de chap 2 lde of order one
DESCRIPTION
Lessons of Linear Differential EquationTRANSCRIPT
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LINEAR DIFFERENTIAL EQUATIONS
OF ORDER ONE
Chapter 3
Prepared by Maria Cristina R. Tabuloc
1
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First Order First Degree ODE’s
◦An ordinary differential equation of first
order and first degree may be expressed
in any of the following forms:
),( yxfdx
dy
0),(),( dyyxNdxyxM
Derivative form:
Differential form:
),(
),(),(
yxN
yxMyxf where
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Examples of first order first degree ODE
Derivative Form Differential Form
xexyy 3233 022 dyeydxx x
824
623
yxyx
yxyx
dx
dy
0432 22 dvwvdwvw
)( mTTkdt
dT
0)cos()sin( dttuedutte uu
dyyxy )tan(
θdrrdθr tan)1( 2
dx
dTkAq
kdtQ
dQ
3
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Two principal goals of studying DE
The study of differential equations has two principal goals:
To discover the differential equation
model that describes a specified physical situation
To find the appropriate solution of that differential equation, that is, to obtain a function defined either explicitly or
implicitly free from derivatives and that satisfies the differential equation.
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Types of First Order First Degree ODE’s
◦Variables Separable DE
◦DE with Homogeneous Coefficients
◦Exact DE
◦Non-exact DE
◦DE with Integrable Combinations
◦Linear DE of Order One
◦Bernoulli’s DE
◦DE Solvable by Simple Substitution
◦DE with Coefficients Linear in Two Variables5
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VARIABLE SEPARABLE DIFFERENTIAL EQUATIONS
Chapter 2
Section 1
6
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Variables Separable DE’s
◦The simplest type of first order first degree
ordinary differential equations is the
variables separable differential equation.
This type of a differential equation can be
solved by direct integration of terms of the
same variable type.
◦Definition 2.1: A variables separable
DE is one with all nonzero terms
expressible with one type of variable
only.7
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Variables Separable DE’s◦Consider a differential equation of the form
◦ If it is possible to collect all x terms with dx and all
y terms with dy, then by algebraic manipulation,
the above equation may be written as
so that, by dividing both sides by , it becomes
8
0),(),( dyyxNdxyxM
0)()()()( 1221 dyyBxAdxyBxA
)()( 22 yBxA
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Variables Separable DE’s◦ If we put and , then we get
◦ If we let and
it becomes
From which the general solution can be obtained as follows.
9
0)(
)(
)(
)(
2
1
2
1 dyyB
yBdx
xA
xA
)(
)()(
2
1
xA
xAxA
)(
)()(
2
1
yB
yByB
0)()( dyyBdxxA
CdyyBdxxA )()( +
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Examples
◦Find the general solution of
10
0cotsin)1 2 ydyydxx
01
sin)2()2 22
dyx
ydxxyy
0)323 dyedxxy x
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Exercises
◦Find the general solution of the ff DEs
11
yxy 42)1
21)2 ydx
dy
02
)4(
1)3
2
2
xx
dyy
x
ydx
dyxyxdx 32 4ln)4
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Exercises◦Find the general solution of ff DEs
12
0seccossin)5 22 xdyydxx
041)6 42 drtdtrt
yy
xy
dx
dyx
)4ln(
ln)4()7
2
2
0)8 2 dyeyxdx yx
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Examples: Initial-Value Problems
◦Determine the particular solution
13
1 = /4)(rdθ
drθrr ,tan1)1 2
,03sec2csc)2 dyxdxy 3/ /2)(y
,824
623)3
yxyx
yxyx
dx
dy23 )y(
,sin)cos1()4 θdθxdxθ
,0)52
x
dydxe xy
4ln2ln )( y
;
02 whenx
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Assignment
◦Solve Exercises from the text book for
your review
◦See attached Exercises
14
Exercise 2.1
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HOMOGENEOUS LINEAR
DIFFERENTIAL EQUATIONS
15
Chapter 2Section 2
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Homogeneity of a Function
◦Definition
A function f(x, y) is said to be
homogeneous if and only if
, k > 0
for some real number n where n is the
degree of homogeneity of f.
