ordinary differential equation
DESCRIPTION
Introduction to ordinary differential equationTRANSCRIPT
PART3:
INTRODUCTION TO ORDINARY
DIFFERENTIAL EQUATIONS
108
Take-off run of an aircraft (GJ 10.2)
G = mg − L
L = αv2, D = βv2, R = µG
Newton:
ma = T −D −R = T −D − µ(mg − L)
With v = dsdt :
md2s
dt2= T − β
(ds
dt
)2− µmg + µα
(ds
dt
)2So:
md2s
dt2− (µα− β)
(ds
dt
)2= T − µmg
Question: what should be the minimum length
of the runway?
109
Vibrations of a pre-tensioned string
Note: you do not need to know the derivationthat follows!
x
y(x,t)
String of length ℓ with constant mass densityρ and string tension T .
y(x, t): y-coordinate of string at position x andtime t.
Assume y(x, t) is small.
110
T
T y(x,t) y(x+dx,t)
α(x,t)
α(x+dx,t)
dx
Newton’s Second Law:
ρdx∂2y
∂t2= T (sinα(x+ dx)− sinα(x))
Taylor:
sinα(x+ dx) ≈ sinα(x) +d
dx(sinα(x))dx =
sinα(x) + cosα(x)∂α(x)
∂xdx
111
Have:
tanα =∂y
∂xso
1
cos2α
∂α
∂x=
∂2y
∂x2
Gives:
sinα(x+ dx) ≈ sinα(x) + cos3α(x)∂2y
∂x2dx ≈
sinα(x) +∂2y
∂x2dx
because y small ⇒ α small ⇒ cosα ≈ 1.
112
So:
ρdx∂2y
∂t2= T (sinα(x+ dx)− sinα(x)) ≈
T (sinα(x) +∂2y
∂x2dx− sinα(x)) =
T∂2y
∂x2dx ⇔
∂2y
∂t2= c2
(∂2y
∂x2
), c =
√T
ρ
A partial differential equation of this form is
known as a wave equation.
113
Classification of differential equations (GJ
10.3)
Differential equation: an equation involving vari-
able(s) and derivative(s) of variables.
Ordinary differential equation: a differential
equation only involving ordinary derivatives.
Partial differential equation: a differential equa-
tion involving partial derivatives.
Independent variables: variables with respect
to which differentiation occurs.
Dependent variables: variables that are being
differentiated.
114
Examples
∂f
∂x+
∂f
∂y= 4x2 +2y
is a partial differential equation.
Independent variable(s): x, y.
Dependent variable(s): f .
d2f
dx2− 4x
df
dx= cos2x
is a ordinary differential equation.
Independent variable(s): x.
Dependent variable(s): f .
115
4dxdt +3dy
dt − x+2y = cos t
6dxdt − 2dy
dt − 2x+ y = 2sin t
is a ordinary differential equation.
Independent variable(s): t.
Dependent variable(s): x, y.
116
Order of a differential equation: the degree of
the highest derivative occurring in the differ-
ential equation.
Examples
∂3f
∂x∂y2+
∂2f
∂x2− x
∂f
∂y= 4x2 +2y
is a third-order partial differential equation.
d2f
dx2− 4x
df
dx= cos2x
is a second-order ordinary differential equation.
117
(dx
dt
)2+4
dx
dt= 0
is a first-order ordinary differential equation.
118
Linear differential equation: a differential equa-tion in which the dependent variable(s) andtheir derivatives do not occur as products, raisedto powers or in nonlinear functions.
Nonlinear differential equation: a differentialequation that is not linear.
Examples
∂3f
∂x∂y2= 4x2 +2y
is a linear partial differential equation.
(dx
dt
)2+4
dx
dt= 0
is a nonlinear ordinary differential equation.
119
d2x
dt2+ x
dx
dt= 4sin t
is a nonlinear ordinary differential equation.
d2f
dx2− 4x
df
dx= cos2x
is a linear ordinary differential equation.
120
Homogeneous linear differential equation:a linear differential equation that doesn’t con-tain any terms depending on the independentvariable(s) only.
Nonhomogeneous linear differential equation:a linear differential equation that is not homo-geneous, i.e., contains terms that depend onthe independent variable(s) only.
Examples
4dx
dt+ (sin t)x = 0
is a homogeneous linear differential equation.
d2x
dt2+ t
dx
dt= 4sin t
is a nonhomogeneous linear differential equa-tion.
121
Exercise Classify the following differential equa-tions.
∂2f
∂x2+ y
(∂f
∂x
)(∂f
∂y
)+ sin y = 0
d2s
dt2+ (sin t)
ds
dt+ (t+ cos t)s = et
dr
dz+ z2 = 0
122
dy
dx= y + xy2
dz
dx= −z − x
Is there any relationship between the last two
differential equations?
