) is a transformation of the unit step-function
uc (t) = u(t − c) =



0, t − c < 0


1, t − c > 0
43
(3.4)
Example 3.15 For a < b, find u(t − a) − u(t − b).
u(t − a) − u(t − b) =



0, (t < a) ∪ (t > b)


1,
a<t<b
Example 3.16
Draw f (t) =




3,







1,
t<2
2<t<5



t,
5<t<8







t2/10, t > 8
Theorem 3.17 For C ≥ 0
e−cs
, s>0
L {uc (t)} = L {u(t − c)} =
s
44
Proof
Z
∞
L {uc (t)} =
Recall uc (t) =
uc (t)e−st dt
0


0, t < c


1, t > c
Z c
Z ∞
0−st
*
*1
=
uc
(t)e dt +
uc
(t)e−st dt
0
c
Z ∞
e−st dt
=0+
c
−st
∞
e
= −
s c
1
c = − 0 − e−sc
s
e−sc
, s>0
=
s
Example 3.18
L {uc (t)f (t − c)} = e−cs F (s), s > 0
Z ∞
=
uc (t)f (t − c)e−st dt
Z ∞
Z0 c
*0
*1
−st
u
(t)f
(t
−
c)e
dt
+
u
(t)f
(t − c)e−st dt
=
c
c
c
0
Z ∞
f (t − c)e−st dt
=
c
Let v = t − c
Z ∞
=
f (v)e−s(v+c) dv
Z0 ∞
=
f (v)e−sv e−sc dv
0
Z ∞
−sc
=e
f (v)e−sv dv
0
= e−sc L {f (v)}
= e−sc F (s)
45
Example 3.19 For homework:
L t2 u(t − 1) = L t2 u1 (t)
MISSING NOTES HERE
Example 3.20
t
Z
F (s) = L
e
−(t−τ )
sin τ dτ
0
= L e−t ∗ sin t
= L e−t · L {sin t}
1
1
=
s+1
s2 + 1
1
, s>0
=
(s + 1)(s2 + 1)
Example 3.21
Z
t
sin(t − τ ) cos τ dτ
F (s) = L
0
= L {sin t ∗ cos t}
1
s
=
s2 + 1
s2 + 1
s
= 2
, s>0
(s + 1)2
46
Example 3.22
−1
1
2
2
s (s + 1)
f (t) = L
1
−1
L
=t
s2
1
−1
= sin t
L
s2 + 1
1
L {t} = 2
s
1
L {sin t} = 2
s +1 1
1
L {t} L {sin t} =
s2
s2 + 1
1
L {t ∗ sin t} = 2 2
s (s + 1)
1
−1
t ∗ sin t = L
s2 (s2 + 1)
Z t
(t − τ ) sin τ dτ
=
 0


u = t − τ,
u0 = −dτ
v 0 = sin τ
Z t
t
cos τ dτ
= [− cos τ (t − τ )]0 −


v = − cos τ,
0
= t − [sin τ ]t0 = t − sin t
Example 3.23
G(s)
L
s2 + 1
1
F (s) = 2
s +1
−1
f (t) = sin t
−1
G(s)
s2 + 1
sin t ∗ g(t) = L
Z t
=
sin(t − τ )g(τ ) dτ
0
47
Example 3.24
y 00 + 4y = g(t); y(0) = 3, y 0 (0) = −1
L {y 00 } + L {4y} = L {g(t)}
s2 L {y} − 3s + 1 + 4 L {y} = G(s)
L {y} (s2 + 4) − 3s + 1 = G(s)
L {y} =
Let F (s) =
f (t) =
L−1 {G(s)F (s)} =
y=
=
=
3.4
G(s) + 3s − 1
s2 + 4
1
2
s +4
1
sin(2t)
2
1
sin(2t) ∗ g(t)
2
1
1
3s
−1
sin(2t) ∗ g(t) + L
−
2
s2 + 4 s2 + 4
1
1
sin(2t) ∗ g(t) + 3 cos(2t) − sin(2t)
2Z
2
1 t
1
g(t − τ ) sin(2τ ) dτ + 3 cos(2t) − sin(2t)
2 0
2
Gamma Function
Z
Γ(a + 1) =
∞
ta e−t dt, a > −1
0
Theorem 3.25
Γ(a + 1) = aΓ(a)
48
(3.5)
Example 3.26
Γ(4) = 3Γ(3)
= 3 · 2Γ(2)
= 3 · 2 · 1Γ(1)
= 3 · 2 · 1 · 1 = 6 = 3!
So for positive integers, Γ(a + 1) = a!
Proof
Z
∞
Γ(a + 1) =
ta e−t dt
 0


