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Transcript of BERN1
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First-Order Linear O.D.E. and Bernoulli Equations
a- Linear Equations
Definition: A first-order O.D.E. is linearin the dependent variable y and the independent
variable x if it can be expressed as
A(x)dy
dx+B(x)y =C(x)
Remark: If we divide each term by A(x), we getdy
dx+P(x)y =Q(x), whereP(x) = B(x)
A(x)andQ(x) = C(x)
A(x)
Example:
The equation xdy
dx+ (x +1)y=x
3 is linear where A(x) = x, B(x) = x + 1 and C(x) = x3.
We can rewrite the equation in the form
dydx
+ (1+ 1x)y =x
2,whereP(x) = 1+ 1
xandQ(x) =x2
Lets rewritedy
dx+P(x)y =Q(x) in differential form
[P(x)y Q(x)] dx + dy = 0 where M(x,y) = P(x)y Q(x) and N(x,y) = 1
since!M(x,y)
!y=P(x) and !N(x,y)
!x=0 , the equation is not exact unless P(x) = 0.
Lets find an integrating factor.Assume that the integrating factor depends on x, only, lets call it n(x).
Then, multiplying the equation by n(x), we getn(x)[P(x)y Q(x)] dx + n(x) dy = 0
and it is an exact equation,Therefore,
!
!yn(x)P(x)y"n(x)Q(x)[ ] =n(x)P(x) =
!
!xn(x)[ ]
or
n(x)P(x) =dn
dx, where P(x) is a known function of x, but n(x) is an unknown function.
Solving the differential equation we can determine n(x).
The equation n(x)P(x) =dn
dx is separable O.D.E. and it can be expressed as
P(x)dx =dn
n
integrating, we get
P(x)dx =dn
n!!
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ln n(x) = P(x)dx!
n(x) =exp( P(x)dx) =e P(x)dx!! istheint egratingfactor
Then e P(x)dx! dy
dx+e
P(x)dx!P(x)y =e
P(x)dx!Q(x)
Notice thatd
dxe
P(x)dx! y"#$
%&'= e
P(x)dx! dy
dx+e
P(x)dx! P(x)y
Therefore,d
dxe P(x)dx! y"
#$%&' =e
P(x)dx! Q(x)
Integrating both sides, we get
d
dxe
P(x)dx! y"#$
%&'! dx =e
P(x)dx! y = e P(x)dx! Q(x)! dx +c
solvingforyy =e
( P(x)dx! e P(x)dx! Q(x)! dx +c
"#$
%&'
Theorem: Given the first-order linear O.D.E.dy
dx+P(x)y =Q(x) , it has an integrating
factor of the form n(x) =exp( P(x)dx)=e P(x)dx!! and a one-parameter family of
solutions y =e! P(x)dx" e
P(x)dx" Q(x)" dx +c#$%&'(
.
Examples: Solve the equations
1)dy
dx+2xy =4x
P(x) = 2x and Q(x) = 4x
The integrating factor is n(x) = e P(x)dx!
=e 2xdx!
=ex
2
Multiply the equation by n(x)
ex2 dy
dx+e
x22xy =4xe
x2
d
dx
ex2y!
" #
$= 4xe
x2
d
dxex2y!
" #
$dx =% ex2y = 4xex2dx% =2ex2 +c
y =e&x2
2ex2
+c( ) =2 +ce&x2
2) 2 y !4x2( )dx +xdy = 0
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It can be written as
dy
dx+2y
x!
8x2
x=0 or dy
dx+2
xy = 8x where P(x) = 2/x and Q(x) = 8x
The integrating factor is n(x) =e P(x)dx!
=e
2dx
x!
=e2ln x
=eln x
2
=x2
Multiply the equation by n(x)
x2 dy
dx+x
2 2
xy =8x
3
d
dxx2y!" #
$ =8x3
d
dxx2y!" #
$dx = x2y = 8x
3dx = 2x
4+c%%
y =x&2
2x4+c( ) =2x2 +cx&2
3) y ln y dx + (x ln y) dy = 0If we consider y the dependent variable, then the equation is not linear since we have a
transcendental function of y. But , if we consider x the dependent variable, the equationcan be expresses as
dx
dy+
1
ylnyx =
1
y
then, it is linear in the dependent variable x and the independent variable y.
The integrating factor is n(y) = exp dy
ylny!"#$
%&' = exp ln lny( )( ) =ln y .
Multiply the equation by n(y)
lnydydx
+ lnyylny
x = lnyy
d
dxxlny[ ] =
lny
y
d
dxxlny[ ]dy = xlny =
lny
ydy =
1
2ln
2y +c!!
x = lny( )"1 1
2ln
2y +c
#$%
&'(
b- Bernoulli EquationDefinition: Bernoulli Equationare non-linear differential equations that can betransformed into linear O.D.E.. They have the form:
dy
dx+P(x)y =Q(x)y
n
where n is a real number, n!1 or 0.
If we multiply the Bernoulli equation by y-n
, we get
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y!n dy
dx+P(x)y
1!n=Q(x)
Taking the transformation v = y1-n
, then
dv
dx=(1!n)y
!ndy
dx or 1
1!nyn dv
dx=
dy
dx
and the Bernoulli equation is transformed1
1!n
dv
dx+P(x)v =Q(x) or dv
dx+ (1!n)P(x) =(1!n)Q(x)
calling P1(x) = (1-n)P(x) and Q1(x) = (1-n)Q(x), the equation reads
dv
dx+P1(x)v =Q1(x)
that is a linear equation in the dependent variable v and the independent variable y.
Example: Solve the equation
1)dy
dx
+2xy +xy4=0
Multiply by y-4
,
y!4 dy
dx+2xy
!3=!x
The transformation v = y-3
, !3y!4dy
dx=
dv
dxreduces the equation to
!
1
3
dv
dx+2xv =!x or dv
dx!6xv = 3x a linear equation with P(x) = -6x.
The integrating factor is n(x) = e !6xdx"
= e!3x2
Then
e!3x2 dv
dx!6e!3x
2
xv =3e!3x2
x
d
dxe!
3x2v"#
$% = e
!3x2 dv
dx!6e!3x
2
xv =3e!3x2
x
d
dxe!
3x2v"#
$%dx =& e!3x2v = 3e!3x2xdx =!12 e!3x
2
& +c
v=e3x2 !12e!
3x2+c
'()
*+, =!
1
2+ce
3x2
y!3=!
1
2
+ce3x2
2) x dy {y + xy3(1 + ln x)}dx = 0
It can be expressed as
dy
dx!
1
xy =(1+lnx)y
3 ory!3dydx
!
1
xy!2
=(1+lnx)
The transformation v = y-2
, !2y!3dy
dx=
dv
dxreduces the equation to
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!
1
2
dv
dx!
1
xv = 1+ln x( ) or dv
dx+ 2
xv =!2 1+ln x( )
is a linear differential equation in the dependent variable v and the independent variable
with P(x) = 2/x.
The integrating factor is n(x) = e
2
xdx!=
e
2lnx=
x
2
Then
x2 dv
dx+
2x2
xv =!2x2 1+ln x( )
d
dxx2v"# $% = x
2 dv
dx+2xv =!2x2 1+ln x( )
d
dxx
2v"# $%dx =x
2v =!2 x2 +x2 ln x( )&& dx =!2
x3
3+
x3
3ln x!
x3
9
'
()*
+, +c =!
4
9x
3 !2
3x
3ln x +c
v =x!2 !
4
9
x3 !2
3
x3 ln x +c'
()
*
+,
y!2 =!2
3x
2
3!ln x'
() *
+,+c