FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS

G(x, y, y′) = 0

♦ in normal form:

y′ = F (x, y)

♦ in differential form:

M(x, y)dx+N(x, y)dy = 0

• Last time we discussed first-order linear ODE: y′ + q(x)y = h(x).We next consider first-order nonlinear equations.

NONLINEAR FIRST-ORDER ODEs

• No general method of solution for 1st-order ODEs beyond linear case;rather, a variety of techniques that work on a case-by-case basis.

Examples:

i) Bring equation to separated-variables form, that is, y′ = α(x)/β(y);

then equation can be integrated.

Cases covered by this include y′ = ϕ(ax+ by); y′ = ϕ(y/x).

ii) Reduce to linear equation by transformation of variables.

Examples of this include Bernoulli’s equation.

iii) Bring equation to exact-differential form, that is

M(x, y)dx+N(x, y)dy = 0 such that M = ∂φ/∂x, N = ∂φ/∂y.

Then solution determined from φ(x, y) = const.

• Useful reference for the ODE part of this course(worked problems and examples)

Schaum’s Outline Series

Differential Equations

R. Bronson and G. CostaMcGraw-Hill (Third Edition, 2006)

♦ Chapters 1 to 7: First-order ODE.

First order nonlinear equations

Although no general method for solution is available, there are several cases of

physically relevant nonlinear equations which can be solved analytically :

Separable equations

d ( )

d ( )

y f x

x g y=

( ) ( )g y dy f x dx=! !Solution :

2de

d

xyy

x=Ex 1

2e dxdyx

y=! " "

1exc

y

!= +i.e

1

(e )xy

c

!=

+or

Almost separable equations

d( )

d

yf ax by

x= +

z ax by= +dd

d d

yz

x xa b= +Change variables :

1d

( ( ))x z

a bf z= .

+!d

( )d

za bf z

x= + !

2d( 4 )

d

yx y

x= ! + 4z y x= !

2d d4 4

d d

z yz

x x! = " + = "

1 24 2ln( ) Cz

zx

!

+= +

4

4

(1 e )

(1 e )4 2

x

x

k

ky x

+

!" = +

Ex 2

k a constant

z ax by= +

d( )

d

za bf z

x= + !

d( )

d

yf y x

x= / .

Homogeneous equations

The equation is invariant under , .. h.. omogeneousx sx y sy! !

v v' vy x y x!= " = + .Solution

d(v) v

d ln constantv

f

xx

x! = = + ." "

1. . v ' ( (v) v)i e f

x= !

2 2d( ) e

d

y xyxy y x y

x

! /! = +Ex 3

2 vv

2

(1 v) e vdv(v v) v e ln

v (1 v)x x

!+" + ! = # = .

+$

1 vu ! +

1 1

2

1 1 ee ( )e d e [ ]

u

uu

u u u

! !! = ."

v v vy x y x! != " = + .Change variables

To evaluate integral change variables

. . ln1

y

x

y

x

ei e x =

+

Homogeneous

Homogeneous but for constants

2 1

2

dy x y

dx x y

+ +=

+ +

' , 'x x a y y b= + = +' ' ' '

.' '

dy dy dy dx dy

dx dx dx dx dx! = = =

' ' 2 ' 1 2

' ' ' 2

dy x y a b

dx x y a b

+ + + +=

+ + + +

1 2 0a b+ + =

2 0a b+ + =

3, 1a b= ! =

' ' 2 '

' ' '

dy x y

dx x y

+=

+Homogeneous

The Bernoulli equation

d( ) ( ) , 1

d

nyP x y Q x y n

x+ = !

To solve, change variable to 1 n

z y!

= (1 ) ndz dyn y

dx dx

!" = !

(1 ) ( ) (1 ) ( )dz

n P x z n Q xdx

+ ! = !Gives the equation

Ex 42/3

'y y y+ =1 1/3n

z y y!

= = 1

3'3

zz! + =

x/3Integrating factor e

1/3 /31

xz y ce

!= = +

/ 3 /3/ 3

x xze e dx! = "

1st order Linear

1st order linear

Exercise:

Solve the equation 2 y ′ = y/x+ x2/ywith initial condition y(1) = 2.

• This equation is Bernoulli with n = −1.

• Set z = y2. Then z′ − z/x = x2.

• Integrating factor I(x) = 1/x

⇒ z(x) = x [

∫

dx x2/x+ const.] = x3/2 + const. x

Thus y = z1/2 = ±√

x3/2 + const. x

• Initial condition y(1) = 2

⇒ y(x) =

√

x3 + 7x

2

Homework

1. Solve the differential equation

2dy

dx=

y(x+ y)

x2“homogeneous”

with y(1) = −1.

[Answ.: y = x/(1− 2√x)]

2. Solve the differential equation

dy

dx+ xy = xy2 “Bernoulli”

with y(0) = 1/2.

[Answ.: y = 1/(1 + ex2/2)]

Exact equations

• A first-order ODE

M(x, y)dx+N(x, y)dy = 0

is exact if there exists a function φ(x, y) such that

∂φ

∂x= M ,

∂φ

∂y= N .