u = ta ,
u0 = a ta−1


v = −e−t , v 0 = e−t
Z ∞
a −t
∞
0
:
= −t
ta−1 e−t dt
e 0 + a
0
= 0 + aΓ(a)
49
Example 3.27
Γ(1) = Γ(0 + 1) ⇒ a = 0
Z ∞
t0 e−t dt
=
Z0 ∞
=
e−t dt
0
∞
= −e−t 0
=1
Γ(3) = Γ(2 + 1)
= 2Γ(2)
=2·1=2
Γ(4) = Γ(3 + 1)
= 3Γ(2 + 1)
= 3 · 2Γ(1 + 1)
= 3!
Γ(n + 1) = n!, n ∈ Z+
50
Proof
1
1
Γ
=Γ − +1
2
2
Z ∞
−1
t /2 e−t dt
=
0
2
2 Z ∞
1
−1/2 −t
t e dt
=
Γ
2
0
Let u2 = t, 2u du = dt
2 Z ∞
2
1
−1 −u2
Γ
=
u e
· 2u du
2
0
Z ∞
2
−u2
= 2
e
du
0
Z ∞
2
−u2
e
=
du
−∞
Z ∞
Z ∞
−y 2
−x2
=
e dy
e
dx
−∞
−∞
Z ∞Z ∞
2
2
=
e−(x +y ) dx dy
−∞
−∞
In cylindrical coordinates: x2 + y 2 = r2
Z 2π Z ∞
2
=
e−r r dr dθ
0
0
(Let w = r2 , dw = 2r dr)
Z Z
1 2π ∞ −w
e dw dθ
=
2 0
0
Z
1 2π
=
1 dθ
2 0
2
1
1
Γ
= (2π) = π
2
2
√
1
Γ
= π
2
51
Example 3.28
3
1
Γ
=Γ
+1
2
2
1
1
= Γ
2
2
√
π
=
2
7
5
Γ
=Γ
+1
2
2
3
5
+1
= Γ
2
2
5 3
1
= · Γ
+1
2 2
2
15 1
1
=
· Γ
4 2
2
√
15 π
=
8
Example 3.29
5
Γ −
3
8
=Γ − +1
3
8
−11
=− Γ
+1
3
3
88
−14
= Γ
+1
9
3
Notice that it’s just getting more and more complicated. So we need to do the opposite of what we’ve
52
been doing
5
5
5
2
− Γ −
=Γ − +1 =Γ −
3
3
3
3
3
2
5
=− Γ −
Γ −
3
5
3
2
2
2
1
− Γ −
=Γ − +1 =Γ
3
3
3
3
2
3
1
Γ −
=− Γ
3
2
3
5
9
1
Γ −
=
Γ
3
10
3
Example 3.30
13
Γ −
7
4
7
6
=− Γ −
13
7
1
49
Γ
=
78
7
Systems of Differential Equations
Example 4.1



dx

= 3x + 5y

dt



dy
= 3x + y
dt






x(0) = 1, y(0) = 2
53



L {x0 (t)} = 3 L {x} + 5 L {y}


L {y 0 (t)} = 3 L {x} + L {y}



*1
s L {x} − x(0)
= 3 L {x} + 5 L {y}


*2
s L {y} − y(0)
= 3 L {x} + L {y}



(s − 3) L {x} − 5 L {y} = 1


−3 L {x} + (s − 1) L {y} = 2


  
−5  L {x} 1
(s − 3)


= 
−3
(s − 1)
L {y}
2
−1  

 
−5  1
L {x} (s − 3)
  

