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PERTEMUAN II

Homogeneous Equations

The differential equation

is homogeneousif the functionf(x,y) is homogeneous, that is-

f (tx , ty ) = f( x , y ) for any number t

Check that the functions

.

are homogeneous.

In order to solve this type of equation we make use of a sustitution (as we did

in case of !ernoulli equations). Indeed, consider the sustitution

x

yv = atau

.

Iff(x,y) is homogeneous, then we have

"ince y' = xz' + z,the equation (H) ecomes

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which is a separale equation. $nce solved, go ack to the old varialeyvia

the equationy = x z.

%et us summari&e the steps to follow'

(1)ecogni&e that your equation is an homogeneous equation that is, you need

to check that f(tx,ty) =f(x,y),meaning thatf(tx , ty)is independent of

the variale t;

(2)*rite out the sustitutionz =y/x;

(3)Through easy differentiation, find the new equation satisfied y the new

functionz.

+ou may want to rememer the form of the new equation'

()"olve the new equation (which is always separale) to findz

(!)o ack to the old functionythrough the sustitutionyx z

(")If you have an I/, use the initial condition to find the particular solution.

"ince you have to solve a separale equation, you must e particularly

careful aout the constant solutions

E#am$%e&0ind all the solutions of

'o%ution&0ollow these steps'

(1)It is easy to check that

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is homogeneous

(2)Consider

(3)*e have

,

which can e rewritten as

This is a separale equation. If you don2t get a separale equation at this

point, then your equation is not homogeneous, or something went wrong

along the way.

()3ll solutions are given implicitly y

(!)!ack to the functiony, we get

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5ote that the implicit equation can e rewritten as

onto *

#. "elesaikan /.6 : ( 2x y ) dy = ( x 2y ) dx ruas kiri dan kanan

masing-masing diagi denganx

dx)x

y2-1(dy)

x

y-2(dx

x

2yx -dy

x

y2x -==

sustitusi y = vx dan dy = x dv + v dx

( 2 v ) ( x dv + v dx ) = ( 1 2 v ) dx

x ( 2 v ) dv = ( ( 1 - 2 v ) - v ( 2 v ) ) dx

x ( 2 v ) dv = ( 1 - 2 v - 2v + v2 ) dx

7dv

8#

v-11

dx

vv=

+ - ln ( 1 4v + v2) = ln x +

ln c1

x2( 1 - 4y/x + y2/x2 ) = c atau

x2 - 4 xy + y2 = c

1. "elesaikan /.6 ' 2 xy dy = ( 4 x2+ 3 y2) dx ruas kiri dan kanan

diagi denganx2

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d7)7

y48(dy)

7

y1(

7

4y87dy

11

1

1

11

1 +=

+= dx

x

xy

2v dy = ( 4 + 3 v 2) dx 2v ( v dx + x dv ) = ( 4 + 3 v2)

dx

2vx dv = ( 4 + 3v2 2 v2) dx

7

dv8

v11

dx

v=

+ - ln ( 4 + v2) = ln x + ln

c1

4 + v2 = cx 4 + y2/x2 = cx atau 4 x2+ y2 = cx3

4. "elesaikan /.6 'y97ln7

y

7 += y

dx

dysustitusi y = vx

( xdv + v dx ) = ( v + v / lnv ) dx

d7)v-ln v

v( += vdvx

7

dvln dxvv = ln v d lnv

xdx

: ln1v ln 7 ; ln c

ln1v 1 ln 7 ; c sustitusi y v7 didapat

ln2(y/x) = 2 ln x + c

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E#a+t an None#a+t Equations

3ll the techniques we have reviewed so far were not of a general nature since

in each case the equations themselves were of a special form. "o, we may ask,

what to do for the general equation

%et us first rewrite the equation into

This equation will e called e#a+tif

,

and none#a+totherwise.

!ila =7act maka (E) mempunyai penyelesaian >mum '

=+=+ cdyy)5(7,d7a)?(7,ataucdyy)5(a,d7),( yxM

onto soa%*

#. "elesaikan /.6 : ( x + y 10 ) dx + ( x y 2 ) dy =

0

@awa. M (x , y ) = ( x + y 10 )