For a polynomial function, the
degree of homogeneity is determined by
the polynomial degree of each term.
),(),( yxfkkykxf n
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Homogeneity of a Function
If all the terms of a polynomial function have
the same degree, it is homogeneous. For
instance, the polynomial
is homogeneous of degree 4.
2243 35 yxxyx
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Example◦Decide if the function defined in each of the
following is homogeneous or not. Give its degree
If it is a homogeneous function.
434 )())((7)(4),( kykykxkxkykxf 443344 ))((74 ykykkxxk
4344 74 yxyxk
),(),( 4 yxfkkykxf
434 74),( yxyxyxf 1)
Therefore 434 74),( yxyxyxf
is homogeneous of degree 4 18
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Example
y
xy
x
yxyxf lntan),( 2/112/12)
ky
kxky
kx
kykxkykxf ln)(tan)(),( 2/112/1
y
xyk
x
yxk lntan 2/12/112/12/1
),(),( 2/1 yxfkkykxf
Therefore,
is a homogeneous function of degree 1/2.
y
x y xy
/xk ln2/11tan212/1
y
xy
x
yxyxf lntan),( 2/112/1
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Example
)( lnln),()4 22 xyxyyxf
yxyxf 74),( 3)
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Exercise
◦Test the given function for homogeneity.
Determine its degree if the function is
homogeneous
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Homogeneous DE
◦Definition:
If the functions M and N are both
homogeneous functions of the same
degree in x and y, then
is said to be a homogeneous DE (DE
with homogeneous coefficients)
0dyyx,Ndxyx,M )()(
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Solution of Homogeneous DE
An ordinary differential equation
with homogeneous coefficients can
be reduced to a separable DE by
using the appropriate set of
substitutions.
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Solution of Homogeneous DE
Rules:
stands out, i) if N is simpler or xy /
stands out, ii) if M is simpler or
let y = vx ; dy = vdx + xdv
let x = uy ; dx = udy + ydu
24
yx /
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Example 1
3
44 2
xy
yxy'
Solve
02 344 dyxydxyx
Solution: Write the equation in Mdx + Ndy = 0 form
Since N is simpler,
let y = vx, and dy = vdx + xdv
0)(2 33444 xdvvdxxvxdxxvx
02 3544444 dvvxdxxvdxxvdxx
035444 dvvxdxxvdxx
0)1( 3544 dvvxdxvx
Variable separable
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Example 1
54
3544
)1(
0)1(
xv
dvvxdxvx
0)1( 4
3
v
dvv
x
dx
cvx ln)1ln(4
1ln 4
cv
xln
)1(ln
4/14
cv
x
4/14 )1(
4/14
1
xy
cx
4
14 1
xy
cx
4
44
14
x
yxcx
44
1
8
yxc
x
482
4 xxcy
4/1482 xxcy
Since v = y/x
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Example 2◦Solve the initial-value problem
27
0222 xydydxyx
Let y = vx ; dy = vdx + xdv so that the DE becomes
0)(2])( 22 dyvxxdxvxx[
0)(2) 2222 xdvvdxvxdxxvx(
.
Factoring out x2 from the first term of the latter equation gives
0)(2) 222 xdvvdxvxdxvx (1
0)(2)2 xdvvdxvdxv(1
02)2 22 vxdvdxvv(1
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By separation of variables, we get
28
031
22
v
vdv
x
dx
1231
2C
v
vdv
x
dx
1231
6
3
1||ln C
v
vdvx
12 )31ln(
3
1||ln Cvx
12 3)31ln(||ln3 Cvx
123 3)31(ln Cvx
Example 2
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Example 2
◦ Solve the initial-value problem
29
)(cossin1 xdyydxx
ydx
x
yx
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EXACT DIFFERENTIAL EQUATIONS
Chapter 2Section 3
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Definition: Exact Differential Equation
◦ DE M(x, y) dx + N(x, y) dy = 0 is said to be exact if and only if there exists a function f such that
throughout some region.