=
2
−3
(s − 1)
L {y}

 
5  1
1
(s − 1)
= 2

 
s − 4s − 12
3
(s − 3) 2


1
 (s + 9) 
= 2


s − 4s − 12 (2s − 3)
s+9
− 4s − 12
2s − 3
L {y} = 2
s − 4s − 12
1
x = e−2t 15e8t − 7
8
1 −2t
y= e
9e8t + 7
8
L {x} =
s2
Example 4.2




x0 (t) = 3x − 5y




0
y (t) = x + 5y






x(0) = −3, y(0) = 4
54



L {x0 (t)} = 3 L {x} − 5 L {y}


L {y 0 (t)} = L {x} + 5 L {y}



* −3
s L {x} − x(0)
= 3 L {x} − 5 L {y}


*4
s L {y} − y(0)
= L {x} + 5 L {y}



(s − 3) L {x} + 5 L {y} = −3


− L {x} + (s − 5) L {y} = 4


  
5  L {x} −3
(s − 3)


= 
−1
(s − 5)
L {y}
4

 


−5  −3
1
(s − 5)
L {x}

 

= 2
s − 8s + 20
1
(s − 3)
4
L {y}


1
−3s − 5
= 2


s − 8s + 20 4s − 15
−3s − 5
− 8s + 20
4s − 15
L {y} = 2
s − 8s + 20
1
x = − e4t (17 sin(2t) + 6 cos(2t))
2
1 4t
y = e (sin(2t) + 8 cos(2t))
2
L {x} =
s2
Example 4.3
Z
2
ex dx
We can’t integrate this using the techniques we learned in Calculus, but we can approximate it as a
power series
f (x) =
∞
X
an (x − a)n , an =
n=0
f (n) (a)
n!
So we can write f (x) as
f (x) = a0 + a1 (x − a) + a2 (x − a)2 + a3 (x − a)3 + . . . + an (x − a)n
55
Example 4.4
y 00 − xy = 0, y(0) = 1, y 0 (0) = 2
y(x) =
∞
X
an x n
n=0
y 0 (0)
y 00 (0) 2 y 000 (0) 3
= y(0) +
x+
x +
x + ...
1!
2!
3!
Now, we know y(0) and y 0 (0), but nothing higher, so first, let’s plug in our initial conditions to the
original equation to find y 00 (0)
y 00 (x) − xy(x) = 0
*1
y(0)
=0
y 00 (0) − 0 · y 00 (0) = 0
Now let’s take the derivative of the equation, and plug in our initial conditions to find y 000 (0)
y 000 (x) − y(x) − xy 0 (x) = 0
*2
*1
y 0 (0)
=0
y(0)
− 0 ·
y 000 (0) − y 000 (0) = 1
Now let’s do it for the fourth derivative
y (4) (x) − y 0 (x) − y 0 (x) − xy 00 (x) = 0
y (4) (x) − 2y 0 (x) − xy 00 (x) = 0
*1
*2
y (4) (0) − 2
y 0 (0)
− 0 ·
y 00
(0) = 0
y (4) (0) = 4
56
We can keep going, but we’ll stop here with our approximation, giving us
1
4
y(x) = 1 + 2x + x3 + x4 + . . .
6
24
Example 4.5
y 00 − x2 x = 0, y(0) = 1, y 0 (0) = 0
0
y (x) =
y 00 (x) =
∞
X
n=1
∞
X
n an xn−1
n(n − 1)an xn−2
n=2
00
2
y (x) − x y =
