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Engineering Analysis I

Basics Concepts and Ideas First Order Differential Equations

Dr. Omar R. Daoud

Many Physical laws and relations appear mathematically in the form

of Differentia Equations

They are one of the important fundamentals in engineering mathematics.

An ordinary differential equation is an equation that contains one or

several derivatives of unknown functions

A differential equation is a relationship between an independent variable x, a dependent variable y and one or more derivatives of y with respect to x.

If the unknown function depends only on one independent variable,

then its called by Ordinary D.E (O.D.E.), while its denoted by Partial

D.E. (P.D.E) if the function depends on two or more independent

variables.

Differential Equations

6/24/2018 Part I 2

2

22

2

34

3

0 is an equation of the 1st order

sin 0 is an equation of the 2nd order

0 is an equation of the 3rd orderx

dyx y

dx

d yxy y x

dx

d y dyy e

dx dx


6/24/2018 Part I 3

The order of a differential equation is given by the highest

derivative involved.

A function which satisfies the equation is called a solution to the

differential equation.

Solving a differential equation is the reverse process to the one

just considered. To solve a differential equation a function has to be

found for which the equation holds true.

The solution will contain a number of arbitrary constants – the

number equalling the order of the differential equation.

• Linear vs. nonlinear differential equations

– A linear differential equation contains only terms that are linear in both the dependent variable and its derivatives.

– A nonlinear differential equation contains nonlinear function of the dependent variable.

6/24/2018 Part I 4


xxyxdx

xdy4)(

)( 2

ydx

xdyx

dx

xydxy

dx

xdyx

dx

xydxyx

dx

xydsin

)()( ,0)(

)()( ,0)(

)( 2

2

22

2

222

2

2

Definition:

A differential equation is an equation containing an unknown function

and its derivatives.

32 xdx

dy

032

2

aydx

dy

dx

yd

36

4

3

3

y

dx

dy

dx

yd

Examples:.

y is dependent variable and x is independent variable,

and these are ordinary differential equations

1.

2.

3.

ordinary differential equations

Partial Differential Equation

Examples:

02

2

2

2

y

u

x

u

04

4

4

4

t

u

x

u

t

u

t

u

x

u

2

2

2

2

u is dependent variable and x and y are independent variables,

and is partial differential equation.

u is dependent variable and x and t are independent variables

1.

2.

3.

Order of Differential Equation

The order of the differential equation is order of the highest derivative in the differential equation.

Differential Equation ORDER

32 xdx

dy

0932

2

ydx

dy

dx

yd

36

4

3

3

y

dx

dy

dx

yd

1

2

3

Degree of Differential Equation

Differential Equation Degree

03

2

2

aydx

dy

dx

yd

36

4

3

3

y

dx

dy

dx

yd

03

53

2

2

dx

dy

dx

yd

1

1

3

The degree of a differential equation is power of the highest

order derivative term in the differential equation.

Linear Differential Equation

A differential equation is linear, if

1. dependent variable and its derivatives are of degree one,

2. coefficients of a term does not depend upon dependent

variable.

Example:

36

4

3

3

y

dx

dy

dx

yd

is non - linear because in 2nd term is not of degree one.

.0932

2

ydx

dy

dx

ydExample:

is linear.

1.

2.

Example: 3

2

22 x

dx

dyy

dx

ydx

is non - linear because in 2nd term coefficient depends on y.

3.

Example:

is non - linear because

ydx

dysin

!3

sin3y

yy is non – linear

4.

• Homogeneous vs. Inhomogeneous differential equations – A linear differential equation is homogeneous if every

term contains the dependent variable or its derivatives.

• A homogeneous differential equation can be written as where L is a linear differential operator.

• A homogeneous differential equation always has a trivial solution y(x) = 0.

6/24/2018 Part I 11


0)(4)()( 2

2

2

xxydx

xdyx

dx

xyd

,0)( xyL

xdx

dx

dx

d4 e.g., 2

2

2

L

6/24/2018 Part I 12

An inhomogeneous differential equation has at least one term that contains no

dependent variable.