Mx
f
N
y
f
and
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Exact Differential Equation
Test for Exactness of a DE
A necessary and sufficient condition that
M(x, y) dx + N(x, y) dy = 0
be an exact DE is
x
N
y
M
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Example 1: Test the DE for exactness
0)sin2()1()1 2 dyyxydxy
1),( 2 yyxM
yy
M2
yxyyxN sin2),(
yx
N2
x
N
y
M
0)sin2()1( 2 dyyxydxy
Since
is an exact DE.
.
Therefore,
Solution: First, identify M(x, y) and N(x, y)
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Example 2: Test the DE for exactness
0)sincos()sin(sin)2 dyxyxyxdxxyyy
xyyyyxM sinsin),( xyxyxyxN sincos),(
xyxyxyyy
Msincoscos
xyxyxyy
x
Nsincoscos
.
x
N
y
M
Since
is an exact DE.
Therefore, 0)sincos()sin(sin dyxyxyxdxxyyy
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Example 3: Test the DE for exactness
35
dwwppwdpwp
w 211
1sectantan2tan 2
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Solve the following
36
1. Find the value of k so that the DE
dyxkyxydxxyyx )cos6()sin2( 223
becomes exact.
2. Find the most general function M(x, y)so that
0)2(),( 3 dyyxexdxyxM y
is exact.
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Exercise
◦Test for exactness
37
drrpeprpdprpepr rr )62sin()cos3( 3222124
0)ln()14(1
sin)(tan 2221
2
1222
dyyxyxe
y
xdxyyxy
x x
o Find the value of the constant k so that each
of the following DE becomes exact.
0)3()15( 224323 dtrtrdrtrtkr
03)1(ln2)4ln( ][2322 dyyexxxydxxykxxy y
0sin2cossin2 ][)(22 22 dykxyexxdxeyyxxy xyxy
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Method of Solving Exact DE
◦ Test the DE M(x, y) dx + N(x, y) dy = 0 for
exactness
◦ Let
◦ Integrate M or N
Ny
fM
x
f
and
constantiswhere
constantholding
)()()(),(
),(),(
yTyTxQyxF
ydxyxMx
fyxf
constantiswhere
constantholding
)()()(),(
),(),(
xQyTxQyxF
xdyyxNy
fyxf
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Alternative Method of Solving Exact DE:The DIRECT METHOD
◦ Test the DE M(x, y) dx + N(x, y) dy = 0 for
exactness
◦Find f(x, y) which is the solution of the DE
alone in of terms has yyxNNwhere
CdyyNdxyxMyxf
),(
)(),(),(
1
1
alone in of terms has xyxMMwhere
CdyyxNdxxMyxf
),(
),()(),(
1
1
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Example
◦Solve the DE using the direct formula
40
0)2sin2)sin23( 22 dyxyxyxdxxyy
0)2cossin2()4sin(22 22 dyexyyxxdxxyeyy yxyx
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NONEXACTDIFFERENTIAL EQUATION
Chapter 3
Section 4
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Nonexact DE
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INTEGRATING FACTORSFOUND BY
INSPECTION
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Integrating Factors Found by Inspection
◦In the previous section, the
discussion is about the
determination of the Integrating Factor I(x) or I(y). This section is
concern with DE that are simple
enough to enable us to find the
integrating factor by inspection.
The following are some exact
differentials that occurs frequently. 44
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Integrating Factors Found by Inspection(Exact Differential Equations)
ydxxdyxyd
2y
xdxydy
y
xd
2x
ydxxdy
x
yd
45
xy
xdyydxxyd
ln
xy
xdyydxd
yx
ln
xy
ydxxdyd
x
y
ln
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Integrating Factors Found by Inspection
◦More Exact differentials
22
22
yx
ydyxdxyxd
n
n
xy
ydxxdynxyd
1
46
2
1
y
xdyydx
y
xn
y
xd
nn
22
22 22ln
yx
ydyxdxyxd
22tan
yx
xdyydx
y
xd
Arc
22tan
yx
ydxxdy
x
yd
Arc
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Example
◦ Form an integrable combination for each differential
47
)()1 2yxd
3
2
)2y
xd
)][ln()3 2yxd
)sin()4 yxd
4.