∞
X
n(n − 1)an x
n−2
−x
n=2
=
∞
X
an x n
n=0
∞
X
n(n − 1)an xn−2 −
∞
X
an xn+2 = 0
n=0
n=2
=
2
∞
X
(n + 2)(n + 1)an+2 xn −
∞
X
an−2 xn = 0
n=2
n=0
= (2)(1)a2 + (3)(2)a3 x +
∞
X
n
(n + 2)(n + 1)an+2 x −
= 2a2 + 6a3 x +
an−2 xn = 0
n=2
n=2
∞
X
∞
X
xn [(n + 2)(n + 1)an+2 − an−2 ] = 0
n=2
2a2 = 0, 6a3 = 0, (n + 2)(n + 1)an+2 − an−2 = 0, n ≥ 2
an+2 =
an−2
, n≥2
(n + 2)(n + 1)
Example 4.6
Solve (−4x2 − 1)y 00 + 5xy 0 = 0; y(0) = −3, y 0 (0) = −8
Recall
∞
X
f (n) (a)(x − a)n
n=0
n!
=
∞
X
an (x − a)n
n=0
We already know y(0) and y 0 (0), so it would be simplest to let a = 0, as we need to know y(a) and
y 0 (a). We’re also going to need further derivatives, so lets find y 00 (0) by evaluating the equation with
x = 0.
(−1)y 00 (0) + 0 = 0 ⇒ y 00 (0) = 0
57
Now we need the third derivative
−8xy 00 + (−4x2 − 1)y 000 + 5y 0 + 5xy 00 = 0
and evaluate at x = 0
0 + (−1)y 000 (0) + 5(−8) + 0 = 0 ⇒ y 000 (0) = −40
Now let’s use the power series
y(0) y 0 (0) 0 y 00 (0) 2
y(x) =
+
x +
x + ...
0!
1!
2!
20
= −3 − 8x − x3 + . . .
3
Example 4.7
Solve (−3x + 6)y 00 + 8y 0 + (2x − 5)y = 0; y(0) = 5, y 0 (0) = 0
First, evaluate at x = 0 to find y 00 (0)
25
*5
*0
y(0)
= 0 ⇒ y 00 (0) =
y 0 (0)
+ (−5)
6y 00 (0) + 0 − 5
6
Now take the derivative and evaluate at x = 0 to find y 000 (0)
−3y 00 (x) + (−3x + 6)y 000 (x) + 8y 00 (x) + 2y(x) + (2x − 5)y 0 (x) = 0
25/6
25/6
*0
*5
−3
y (0) + 6y (0) + 8
y (0) + 2
y(x)
− 5
y 0 (x)
=0
*
00
000
*
00
−
75
100
+ 6y 000 (0) +
+ 10 = 0
6
3
y 000 (0) = −
58
185
36
We can go further, but if we stop here, we have
y(x) = y(0) +
y 0 (0)
y 00 (0) 2 y 000 (0) 3
x+
x +
x + ...
1!
2!
3!
y(x) = 5 +
25 2 185 3
x −
x + ...
12
216
Example 4.8
Solve xy 00 + y 0 + y = 0; y(0) = −1, y 0 (0) = 1
Radius of Convergence
Definition Let p(x), q(x), and r(x) be analytic functions. A point x0 is a singular point for the
differential equation
p(x)y 00 (x) + q(x)y 0 (x) + r(x)y(x) = 0
if
(4.1)
r(x0 )
q(x0 )
or
is undefined. Otherwise, x0 is called a regular point.
p(x0 )
p(x0 )
Try solving the following:



xy 00 + y 0 + y = 0


y(0) = −1, y 0 (0) = 1
If we try evaluating at x = 0 to find y 00 (0), we end up multiplying the y 00 (0) by zero, so we can’t
get it by that method. We could rewrite this problem by dividing everyting by x
1
1
y 00 + y 0 + y = 0
x
x
Now we see that x = 0 is a singular point.
Find the singular points for the following equations:
x2 y 00 + xy 0 = 0 ⇒ x0 = 0
2y 00 +
y
= 0 ⇒ x0 = 1
x−1
59
Theorem 4.9 Suppose
y(x) =
∞
X
an (x − x0 )n
n=0
is a series solution do Equation 4.1. Then, the radius of convergence for y(x) is at least as large as
the distance from x0 (regular point) to the nearest singular point of the equation.
Example 4.10 Find the radius of convergence about x0
(x + 1)y 00 + y 0 + (x + 1)y = 0, about x0 = 0
So first we divide by (x + 1)
y 00 +
y0
+y =0
x+1
x = −1 is a singular point. So the radius of convergence is at least 1.
Example 4.11 Find the radius of convergence about x0
(x2 − 2x − 3)y 00 + (x + 1)y 0 + 3y = 0, about x0 = 0
First divide by (x+1), to get singular point x = −1. Then divide the original equation by (x2 −2x−3)
to get
00
y +
0
(x+1)y
+
(x − 3)
(x+1)
3y
=0
(x − 3)(x + 1)
which has singular points x = −1 and x = 3. So the radius of convergence is at least 1.
Example 4.12
(x3 + x2 + x + 1)y 00 + 2xy 0 + (x − 1)y = 0, about x0 = 0
Factor the big polynomial first
x3 + x2 + x + 1 = x2 (x + 1) + 1(x + 1) = (x2 + 1)(x + 1)
60
Now divide that out to get
y 00 +
2xy 0
(x − 1)y
+ 2
=0
2
(x + 1)(x + 1) (x + 1)(x + 1)
x = −1, x = ± ι. Radius of convergence is at least
61
√
2
5
Final Examples
Example 5.1
L {t cosh(t)}
cosh(t) =
F (s) =
=
L tn eat , n ∈ Z+ =
F (s) =
=
et + e−t
2
1 t
L te + te−t
2
1 t 1 −t L te + L te
2
2
n!
, s>a
(s − a)n+1
1
1!
1!
1
+
, s>1
1+1
2 (s − 1)
2 (s + 1)1+1
1
1
1
+
, s>1
2 (s − 1)2 (s + 1)2
Example 5.2
t
Z
e2(t−τ ) τ sin(3τ ) dτ
L {f (t)} , where f (t) = 3 cos(2t) +
0
Z
L {f (t)} = 3 L {cos(2t)} + L
t
2(t−τ )
e
τ sin(3τ ) dτ
0
s
2
s +4
s
=3 2
s +4
s
=3 2
s +4
s
=3 2
s +4
=3
+ L e2t ∗ t sin(3t)
+ L e2t L {t sin(3t)}
1
6s
s − 2 (s2 + 9)2
6s
+
(s − 2)(s2 + 9)2
+
62
Example 5.3
L
−1
2s + 9
2
s + 8s + 17
2s + 9
2s + 9
−1
=L
f (t) = L
s2 + 8s + 16 + 1
(s + 4)2 + 12
s
1
−1
−1
= 2L
+ 9L
(s + 4)2 + 12
(s + 4)2 + 12
s+4−4
1
−1
−1
+ 9L
= 2L
(s + 4)2 + 12
(s + 4)2 + 12
1
s+4
−1
−1
+L
= 2L
(s + 4)2 + 12
(s + 4)2 + 12
−1
= 2e−4t cos(t) + e−4t sin(t) = e−4t (2 cos(t) + sin(t))
Example 5.4
y 00 + 6y 0 + 8y = δ(t), y(0) = 0, y 0 (0) = 0
s2 L {y} + 6s L {y} + 8 L {y} = e−0s
L {y} (s2 + 6s + 8) = 1
1
+ 6s + 8
1
=
(s + 2)(s + 4)
1
A
B
=
+
(s + 2)(s + 4)
s+2 s+4
L {y} =
s2
1 = A(s + 4) + B(s + 2)
A = 1/2, B = −1/2
1 1
1 1
−
2s+2 2s+4
1 −2t
y=
e − e−4t
2
L {y} =
63
Example 5.5
y 00 + 2xy 0 + 4y = 0, y(0) = 1, y 0 (0) = 2
Just use:
∞
X
y (n) (0)
n=0
n!
xn
The solution is:
4
y(x) = 1 + 2x − 2x2 − 2x3 + x4 + . . .
3
Example 5.6 Find the radius of convergence for the series solution about x0 = 0 to the equation:
(x2 + 2x + 10)(x2 + 6x + 8)y 00 + 2xy 0 − 4y = 0
y 00 +
2xy 0
4y
−
=0
2
(x + 2)(x + 4)(x + 2x + 10) (x + 2)(x + 4)(x2 + 2x + 10)
You can actually just use the quadratic formula. The roots will be the singular points, so the singular
points are x = −1 ± 3i and x = −4, −2. The closest point is −1 − 3i, and it is 2 away from 0, so
the radius of convergence is at least 2.
Example 5.7
y 00 + 3y 0 + 2y =
1
=g
1 + ex
λ2 + 3λ + 2 = 0
λ = −2, λ = −1
yh = c1 e−2x + c2 e−x = y1 + y2
y1 y2 W =
y 0 y 0 1 2
Z
Z
y2 · g
y1 · g
yp = −y1
+ y2
W
W
64