LDEofOrder1

7/28/2019 LDEofOrder1

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Linear Differential Equationsof Order One

Chapter 2

Reference: DE by R. Marquez

1

First Order First Degree ODEs

A special class of first-order ordinary Des which is

generally a nonexact DE with special integrating

factor is a linear differential equation.

A linear DE of order one takes the form .

)()( xQxyPdx

dy dxxQdxxyPdy )()(

Derivative form

Differential form

dxxpexv )()(Whose integrating factor, v(x)

2

The General Solution

The general solution of

Is found by multiplying the integrating factor,

v(x) to each term and then solving for theexact DE formed.

3

)()( xQxyPdx

dy

dxxPdxxP edxxQdxxyPdye )()( )()(

dxxQeyeddxxPdxxP

)()()(

dxxQedxxPyedyedxxPdxxPdxxP

)()()()()(


Simplify the equation using

Integrating both sides will result to

Thus, the general solution is

4

dxxQxvxyvd )()()(

dxxpexv )()(

dxxQxvxyvd )()()(

CdxxQxvxyv )()()(

CdxxQxvxvy )()()( 1


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It is also possible that the given DE is not linear

in ybut instead linear in other variable.

Linear DE iny:

Linear DE in x:

. Linear in w:

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CdxxQxvxvy )()()( 1

CdyyQyvyvx )()()( 1

CdttQtvtvw )()()( 1

)()( xQxyPdx

dy

)()( yQyxPdy

dx

)()( tQtwPdt

dw

Example: Find the general solution

6

xxxydx

dycossin4cos2)3 2

xxyy cos2cot)1 0)tan(sec)2 3 dyyxydx

2)1(1

)4 xxy

y

0sectan)5 yywdy

dw

xyedy

dx y sin)6

Find the particular solution

when x =1, y = 2

ify(10) =0

when x =1, y = 2

7

xyyx 212 dyyexdx y )sin4( 2

xyx

x

dy

dx

463 3

2

The Bernoullis Equation

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Definition

A differential equation that can be

expressed in the form

is called a Bernoullis equation (BE) in y.

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)()( xQyxyPdxdy n

Definition

The BE

may be made linear upon multiplication of

and substitution of

10

)()( xQyxyPdx

dy n

nyn )1(

dx

dyyn

dx

dzyz nn )1(;1

The General Solution of BE

Thus, the BE becomes

The general solution of this linear DE in

z may be accomplished by the method

of linear DE of order one.

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)()1()()1()1( 1 xQnxPnydx

dyyn nn

)()1()()1( xQnxPnzdx

dz

Example: Solve the following DE

12

23)1 xexyxyy Solution

dxdyy

dxdzyz 32 2; Let

Thus, the given DE becomes

2)13()13( xxezx

dx

dz 2

2)2( xxexzdx

dz which is linear in z


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Example: Solve the following DE

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Continuation of Solution of Ex. 1

So that the integrating factor is2)2( x

xdxee

Thus, Cdxexeze xxx ))(2(

222

)2(2 Cdxxze x

Cxez x 22 2 yz22

2Cxey x

,

2)(1 22 xexCy

Exercise: Solve the given DEs

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xyxydx

dy 33 sectan)1 xxyyyx ln)2 3/13/4

)124(77)3 35 xxxyyyx)1()4 531 yxxy

dy

dx

9

)52

2

xx ee

yy

dx

dy

7.

Exercise: Solve the given DEs

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44534)6 yxxy

dy

dx 0)tancsc(2)7 2 dvvvwdww

37)8 yxyyx 4cot)9 xyx

dy

dx

7.

Differential Equations

Solvable by Simple

Substitution

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Sometimes the DE

may not be reducible at once to any of

the forms discussed so far. This meansthat the previous methods even how

effective they were would not work. But

if a wise change of variable would be

done, the equation could be

transformed into an equation that could

be well handled by one of the previous

methods.