)arctan()5 yxd
)()6 yxed
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Example
◦Evaluate each of the following integrals, omitting the constant of integration.
48
3.
)()()1 3 ydxxdyxy
)3()2 23 ydxxdyx
22
)4yx
ydyxdx
ydxxdy
y
x4
2
)3
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Example
49
0)()()1 32 dyxydxyxy
0)1(3)3 3342 dyyxdxyx
0)()2 3 dyxyydx
0)()2()4 22 dyxyxdxxyy
0)1()1()5 232 dyyxdxyxyy
Solve each of the following differential equations
applying the integrating factors by inspection.
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Example
50
0)1()()6 22232 dyxxdxyxxy
0)1()13()7 2222 dyyxxdxyxy
0)1()8 22 xdydxyxy
0)()()9 33 dyxyydxxyx
0)2()22()10 3223 dyyyxxdxxyxyx
Solve each of the following differential equations
applying the integrating factors by inspection.
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Solution for #1
51
0)()()1 32 dyxydxyxy
032 xdydyyydxdxxy
032 dyydxxyxdyydx
02
ydyxdxy
xdyydx
02
22
yxd
y
xd
Cyx
y
x
2
22
1322 Cyyxx
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Solution for #2
52
0)()2 3 dyxyydx
03 xdydyyydx
03 dyyydxxdy
0)( 441 ydxyd
Cydxyd
441)(
Cyxy 441
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53
0)1(3)3 3342 dyyxdxyx
Solution for #3
03
03
332
33
33
3
42
dyydyxydxx
y
dydy
y
yxdx
y
yx
03 3342 dydyyxdxyx
033 dyyyxd
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54
Solution for # 4
0)()2()4 22 dyxyxdxxyy
02 223 dyxdyxyxydxdxy
0)2()(
0)2()(
02
2
2
2
2
22
223
y
dyxxydx
y
xdyydxy
dyxxydxxdyydxy
dyxxydxdyxydxy
0)2()( 221 dyyxdxxyxdyydx
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Alternative Solution for # 4
0)()2()4 22 dyxyxdxxyy
02 223 dyxdyxyxydxdxy
02 223 dyxxydxdyxydxy
Multiply the equation by xm yn
0)2()( 211213 dyyxdxyxdyyxdxyx nmnmnmnm
)( 31 nm yxd 0)( 12 nm yxd dyyxndxyxm nmnm 223 )3()1( 0)1()2( 211 dyyxndxyxm nmnm
21
3
1
1
nm
nm
42
2221
1
2
2
nm
nmnm
2;0 nm
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56
0)1()1()5 232 dyyxdxyxyy
Solutions
02233 xdydyxydxyxdxyydx
0)(
22
23
22
2
22
yx
dxyx
yx
xdyydxy
yx
xdyydx
0)( 232 dxyxxdyydxyxdyydx
0)(2
2
xdx
x
xdyydxxdyydxxy
0)( 2211
xd
x
ydxyd
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FIRST ORDER FIRST DEGREE ODE
Chapter 3
Section 157
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First Order First Degree ODE’s◦A special class of first-order ordinary DE’s which
is generally a nonexact DE with special
integrating factor is a linear differential
equation.
◦A linear DE of order one takes the form .
58
)()( xQxyPdx
dy
dxxQdxxyPdy )()(
Derivative form
Differential form
dxxp
exv)(
)(Whose integrating factor is
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The General Solution ◦The general solution of
Is found by multiplying the integrating factor, v(x) to each term and then solving
for the exact DE formed.
59
)()( xQxyPdx
dy
dxxPdxxP
edxxQdxxyPdye)()(
)()(
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The General Solution
Continuation
Thus, by the Derivative of a Product:
60
dxxQeyeddxxPdxxP
)()()(
dxxQedxxPyedyedxxPdxxPdxxP
)()()()()(
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The General Solution ◦Simplify the equation using
◦ Integrating both sides will result to
61
dxxQxvxyvd )()()(
dxxp
exv)(
)(
dxxQxvxyvd )()()(
CdxxQxvxyv )()()(
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The General Solution
Thus, the general solution is
62
CdxxQxv
xvy )()(
)(
1
CdxxQxvxvy )()()( 1
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The General Solution ◦ It is also possible that the given DE is not
linear in y but instead linear in other
variable.