123
Solving differential equations (GJ 10.4)
Solution: A function of the independent vari-
able(s) that satisfies the differential equation.
Example
∂f
∂x+
∂f
∂y= 0
Solution:
f(x, y) = x− y
because∂f
∂x+
∂f
∂y=
∂(x− y)
∂x+
∂(x− y)
∂y= 1− 1 = 0
However, e.g. f(x, y) = (x − y)2, f(x, y) =
sin(x − y), f(x, y) = sin(cos
(tan
(ex−y
)))are
also solutions.124
Example
d2x
dt2= e2t
Solution:
x(t) =1
4e2t
because
d2x
dt2=
d
dt
(d
dt
(1
4e2t
))=
d
dt
(1
2e2t
)= e2t
However, x(t) = 1 + 14e
2t, x(t) = t√2 + 1
4e2t,
x(t) = π +5t+ 14e
2t are also solutions.
125
A more constructive way to find solution(s):
d2x
dt2= e2t ⇔
dx
dt=∫
e2tdt =1
2e2t + c ⇔
x(t) =∫ (
1
2e2t + c
)dt =
1
4e2t + ct+ d
for ALL constant values of c and d.
This expression is called THE general solution
of the differential equation.
The number of free parameters in the general
solution of an ordinary differential equation is
always equal to the order of the differential
equation.
Solutions for a particular choice of c and d are
called A particular solution of the differential
equation.
So solutions on the previous page are all par-
ticular solutions.126
From general solutions to particular solu-
tions
Example: Initial value problem
Find the solution of
d2x
dt2= e2t
that satisfies x = 0, dxdt = 1 when t = 1
Solution General solution:
x(t) =1
4e2t + ct+ d ⇒
dx
dt=
1
2e2t + c
Then:
1 =dx
dt(1) =
1
2e2 + c ⇔ c = 1−
1
2e2
127
0 = x(1) =1
4e2 + c+ d ⇔
d = −(1
4e2 + c
)= −
(1−
1
4e2)
So the particular solution sought is:
x(t) =1
4e2t +
(1−
1
2e2)t−
(1−
1
4e2)
128
Example: boundary value problem
Find the solution of
d2x
dt2= e2t
that satisfies x = 0, when t = 0 and x = 1
when t = 1.
Solution General solution:
x(t) =1
4e2t + ct+ d
0 = x(0) =1
4+ d ⇔ d = −
1
4
1 = x(1) =1
4e2+c+d ⇔ c = 1−
1
4e2−d =
5
4−1
4e2
So the particular solution sought is
x(t) =1
4e2t +
1
4
(5− e2
)t−
1
4
129
Solution of first-order ordinary differential
equations: separable equations (GJ10.5.3)
A first-order ordinary differential equation
dx
dt= f(t, x)
is called separable if it can be manipulated into
the form
g(x)dx = h(t)dt
The solution can then be obtained by integra-
tion of both sides of the equality.
130
Example
Find the general solution of
dx
dt= −4x
Solution
Rewrite as
dx
x= −4dt ⇔
∫dx
x= −4
∫dt ⇔
lnx = −4t+ d ⇔ x(t) = ede−4t = ce−4t
131
Example
Find the general solution of
dx
dt= ax(x− 1)
Solution
Rewrite:
dx
x(x− 1)= adt ⇔
∫adt =
∫dx
x(x− 1)=
∫ (1
x− 1−
1
x
)dx ⇔
ln(x− 1
x
)= at+ d ⇔
x− 1
x= ceat ⇔
x(t) =1
1− ceat
132
Example
Find the solution of the following initial value
problem:
dx
dt=
t2 +1
x+2, x(0) = −2
Solution
(x+2)dx = (t2+1)dt ⇔1
2x2+2x =
1
3t3+t+d ⇔
3x2 +12x = 2t3 +6t+ c
From initial condition:
c = 3x(0)2 +12x(0) = −12
133
So
3x2 +12x− 2(t3 +3t− 6) = 0 ⇔
x(t) =−12±
√24(t3 +3t)
6=
−2±1
3
√6(t3 +3t)
134
GJ10.5.5
Consider a differential equation of the form
dx
dt= f
(x
t
)
This can be turned into a more easily solvable
differential equation by defining
y =x
t
Because:
dy
dt=
d
dt
(x
t
)=
1
t
dx
dt−
x
t2=
1
tf
(x
t
)−
x
t2=
1
t(f(y)− y) ⇔
dy
f(y)− y=
1
tdt
so we can now integrate to solve.