•The general solution to a linear inhomogeneous differential equation can be written

as the sum of two parts:

Here yh(x) is the general solution of the corresponding homogeneous equation, and

yp(x) is any particular solution of the inhomogeneous equation.

xxyxdx

xdy4)(

)( 2

)()()( xyxyxy ph


Thus, the first order differential equitation contains y’ and may

contains y and given function of x;

A solution of a given F.O.D.E. on some open interval a<x<b is a

function y=h(x) that has a derivative which satisfies Equation (1) for all

x in that interval.


6/24/2018 Part I 13

),( form theof isequation aldifferenti aWhen yxfdx

dy

dxdyyf

)(

1We can then integrate both sides.

dxyf

dy

)(This will obtain the general solution.

),('or 0)',,( yxfyyyxF

Modelling is the steps that lead from the physical situation to a

mathematical formulation and solution and to the

Physical interpretation of the result; setting up a mathematical

model (Differential Equations) of the physical process.

Solving the D.Es.

Determination of a particular solution from an initial conditions (to transform the general solution to a particular one).

An initial value problem is a differential equation together

with an initial condition.

Checking.


6/24/2018 Part I 14

Formation of differential equations

Differential equations may be formed from a consideration of the

physical problems to which they refer. Mathematically, they can

occur when arbitrary constants are eliminated from a given function.

For example, let:

2

2

2

2

sin cos so that cos sin therefore

sin cos

That is 0

dyy A x B x A x B x

dx

d yA x B x y

dx

d yy

dx


6/24/2018 Part I 15

Here the given function had two arbitrary constants:

and the end result was a second order differential equation:

In general an nth order differential equation will result from

consideration of a function with n arbitrary constants.

sin cosy A x B x

2

20

d yy

dx


Formation of differential equations

6/24/2018 Part I 16

Various methods of solving F.O.D.E. will be

discussed. These include:

• Variables separable

• Homogeneous equations

• Exact equations

• Equations that can be made exact by multiplying by an

integrating factor

6/24/2018 Part I 17


6/24/2018 Part I 18

y‘=f(x,y)

Linear

Integrating Factor

Non-Linear

Separable Homogeneous

Change to Separable

Exact Integrating

Factor

Change to Exact

6/24/2018 Part I 19


Solution of linear differential equations

)( form theof isequation aldifferenti a When 1. yfdx

dy

dxdyyf

)(

1We can then integrate both sides.

dxyf

dy

)(This will obtain the general solution.

6/24/2018 Part I 20



When the equation is of the form ( ) ( ), thendy

f x g ydx

1( )

( )dy f x dx

g y

( )( )

dyf x dx

g y

2.

6/24/2018 Part I 21



3. A first order linear differential equation is an equation of the form

( ) ( )dy

P x y Q xdx

To find a method for solving this equation, lets consider the simpler equation

( ) 0dy

P x ydx

Which can be solved by separating the variables.

1

( ) 0dy

P x ydx

( )dy

P x ydx

( )dy

P x dxy

ln ( )y P x dx c ( )P x dx c

y e

( )P x dx cy e e

( )P x dx

y Ce

or ( )P x dx

ye C

Using the product rule to differentiate the LHS we get:

6/24/2018 Part I 22

( )P x dxdye

dx

( ) ( )

( )P x dx P x dxdy

e yP x edx

( )

( )P x dxdy

P x y edx

Returning to equation 1, ( ) ( )dy

P x y Q xdx

If we multiply both sides by ( )P x dx

e

( ) ( )

( ) ( )P x dx P x dxdy

P x y e Q x edx

( ) ( )

( )P x dx P x dxd

ye Q x edx

Now integrate both sides.