◦ Linear DE in y:
◦ Linear DE in x:
63
CdxxQxvxvy )()()( 1
CdyyQyvyvx )()()( 1
)()( xQxyPdx
dy
)()( yQyxPdy
dx
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The General Solution ◦ It is also possible that the given DE is not
linear in y but instead linear in other
variable.
◦Linear in w:
64
CdttQtvtvw )()()( 1
)()( tQtwPdt
dw
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Example: Find the general solution
65
xxydx
dycos4cos2)3
xxyy cos2cot)1
0)tan(sec)2 3 dyyxydx
2)1(1
)4
xx
yy
0sectan)5 yywdy
dw
xyedy
dx y sin)6
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Example I: Find the general solution
66
xxyy cos2cot)1
xdxxdxydy cos2cot
xxQxxP cos2)(;cot)(
xe
eexv
x
xdxdxxP
sin
)(
sinln
cot)(
dxxxxy cos2sinsin
Cxxy 2sinsin
Solving for the general solution:
xCxy cscsin
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Example I: Find the general solution
67
0)tan(sec)2 3 dyyxydx
dyydyyxdx 3sectan
yyQyyP 3sec)(;tan)(
w/c is linear in x
yy
eeyv
y
ydyy
seccos
)(
cos11
coslntan
dyyyyx 3secsecsec
Solving for the gen solution:
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Exercise I: Find the particular solutionShow how P(x) & Q(x) were derived.
when x = 1, y = 2
if y(10) = 0
when x = 1, y = 2
68
xyyx 21)1 2
dyyexdx y )sin4()2 2
xyx
x
dy
dx
463)3
3
2
Cxyx 22
1)(;
2)(
xxQ
xxP Gen solution:
yeyQyP y 2sin4)(;1)( Cyyxe y
2sin2Gen sol’n:
2
363)(;
4)(
x
xxQ
xxP
Cxxyx 33 35
Gen sol’n:
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Exercise I: Find the particular solution
if x(0) = 2
if T(0) = T0
where k, Tm & T0 are constants
69
xyedy
dx y sin)4
yeyQyP y sin)(;1)(
yyCeyyex
)cos(sin5
Gen solution:
)()5 mTTkdt
dT
mkTtQktP )(;)( ktm CeTT Gen solution:
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Exercise: Find the particular solution
when x = 1, y = 2
if y(10) = 0
when x = 1, y = 2
70
xyyx 21)1 2
dyyexdx y )sin4()2 2
xyx
x
dy
dx
463)3
3
2
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THE BERNOULLI’S EQUATION
Chapter 3Section 3
71
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Definition
◦A differential equation that can
be expressed in the form
is called a Bernoulli’s equation
(BE) in y.
72
)()( xQyxyPdx
dy n
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Definition
The BE
may be made linear upon
multiplication of
and substitution of
73
)()( xQyxyPdx
dy n
nyn )1(
dx
dyyn
dx
dzyz nn )1(1 ;
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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.74
)()1()()1()1( 1 xQnxPnydx
dyyn nn
)()1()()1( xQnxPnzdx
dz
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Example: Solve the following BE
75
23)1 xexyxyy
Solution: n = 3
dx
dyy
dx
dzyz 32 2 ; Let
Thus, the given DE becomes
2)13()13( xxezx
dx
dz
22)2( xxexz
dx
dz which is linear in z
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Example: Solve the following BE
76
Continuation of Solution of Ex. 1
So that the integrating factor is2)2(
xxdx
ee
Thus, Cdxexeze xxx ))(2(
222
Cdxxze x )2(
2
Cxez x 22 2 yz
Cxey x 22 2
,
2)(1 22 xexCy
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Exercise: Solve the given BEs
77
xyxydx
dy 33 sectan)1
xxyyyx ln)2 3/13/4
)124(77)3 35 xxxyyyx
)1()4 531 yxxydy
dx
7.