135
Example Find the general solution of
x2dx
dt=
t3 + x3
t
Solution Divide by x2:
dx
dt=
t3 + x3
x2t=
t2
x2+
x
t= f
(x
t
), f(y) =
1
y2+ y
So with y = xt :
dy(1y2
+ y
)− y
=1
tdt ⇔
∫y2dy =
∫ 1
tdt ⇔
1
3y3 = ln t+ d ⇔ y(t) = 3
√ln t+ c
and
x(t) = ty(t) = t 3√ln t+ c
136
Solution of exact differential equations
(GJ10.5.7)
Consider the differential equation
p(t, x)dx
dt+ q(t, x) = 0
This differential equation is called exact if there
exists a function h(t, x) such that
∂h
∂x= p(t, x),
∂h
∂t= q(t, x)
Then:
0 = p(t, x)dx
dt+ q(t, x) =
∂h
∂x·dx
dt+
∂h
∂t=
d
dt(h(t, x))
and hence
h(t, x) = c
137
Example
Find the general solution of the following dif-
ferential equation:
(x+ t)dx
dt+ x− t = 0
Solution Consider
h(t, x) =1
2x2 + xt−
1
2t2
Then∂h
∂x= x+ t,
∂h
∂t= x− t
Hence1
2x2 + xt−
1
2t2 = c ⇔
x(t) = −t±√2t2 + c
138
How to recognise exactness?
Differential equation:
p(t, x)dx
dt+ q(t, x) = 0
Exactness: want h(t, x) such that
∂h
∂x= p(t, x),
∂h
∂t= q(t, x)
Then:
∂p
∂t=
∂2h
∂t∂x=
∂2h
∂x∂t=
∂q
∂x
139
How to find h(t, x)
Example
Find the general solution of
textdx
dt+ t+ xext = 0
Solution We have
p(t, x) = text, q(t, x) = t+ xext
∂p
∂t= ext + xtext
∂q
∂x= ext + xtext
so the differential equation is exact.
140
Now need h(t, x) such that
∂h
∂x= text,
∂h
∂t= t+ xext
Integrate the first equality:
h(t, x) =∫
textdx = ext + c(t)
Note: integration variable is x, so t is to be in-
terpreted as constant. Hence: the integration
”constant” should be taken as a function of t.
Next integrate the second equality:
h(t, x) =∫(t+ xext)dt =
1
2t2 + ext + d(x)
Note: integration variable is t, so x is to be in-
terpreted as constant. Hence: the integration
”constant” should be taken as a function of x.
Compare: need c(t) = 12t
2, d(x) = 0.
141
So
h(t, x) = ext +1
2t2 = C ⇔
x(t) =1
tln(C −
1
2t2)
142
Linear homogeneous differential equations
(GJ10.5.9)
dx
dt+ p(t)x = 0
Integrating factor:
g(t) = exp(∫
p(t)dt)
Note:
g′(t) = exp((∫
p(t)dt)p(t) = g(t)p(t)
Multiply differential equation by g(t):
g(t)dx
dt+ g(t)p(t)x = 0 ⇔ g(t)
dx
dt+ g′(t) = 0 ⇔
d
dt(g(t)x) = 0 ⇔ g(t)x = c ⇔
x(t) =c
g(t)= c exp
(−∫
p(t)dt)
143
Linear nonhomogeneous differential equa-
tions (GJ10.5.9)
dx
dt+ p(t)x = r(t)
Again multiply by integrating factor g(t):
g(t)dx
dt+ g(t)p(t)x = g(t)r(t) ⇔
d
dt(g(t)x) = g(t)r(t) ⇔ g(t)x =
∫g(t)r(t)dt+c ⇔
x(t) =1
g(t)
(∫g(t)r(t)dt+ c
)
144
Example
Find the general solution of
dx
dt+
x
t= 0
Solution Integrating factor:
g(t) = exp(∫ 1
tdt
)= exp (ln t) = t
General solution:
x(t) =c
g(t)=
c
t
145
Example
Find the solution of the initial value problem
dx
dt−
x
t2=
4
t2, x(1) = 0
Solution Integrating factor:
g(t) = exp(−∫ 1
t2
)= exp
(1
t
)
Multiply by integrating factor:
d
dt
(x exp
(1
t
))=
4
t2exp
(1
t
)⇔
x exp(1
t
)=∫ 4
t2exp
(1
t
)dt = −4exp
(1
t
)+c ⇔
x(t) = −4+ c exp(−1
t
)
146
Initial value:
0 = x(1) = −4+ ce−1 ⇔ c = 4e
So solution sought is
x(t) = 4(exp
(1−
1
t
)− 1
)
147