• In this case the differential equation can be recast as

dφ = M(x, y)dx+N(x, y)dy = 0

so that the solution to it is determined by

φ(x, y) = constant .

Example: Solve the equation xy′ = −2 tan y.

• This equation can be rewritten as

2x sin y dx+ x2 cos y dy = 0 ,

i.e., M(x, y) = 2x sin y , N(x, y) = x2 cos y ,

which is exact because∂φ

∂x= 2x sin y ⇒ φ(x, y) = x2 sin y + α(y)

∂φ

∂y= x2 cos y ⇒ x2 cos y + α′(y) = x2 cos y ⇒ α = constant

• Therefore φ(x, y) = x2 sin y + c,and the general solution is determined by x2 sin y = const.:

⇒ y(x) = arcsin(

const./x2)

DIFFERENTIAL EQUATIONS AND FAMILIES OF CURVES

• General solution of a first-order ODE y′ = f(x, y)

contains an arbitrary constant: y = (x, c)

⊲ one curve in x, y plane for each value of c⊲ general solution can be thought of as one-parameter family of curves

Example: y′ = −x/y.

separable equation ⇒

∫

y dy = −

∫

x dx ⇒ y2/2 = −x2/2 + c

i.e., x2 + y2 = constant : family of circles centered at origin

Fig.1

x

y

Orthogonal trajectories

• Given the family of curves representing solutions of ODE y′ = f(x, y),

orthogonal trajectories are given by a second family of curves which are

solutions of

y′ = −1/f(x, y).

♦ Then each curve in either family is perpendicularto every curve in the other family.

Example:

Find the orthogonal trajectories to the family of circles y′ = −x/y.

• Solve y′ = y/x .

⇒

∫

dy

y=

∫

dx

x⇒ ln y = ln x+ constant

i.e., y = cx : family of straight lines through the origin

Homework

a) Find the family of curves corresponding to solutions of the ODE

y′ = (y2 − x2)/(2xy).

b) Find the orthogonal trajectories to the above family of curves.

• homogeneous equation y′ = f(y/x) with f(y/x) = (y/x− x/y)/2

solvable by y→v = y/x and separation of variables

⇒ x2 + y2 = cx : family of circles tangent to y − axis at 0

y

x

Fig.2

• orthogonal trajectories found by solving y′ = −2xy/(y2− x2)

⇒ x2 + y2 = ky : family of circles tangent to x− axis at 0

EXPLOITING FIRST-ORDER METHODS TO TREAT EQUATIONS OF

HIGHER ORDER IN SPECIAL CASES

♣ y not present in 2nd-order equation F (x, y, y′, y′′) = 0

⇒ setting y′ = q yields 1st-order equation for q(x).

♣ x not present in 2nd-order equation F (x, y, y′, y′′) = 0

⇒ setting y′ = q, y′′ = dq/dx = q(dq/dy) yields G(y, q, dq/dy) = 0.

2

x

y

T

θT

s

Example: homogeneous, flexible chain hanging under its own weight

ρ = linear mass density

Using Newton’s law, the shape y(x) of the chain obeys the 2nd−order nonlinear differential equation

y = a 1 + (y )2 , a ρ g / T

Setting y = q q = a 1 + q

• Separation of variables ⇒

∫

1√

1 + q2dq = a

∫

dx

• Using q = dy/dx = 0 at x = 0 ⇒ ln(q +√

1 + q2) = ax

• Solving for q ⇒ q = dy/dx = (eax − e−ax)/2

Thus y(x) =1

a

eax + e−ax

2+ constant =

1

acosh ax+ constant

This curve is called a catenary.

Historical note. The problem of the catenary was the subject of a challenge

posed by Jakob Bernoulli in 1690, in response to which the problem was solved

the following year indipendently by Johann Bernoulli, Leibniz and Huygens.

Homework

1. Find the function y(x) obeying the differential equation

y′2 = x2y′′

and the conditions y(0) = 2, y′(1) = 2.

[Hint: set y′ = q and apply separation of variables.]

[Answ.: y(x) = 2(1− x)− 4 ln(1− x/2)]

2. Find the function y(x) obeying the differential equation

y′′ = y′ey

and the conditions y(0) = 0, y′(0) = 1.

[Hint: y′ = q; y′′ = dq/dx = q(dq/dy); solve equation for q(y).]

[Answ.: y(x) = − ln(1− x)]

Summary

♦ No general method of solution for 1st-order ODEs beyond linear case;

rather, a variety of techniques that work on a case-by-case basis.

Main guiding criteria:• methods to bring equation to separated-variables form

• methods to bring equation to exact differential form

• transformations that linearize the equation

♦ 1st-order ODEs correspond to families of curves in x, y plane⇒ geometric interpretation of solutions

♦ Equations of higher order may be reduceable to first-order problems inspecial cases — e.g. when y or x variables are missing from 2nd order

equations