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Exercise: Solve the given BEs
78
9)5
2
2
xx ee
yy
dx
dy7.
44534)6 yxxy
dy
dx
0)tancsc(2)7 2 dvvvwdww
37)8 yxyyx
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Exercise: Solve the given BEs
79
4cot)9 xyxdy
dx7.
2142 )1(;32)10 yyxy
dx
dyx
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DIFFERENTIAL EQUATIONS
SOLVABLE BY SIMPLE SUBSTITUTION
Chapter 3 Section4
80
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◦Sometimes the DE
may not be reducible at once to any of the forms discussed previously. This means that the previous methods even how effective they were would not work.
81
0),(),( dyyxNdxyxM
DE’s Solvable by Simple Substitution
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◦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.
82
DE’s Solvable by Simple Substitution
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◦Solve the following
83
DE’s Solvable by Simple Substitution
021 yxxe
dx
dy
Let w = x ydx
dw
dx
dy
dx
dy
dx
dw 11
Substituting these to the given DE, we get
0211 wxedx
dw
0202 ww xedx
dwxe
dx
dw
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◦Continuation of solution
84
DE’s Solvable by Simple Substitution
But w = x y
Variable Separable DE
02 wxe
dx
dw
02 xdxdwe
w
02
xdxdwe
w
Cxew 2
xyyxeCxCxe
22)(
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Exercise: Solve the following
85
7.
0)733()4()1 dyyxdxyx
0)74()22 yx
dx
dy
0)tan3(sec)3(tan)3222 dyyxxdxyx
0tan25cos3)42 dyyxdxxyx
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Exercise Solve the following
86
7.
021)71
xdx
dye
y
0tan
])(tan1)[(tan)8
2
22
ydy
dxeyyyyxx
xxyxdy
dxy ln)1(ln)6
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DE’S WITH COEFFICIENTS LINEAR
IN TWO VARIABLES
i87
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DE w/ Coefficient Linear in Two Variables ◦Consider the DE whose form is
(1)
◦It is noted that the coefficients of dx
and dy are both linear in the variables
x and y. If these coefficients are each
equated to zero, we obtain two
equations of the lines of the forms
and 88
0)()( 222111 dycybxadxcybxa
0111 cybxa 0222 cybxa
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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 89
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
equations of the lines are parallel and the
DE is reducible to
which can be handled by simple
substitution. 90
2
1
2
1
2
1
c
c
b
b
a
a
0)()( 222122 dycybxadxcybxak
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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
91
2
1
2
1
b
b
a
a
0
0
222
111
ckbha
ckbha
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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.
92
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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, we
have
The solution point of this is (h, k) = (2, 4)
If we let x = u + h = u + 2 ; dx = du and
y = w + k = w 4 ; dy = dw
then, upon substitution we obtain (u + 2w)du (2u +w)dy = 0. 93
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
94
01
2
2
dz
z
z
w
dw
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Example
95
01
23
1
21 dz
zzw
dw
Czzw ln|1|ln|1|ln||ln23
21
Cz
zwln
1
)1(ln
32
Cz
zw
1
)1(32
11)4(423
422
yx
yx Cy
)2()6(3 yxCyx
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ExerciseSolve the following differential
equations.
1. (3x + y 8)dx + (x 2y + 2)dy = 0
2. (16x + 5y 6)dx + (3x + y 1)dy= 0
3. (x y 2)dx + (3x + y 10)dy = 0
4. (2x + 3y + 3)dx + (3x 4y + 13)dy= 0
5. (3x + y 9)dx + (x 4)dy = 0
96
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ExerciseSolve the following differential
equations.
6. (x + y + 1)dx + (x y 5)dy = 0
7. (6x + y 9)dx + (x 2y + 5)dy = 0
8. (y 1)dx 2(x + y + 1)dy = 0
9. (3x + y 10)dx + (x + 3y + 2)dy = 0
10.(7x + y 8)dx + (x 2y + 5)dy = 0
97
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References
◦ Differential Equations by D. Zill
◦ Differential Equations by Bedient
◦ Differential Equations by R. Marquez
98