( ) ( )

( )P x dx P x dx

ye Q x e dx For this to work we need to be able to find

( )

( ) and ( )P x dx

P x dx Q x e dx

6/24/2018 Part I 23

2Solve the differential equation

dyy x

dx x

2Step 1: Comparing with equation 1, we have ( )P x

x

2( )P x dx dx

x 2ln x

2ln x 2ln x

2( ) lnStep 2: P x dx xe e 2x ( )

is called the integrating factorP x dx

e

Step 3: Multiply both sides by the integrating factor.

2 22dyx y x x

dx x

2 3dyx x

dx

6/24/2018 Part I 24

2 3yx x dx 4

2

4

xyx c

2 21

4y x cx

Solve the differential equation 2 3dy

xy xdx

Step 1: Comparing with equation 1, we have ( ) 2 and ( ) 3P x x Q x x

( ) 2P x dx xdx

2x

2( )

Step 2: Integrating Factor P x dx xe e


2 2

2 3x xdye xy e x

dx

2 2

3x xdye xe

dx

6/24/2018 Part I 25

2 2

3x xdye xe

dx

2 2

3x xye xe dx To solve this integration we need to use substitution.

2Let t x 2dt xdx

2 33

2

x txe dx e dt 3

2

te 23

2

xe

2 23

2

x xye e c

23

2

xy ce

6/24/2018 Part I 26

2 3Solve the differential equation 0dy

x x xydx

1Step 1: Comparing with equation 1, we have ( ) and ( )P x Q x x

x

( )dx

P x dxx

ln x

( ) lnStep 2: Integrating Factor

P x dx xe e x


2dyx x

dx

1rewriting in standard form:

dyy x

dx x

(note the shortcut I have taken here)

2yx x dx 2 11

3y x cx

The modulus vanishes as

we will have either both

positive on either side or

both negative. Their effect

is cancelled.

6/24/2018 Part I 27

2Solve the differential equation cos sin cosdy

x y x xdx

Step 1: Comparing with equation 1, we have ( ) tan and ( ) cosP x x Q x x

( ) tanP x dx xdx ln sec x

( ) ln secStep 2: Integrating Factor sec

P x dx xe e x


sec 1d

y xdx

rewriting in standard form: tan cosdy

y x xdx

(note the shortcut I have taken here)

sec 1y x dx cos cosy x x c x

The modulus vanishes as

we will have either both

positive on either side or

both negative. Their effect

is cancelled.

6/24/2018 Part I 28

2Solve the differential equation given 1 when 0dy

x y x x ydx

1Step 1: Comparing with equation 1, we have ( )P x

x

1( )P x dx dx

x ln x

( ) ln 1Step 2: Integrating Factor

P x dx xe e

x


1d y

dx x

1rewriting in standard form:

dyy x

dx x

6/24/2018 Part I 29

2y x cx given 1 when 0x y

0 1 c

1c

Hence the particular solution is 2y x x

1y

dxx

2y x cx

6/24/2018 Part I 30

6/24/2018 Part I 31


Solution of Separable differential equations

F.O.D.E. can be reduced to the form of

This equation is called separable because that the variables x and

y could be separated, so that x appears only on one side while y on

the other one.

)2.1()(

then,' since

)1.1()(')(

xxfyg(y)

x

yy

xfyyg

6/24/2018 Part I 32



Direct integration

If the differential equation to be solved can be arranged in the form:

the solution can be found by direct integration. That is:

( )y f x dx

( )dy

f xdx

6/24/2018 Part I 33



Direct integration

For example:

so that:

This is the general solution (or primitive) of the differential

equation. If a value of y is given for a specific value of x then a value

for C can be found. This would then be a particular solution of the

differential equation.

23 6 5dy

x xdx

2

3 2

(3 6 5)

3 5

y x x dx

x x x C

6/24/2018 Part I 34



Separating the variables

If a differential equation is of the form:

Then, after some manipulation, the solution can be found by direct

integration.

