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Introduction to ordinary differential equation

40

PART3: INTRODUCTION TO ORDINARY DIFFERENTIAL EQUATIONS 108

Transcript

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

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