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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