( )

( )

dy f x

dx F y

( ) ( ) so ( ) ( )F y dy f x dx F y dy f x dx

6/24/2018 Part I 35




For example:

so that:

That is:

Finally:

2

1

dy x

dx y

( 1) 2 so ( 1) 2y dy xdx y dy xdx

2 2

1 2y y C x C

2 2y y x C

6/24/2018 Part I 36




For example:

so that:

Let y=zx; that is:

02)3( 22 dxxydyyx

)3(

222 yx

xy

dx

dy

)3(

2

)3(

22222

2

z

z

xzx

xz

dx

dzxz

6/24/2018 Part I 37


For example:

323 ln 0

xx y dx dy

y

Separating the variables, we get

dxx

dyyy

3

ln

1

Integrating we get the solution as

kxy ln3|ln|ln3

lnx

cy or



6/24/2018 Part I 38


Solution of Inseparable differential equations

Certain Des are inseparable, however they could be transferred to

separable DE by the introduction of a new unknown function,

.

x

yg

For example:

Divide by 2xy, then

22'2 xyxyy

u

u

uuxu

uuuxuy

xy

uxy

x

xy

yy

2

11

2

1'

1

2

1''

that assume 22

'

2

22

x

u

6/24/2018 Part I 39



Cont. For example:

Cxyx

xy

uuse

x

Cu

Cxu

uu

u

x

x

22

2

2

2

Finally,

1

lexponentia the take,)ln(1ln

)(Integrate 1

2

22'2 xyxyy

6/24/2018 Part I 40



For example:

so that:

Let y=zx; that is:

02)3( 22 dxxydyyx

)3(

222 yx

xy

dx

dy

)3(

2

)3(

22222

2

z

z

xzx

xz

dx

dzxz

or

)3()3(

22

3

2 z

zzz

z

z

dx

dzx

6/24/2018 Part I 41



Cont. For example:

02)3( 22 dxxydyyx


x

dxdz

zz

z

3

2 )3( , Integrating we get

x

dxdz

zz

z3

2 )3(We express the LHS integral by partial fractions. We get

cx

dxdz

zzzln

1

1

1

13

cxzzz )1)(1(3

or

6/24/2018 Part I 42



Cont. For example:

02)3( 22 dxxydyyx

Noting z = y/x, the solution is:

3))(( ycxyxy or 322 ycyx

6/24/2018 Part I 43


Homogenous differential equations

The general form of the L.D.E. is

If then the L.D.E. is called Homogenous

otherwise it is not.

The solution of Homogenous D.E. could be attained by

using the separable method.

The Integrating Factor (IF) will be used to change the

Inhomogeneous D.E. to a homogenous D.E.

)()(' xryxpy

0)( xr

6/24/2018 Part I 44


Solution of Homogeneous differential equations

For example:

3 3 3

03'

xxy

yxxyyxy

x

y

xyy

Solution:

Integrate

xCey

Cx

y

xxy

y

5.1

2

lexponentia the take,2

3ln

3

6/24/2018 Part I 45


Solution of Homogeneous differential equations

For example:

)sin(x

xy

x

yy

Solution:

Let y=zx

zzdx

dzxz sin or z

dx

dzx sin


x

dxdz

z

sin

1

Integrating

Cxx

y-

x

y

Cxzz-

)cot( )cosec(

cot cosec

6/24/2018 Part I 46


Solution of Inhomogeneous differential equations

)()(')(or )()(' 01 xbyxayxaxryxpy

Two Different ways to solve it

a0(x)=a’1(x) a0(x)≠a’1(x)

6/24/2018 Part I 47



1)

Cxxbxa

xy

xxbxyxaxxbxxyxa

xxbxyxa

xyxaxyxaxyxa

xbxyxaxyxa

xaxa

)()(

1)(

)()()( )()()(

respect to with integrate ),()()(

Then

)()()()()(')(But

)()()()(')(

)()(

1

1

'

1

'

1

'

1'11

'11

'10

6/24/2018 Part I 48



2)

