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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