N ( x , y ) = ( x y 2 )

xyM

==

5#

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Badi e7act., sehingga penyelesaian umumnya adalah '

cdy)1-y-(d7)#C-y7( =++

x2

+ xy 10 x - y2

2y = c

1. "elesaikan /.6 ' ( 2x y ) dx + ( y x ) dy = 0

@awa.xy

M

==

5#- Badi e7act., sehingga

penyelesaian

umumnya cdy)y(d7)y71( =+

x2 xy + y2 = c

4. "elesaikan /.6 ' ( x3 3 xy2) dx + ( y33x2y ) dy =0 , y(0 ) = 1

@awa.xy

M

==

5A7y- e7act

cdy)y(d7)7y47(414

=+

D 78 - 491 71y1 ; D y8 c sustitusi y # dan 7

didapat c D

@adi penyelesaian khusus x4 6 x2y2+ y4 =1

8. "elesaikan /.6 ' ( x3 + 3xy2) dx + ( 3x2y + y3) dy =0

@awa.xy

M

==

5A7y e7act

cdy)y(d7)7y47( #414

=++ x4+ 3/2 x2y2 + y4 =c1

x4+ 6 x2y2 + y4 = c

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the integrating factor is e7tremely difficult e7cept for the following two special

cases'

ase 1&There e7ists an integrating factor u(x) function ofxonly. This happens

if the e7pression

,

is a function ofxonly, that is, the varialeydisappears from the

e7pression. In this case, the function uis given y

ase 2&There e7ists an integrating factor u(y) function ofyonly. This happens

if the e7pression

,

is a function ofyonly, that is, the varialexdisappears from the

e7pression. In this case, the function uis given y

$nce the integrating factor is found, multiply the old equation y uto get a new

one which is e7act. Then you are left to use the previous technique to solve the

new equation.

3dvice' if you are not pressured y time, check that the new equation is in fact

e7actG

%et us summari&e the aove technique. Consider the equation

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If your equation is not given in this form you should rewrite it first.

'te$ 1&Check for e7actness, that is, compute

,

then compare them.

'te$ 2&3ssume that the equation is not e7act (if it is e7act go to step ). Then

evaluate

If this e7pression is a function ofxonly, then go to step 4. $therwise,

evaluate

If this e7pression is a function ofyonly, then go to step 4. $therwise,

you can not solve the equation using the technique developped aoveG

'te$ 3&0ind the integrating factor. *e have two cases'

If the e7pression is a function ofxonly. Then an integrating factor is

given y

3*2

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If the e7pression is a function ofyonly, then an integrating factor is

given y

'te$ &?ultiply the old equation y u, and, if you can, check that you have anew equation which is e7act.

'te$ !&"olve the new equation using the steps descried in the previous section.The following e7ample illustrates the use of the integrating factor technique'

onto

'o%ution&5ote that this equation is in fact homogeneous. !ut let us use thetechnique of e7act and none7act to solve it. %et us follow these steps'

(1)*e rewrite the equation to get

.

(2)*e have

,

which clearly implies that the equation is not e7act.

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(3)%et us find an integrating factor. *e have

.

Therefore, an integrating factor u(x) e7ists and is given y

()The new equation is

,

which is e7act. (Check itG)

(!)%et us findF(x,y). Consider the system

'

(")%et us integrate the first equation. *e get

(0)6ifferentiate with respect toyand use the second equation of the

system to get

,

which implies ( y ) = 0 , that is, ( y ) = C is constant.Therefore, the functionF(x,y) is given y

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*e don2t have to keep the constant Cdue to the nature of the solutions

(see ne7t step).

()3ll the solutions are given y the implicit equation

Rema-&5ote that if you consider the function

,

then we get another integrating factor for the same equation. That is, the new

equation

is e7act. "o, from this e7ample, we see that we may not have uniqueness of the

integrating factor. 3lso, you may learn that if the integrating factor is given to

you, the only thing you have to do is multiply your equation and check that the

new one is e7act.

Solving an Exact Ordinary Differential Equationhttp'99mss.math.vanderilt.edu9cgi-in9?""3gent9Jpscrooke9?""9e7actodenew.def

$6='2*x*y

d7 ;x^2

dy

Solve

2

Your ODE is: [2 x y] dx + [x ] dy = 0.

2

The general solution is: x y = c.

'oa% soa%

1.( y 2x3) dx - x( 1 xy ) dy = 0

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ans. y/x - x2+ y2/2 = C

2.( x sec2y x2cos y ) dy = ( tan y 3x4 )

Ans . 1/x tan y - x3+ siny = C

3.y ( 1 + xy ) dx + x ( 1 xy ) dy = 0

ans xy ln y/x = c xy 1

4.( y3 3 xy2) dx + ( 2x2y xy2) dy = 0

ans y/x + 3 ln x 2 ln y = C

5.x dx + y dy + ( x dy y dx ) / ( x2+ y2) = 0

ans . x2/2 + y2/2 + arc tan y/x = C

Terima kasih

II-#8