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0),(),( dyyxNdxyxMDEs Solvable by Simple Substitution

Solve the following

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DEs Solvable by Simple Substitution

021 yxxedx

dy

w = x ydx

dw

dx

dy

dx

dy

dx

dw 11Substituting these to the given DE, we get

0211 wxedx

dw

0202 ww xedx

dwxe

dx

dw

Continuation of solution

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DEs Solvable by Simple Substitution

But w = x y

Variable Separable DE

02 wxedx

dw

02 xdxdwe w02

xdxdwe w

Cxe w 2xyyx eCxCxe 22)(

Solve the following

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7.

0)733()4( dyyxdxyx0)74( 2 yx

dx

dy

0)tan3(sec)3(tan 222 dyyxxdxyx 0tan25cos3 2 dyyxdxxyx

021 yxxedx

dy

xxyxdy

dxy ln)1(ln


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Solve the following

21

7.

021 1

xdx

dyey

0tan])(tan1)[(tan 222 ydydxeyyyyx x

DES WITH COEFFICIENTS

LINEAR IN TWO VARIABLES

22

DE w/ Coefficient Linear in Two Variables Consider the DE whose form is

(1)

It is noted that the coefficients ofdxand dyare both linear in the variables xandy. If

these coefficients are each equated to zero,

we obtain two equations of the lines of the

forms

and

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0)()( 222111 dycybxadxcybxa

0111 cybxa0222 cybxa

DE w/ Coefficient Linear in Two Variables For these associated equations of the lines,

we consider the following three cases:

Case 1: If

then the graph of the associated equations

of the lines are coincident and the DE is

reducible to

kdx + dy = 0 , k is a constant (2)

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0111 cybxa 0222 cybxa

2

1

2

1

2

1

c

c

b

b

a

a


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DE w/ Coefficient Linear in Two Variables Case 2: If

then the graph of the associated equationsof the lines are parallel and the DE is

reducible to

which can be handled by simple substitution.

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2

1

2

1

2

1

c

c

b

b

a

a

0)()( 222122 dycybxadxcybxak

DE w/ Coefficient Linear in Two Variables Case 3: If

then the associated lines are intersecting

and the DE is reducible to homogeneous

DE using the substitution

x = u + h ; dx = du

y = w + k ; dy = dw

where (h, k) is the point of intersection of

the linear system

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2

1

2

1

b

b

a

a

0

0

222

111

ckbha

ckbha

DE w/ Coefficient Linear in Two Variables In case the DE falls under case 3, try also to

check if it is an exact DE. If so, it is

suggested to use the method for exact DE

since it is much easier to perform.

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Example1) Find the general solution of(x+ 2y + 6)dx

(2x+y )dy = 0

Solution

For the associated system of linear equations, wehave

The solution point of this is (h, k) = (2, 4)

If we let x= u + 2 ; dx = du and

y = w 4 ; dy = dw

then, upon substitution we obtain

(u +2w)du (2u +w)dy =0.28

02

062

kh

kh


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ExampleContinuation:

This is a homogeneous DE with

u = zw ; du = zdw + wdz

Then, by substitution, we get

(zw +2w) (zdw + wdz) (2zw + w)dw =0

(z +2) (zdw + wdz) (2z +1)dw =0

(z2 +2z 2z 1)dw + w(z +2)dz =0

(z2 1)dw + w(z +2)dz =0

29

01

22

dz

z

z

w

dw

Example

30

11

23

1

21 Cdz

zzw

dw

12321 |1|ln|1|ln||ln Czzw 1

32

21

)1(ln C

z

zw

1232

1

)1( Cez

zw

C

y

yx

yx

1

1)4(

42

3

422

)2()6( 3 yxCyx

Exercise

Solve the following differential equations.

1. (3x + y 8)dx + (x2y +2)dy =0

2. (16x +5y 6)dx + (3x + y 1)dy =0

3. (xy

2)dx + (3x + y

10)dy =0

4. (2x +3y +3)dx + (3x4y +13)dy =0

5. (3x + y 9)dx + (x4)dy =0

6. (x + y +1)dx + (xy 5)dy =0

7. (6x + y 9)dx + (x2y +5)dy =0

8. (y 1)dx2(x + y +1)dy =0

9. (3x + y 10)dx + (x +3y +2)dy =0

10.(7x + y 8)dx + (x2y +5)dy =031

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LDEofOrder1