engineering mathematics topic 2

8/13/2019 Engineering Mathematics Topic 2

1/8

2 - FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS

a) Separation of Variables a!"!a! Separable)

1st Order DE can be reduced to the form )()( x f y y g = by algebraic manipulations.

Variables x and y can be separated so that x appears only on RHS and y appears only

on the HS.

!o sol"e this e#n$ integrate both sides %.r.t. x &

+= C dx x f dx dx dy

y g )()(

E'ample 1&Sol"e * =+ x y y

Solution&

+y separating "ariables$ xdx ydy * =

+y integrating both sides$ therefore the #eneral sol$tion%

E'ample ,&

Sol"ey x

y = $ -)1( =y

Solution&

+y separating "ariables$ xdx ydy =

ntegrate both sides$ the general solution %ill be&

C x y += ,,

,

1

,

1

/pply initial condition$ C += ,, 1,1

-,1

$ =C

!herefore parti&$lar sol$tion $ 0,, = x y

, 2irst Order ODE 3

C x y += ,, ,,*


2/8


3/8

E'ample ,&

Sol"e ,, )( y xy dx dy

xy x =+

Solution&

!hat is&

xy x

y xy dx dy

+= ,

,

et vx y = $

differentiate implicitly&dx dv

x v dx dy

+= and substitute vx y = intodx dy

v v v

vx x

x v vx

xy x

y xy

+=

+=

+

1

,

,,

,,,

,

,

E#uate dx dy then&

v v v

dx dv

x v +

=+

1

,

v v

v v v v v

v v v v

dx dv

x +

=

+

=

+

=

1,

11

,,,,

Separate the "ariables

=+

dx x dv v v ,1

,

et =

+ dx

x dv

v v

,1,

C x v

v += ln,1ln

et AC ln= 7

v A x v

1lnln,ln +=+

y x A x

x y +=

lnln ,

y x

Ae xy =

&) E a&t E $ation

, 2irst Order ODE 8

y x

A x x y += lnln 2

xy


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/n e'act e#uation is al%ays in the form of )( x f y dx dy =+ %here is function of x.

!he coefficient of y is the deri"ati"e of the coefficient ofdx dy

.

4hen this is the case HS can be %ritten ( )y dx d .

E'ample 1&

Sol"e this e'act e#uation x ey x dx dy

x ,,- - =+ .

Solution&

!he e#uation can be %ritten as ( ) x ey x dx d ,- =

ntegrating both sides

+== C edx ey x x

x

,

,,-

--

,

, x C

x e

y x

+=

E'ample ,&

Sol"e 1)(sincos = y x dx

d x

Solution&

!he e#uation can be %ritten as 1))((cos =y x dx d

ntegrating both sides&

C x y x +=)(cos

So that x

C x

x y

coscos+=

,) Inte#ratin# Fa&tor

, 2irst Order ODE 0


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Steps of sol"ing 2irst order DE&

1) 4rite the e#uation in the standard form

( ) ( ) x Qy x P dx dy =+

,) 2ind the integrating factor$ ( ) x

( ) ( )[ ] = x P x e'p dx -) 9ultiply the e#uation in standard form by ( ) x and recalling that the left

hand side is :ust ( )[ ]y x dx d

$ obtain

( ) ( ) ( ) ( ) ( )

( )[ ] ( ) ( ) x Q x y x dx

d

x Q x y x x P dx dy

x

=

=+

) ntegrate the last e#uation and sol"e for y

E'ample 1&

Sol"e 1=+ x y

dx dy

Solution&

;omparing to standard 1st order linear e#uation&

x x P 1)( = and 1)( = x Q

!hen x ee x x dx

x == = ln1

)(

!herefore +== C x xdx xy ,,

2inally x C x

y +=,

E'ample&

, 2irst Order ODE *


6/8

Sol"e -),(),( = x y dx dy

x $ 1)( =y

Solution ,&

,),(

,

1 =

x y x dx

dy

,11),ln(),ln(

===

x eeIF x x

= dx x x x y ,1

),(,

1 ,

= dx x x y ),(,1

C x

x y +

=

,),(

,1 ,

!herefore ),(,

),( - += x C x y is the #eneral sol$tion

/pply initial condition$ 1)( =y

),(,0

1 C += $ -=C

!herefore the parti&$lar sol$tion ),(6),( - += x x y

, 2irst Order ODE 1


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e) .erno$lli/s E $ation

!he e#uation is in the form nQy Py dx dy

=+

%here P and Q are function of x.

!o sol"e this e#uation$ di"ide both sides by n

y gi"es&

QPy dx dy

y nn =+ 1

et ny z = 1 $ thendx dy

y ndx dz n= )1(

9ultiply e#uation (1) by (1 n) to con"ert the first term intodx dz

&

QnPy ndx dy

y n nn

)1()1()1( 1

=+


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x y x dx

dy y = 1, 1

so that x z

x dx

dz = 1 %hich is of the form QPz dx

dz =+ that can be sol"ed by

normal

integrating factor ( 2).

x z x dx

dz = 1

2 Pdx

e $ == x dx x Pdx ln1

x x eeIF x x

11)ln(ln 1 ====

= IFdx QIF z $ = dx x x x z 11

C x dx x z +== 1

, x Cx z =

but 1= y z $ ,1 x Cx y =

!herefore 1, )( = x Cx y .

, 2irst Order ODE 1,

engineering mathematics topic 2

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