)()(')()(

asrewritten be could (1) then )()()'(Let

1)()()()()()(')(

)()()()(')(

)()(

get to by the D.E. usinhomogeno heMultiply t

)()(

'10

'10

xrxIFxyxIF

xpxIFxIF

xrxIFxyxpxIFxyxIF

xrxyxpxyxIF

xaxa

IF(x)

xaxa

6/24/2018 Part I 49



2)

)()(

)()()()('

(1)in substitute )(

)()(

)(

then )()()'( since

)()(

)('

)(

)()()(

)(

'10

xrexye

xrexyxpexye

exIF

xpxIF

xIF

xpxIFxIF

xaxa

xxpxxp

xxpxxpxxp

xxp

6/24/2018 Part I 50



2)

hhh

xxp

xxp

xxpxxp

Cexxreexy

hxxp

Cxxre

e

xy

xrexxye

x

xaxa

)()(

by )( Denote

)(1

)(

)()(

respect to with Integrate

)()(

)(

)(

)('

)(

'10

6/24/2018 Part I 51



For example:

xx

xxxx

x

x

Ceexy

Cexeeexy

xxhexrxp

eyy

2

2

2

2

)(

)(

)1( ,)( ,1)(

Solution:

6/24/2018 Part I 52

Differential Equations Solution of Inhomogeneous differential equations

For example:

Cbxbbxaba

kexbxke

Cbxbbxaba

kexbxke

f make use o

Cexxxeeexy

xxhxxexrxp

xxeyy

axax

axax

xxxx

x

x

)sin()cos()cos(

and ,)cos()sin()sin(

))2cos(2)2sin(3()(

2)2( )),2cos(2)2sin(3()( ,2)(

))2cos(2)2sin(3(2

22

22

222

Solution:

6/24/2018 Part I 53


For example:

xx

xxx

xxx

x

eCxexy

CeCxeexy

CeCxeexy

xxeyy

22

21

32

21

32

)2sin()(

)2sin()(

)2sin(13

4

13

9)(

))2cos(2)2sin(3(2

Solution:

6/24/2018 Part I 54


For example:

3

1)0(,0),cos()(cos2)(

)cos()sin(2)cos()(

)cos()sin(22sin

)cos()cos(

)2sin()cos()(

)2sin()(

)cos(ln)tan( ),2sin()( ),tan()(

1)0(for ),2sin()tan(

2

))ln(cos())ln(cos())ln(cos(

C

yat xxCxxy

xCxxxxy

xxx)(fmake use o

xCxx

xxxy

Cexxeexy

xxxhxxrxxp

yxxyy

xxx

Solution:

6/24/2018 Part I 55


Exact differential equations

A F.O.D.E. is called

an Exact if there exits a function f(x, y) such that 0),(),( dyyxNdxyxM

( , ) ( , )df M x y dx N x y dy

Here df is the ‘total differential’ of f(x, y) and equals

f fdx dy

x y

Hence the given DE becomes df = 0

Integrating, we get the solution as

f(x, y) = C

6/24/2018 Part I 56

6/24/2018 Part I 57

Exactness Test

( , )f

N x yy

6/24/2018 Part I 58

STEP:1

STEP:2

6/24/2018 Part I 59

STEP:3

Find k(y)

STEP:4

STEP:5

STEP:6

)(22

),(22

yhyeyx

xyyxf x

6/24/2018 Part I 60

6/24/2018 Part I 61

0)(

0)('

2)('2 22

yh

yh

eyxxyheyxxy

f xx

Constant22

)

Constant)

solution general for the

22

),(

22

22

x

x

yeyx

xf(x,y

f(x,y

But

yeyx

xyxf

The integral of zero is

ZERO, simple. Although

derivative of a constant

would be zero, but integral

of zero would always be

zero.

One thing to note: Integral is NOT antiderivative in strict sense. Its an area under

graph f(x) in Cartesian

system, where it is

ofcourse in a two

dimensional plane.

Example 6 Test whether the following DE is exact. If exact, solve it. 2

( ) 0x dy y dxy

Here 2, ( )M y N x

y

1M N

y x

Hence exact.

