FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS

[Pages:20]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 = (x)/(y); then equation can be integrated.

Cases covered by this include y = (ax + by); y = (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 = /x, N = /y.

Then solution determined from (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. Costa McGraw-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

dy = f (x) dx g( y)

Solution :

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

Ex 1

dy = y2ex dx

!

"

dy y2

=

"

exdx

i.e !1 = ex + c or y

y = !1 (ex + c)

Almost separable equations

dy = f (ax + by) dx

Change variables : z = ax + by

dz dx

=

a

+b

dy dx

! dz = a + bf (z) ! x =

1

dz.

dxx

(a + bf (z))

Ex 2

dy = (!4x + y)2

dx

x

=

1 4

ln(

z z

!2 +2

)

+

C

z = y ! 4x

! dz = "4 + dy = z2 " 4

dx

dx

"

y = 4x + 2 (1+ke4x ) (1!ke4 x )

k a constant

Homogeneous equations

dy = f ( y/x). dx

The equation is invariant under x ! sx, y ! sy .... homogeneous

Solution

y = vx " y! = v'x + v.

i.e. v ' = 1 ( f (v) ! v) x

" "= dv f (v)!v

d x x

=

ln

x

+ constant.

Ex 3

xy dy ! y2 = (x + y)2 e! y/x dx

Homogeneous

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

$ (v"x + v) ! v = (1+ v)2 e!v v

#

ln x =

e v vdv .

(1+ v)2

To evaluate integral change variables u ! 1 + v

" e!1

(

1 u

!

1 u2

)eudu

=

e!1[

eu u

].

y

i.e. ln x = e x

1+

y x

Homogeneous but for constants

dy = x + 2 y +1 dx x + y + 2

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

! dy = dy ' = dy ' . dx ' = dy ' dx dx dx ' dx dx '

dy ' = x '+ 2 y '+1+ a + 2b dx ' x '+ y '+ 2 + a + b

1+ a + 2b = 0 2+a+b = 0

a = !3, b = 1

dy ' = x '+ 2 y ' dx ' x '+ y '

Homogeneous

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