Now ( , ) ( )

x x

f x y M dx y dx xy g y

6/24/2018 Part I 62

Differentiating partially w.r.t. y, we get

2( )

fx g y N x

y y

Hence

2( )g y

y

Example 6 Test whether the following DE is exact. If exact, solve it. 2

( ) 0x dy y dxy

6/24/2018 Part I 63

Integrating, we get ( ) 2ln | |g y y

(Note that we have NOT put the arb constant )

Hence ( , ) ( ) 2ln | |f x y xy g y xy y

Thus the solution of the given D.E. is

( , )f x y c 2ln | | ,xy y c or

Example 7 Test whether the following DE is exact. If exact, solve it.

4 2 3(2 sin ) (4 cos ) 0x y y dx x y x y dy

Here 4 2 32 sin , 4 cosM xy y N x y x y

38 cosM N

xy yy x

Hence exact.

Now 4( , ) (2 sin )x x

f x y M dx xy y dx 2 4 sin ( )x y x y g y

6/24/2018 Part I 64

Differentiating partially w.r.t. y, we get

2 34 cos ( )f

x y x y g yy

Hence ( ) 0g y 2 34 cosN x y x y

Integrating, we get

( ) 0g y

Hence

2 4 2 4( , ) sin ( ) sinf x y x y x y g y x y x y

Thus the solution of the given d.e. is

( , )f x y c2 4 sin ,x y x y c or c an arb const.

6/24/2018 Part I 65

In the above problems, we found f(x, y) by integrating M partially w.r.t. x

and then equated

.f

to Ny

We can reverse the roles of x and y. That is we can find f(x, y) by

integrating N partially

.f

to Mx

w.r.t. y and then equate

6/24/2018 Part I 66

Example 8 Test whether the following DE is exact. If exact, solve it.

2 2(1 sin2 ) 2 cos 0y x dx y xdy

Here M

2 sin 2 ;M

y xy

Hence exact.

Now 2( , ) 2 cos

y y

f x y N dy y xdy 2 2cos ( )y x g x

N 21 sin 2 ,y x

22 cosy x

4 cos sin 2 sin 2N

y x x y xx

6/24/2018 Part I 67

Differentiating partially w.r.t. x, we get

22 cos sin ( )f

y x x g xx

gives ( ) 1g x

Integrating, we get

( )g x xHence

2 2 2 2( , ) cos ( ) cosf x y y x g x y x x

Thus the solution of the given D.E. is

( , )f x y c2 2cos ,x y x c or c an arb const.

21 sin 2M y x

2 sin 2 ( )y x g x

6/24/2018 Part I 68

6/24/2018 Part I 69

N , DE is not exact.

x

M

y

Making Exact

IF

Integrating Factor

To make DE Exact

6/24/2018 Part I 70

6/24/2018 Part I 71

Integrating Factor

To make DE Exact

6/24/2018 Part I 72

6/24/2018 Part I 73

6/24/2018 Part I 74

Example 3 Find an I.F. for the following DE and hence solve it.

2( 3 ) 2 0x y dx xydy Here

6 2M N

y yy x

2( 3 ); 2M x y N xy

Hence the given DE is not exact.

6/24/2018 Part I 75

Now

M N

y x

N

6 2

2

y y

xy

2( ),g x

x a function of x alone. Hence

( )g x dxe

2

2dx

xe x

is an integrating factor of the given DE

3 2 2 3( 3 ) 2 0x x y dx x ydy

which is of the form

0M dx N dy

Note that now

3 2 2 3( 3 ); 2M x x y N x y

Integrating, we easily see that the solution is 4

3 2 ,4

xx y c

6/24/2018 Part I 76

Multiplying by x2, the given DE becomes

Example 4 Find an I.F. for the following DE and hence solve it.

( cot 2 csc ) 0x xe dx e y y y dy

Here

0 cotxM Ne y

y x

; cot 2 cscx xM e N e y y y


6/24/2018 Part I 77

M N

y x

M

Now

0 cotx

x

e y

e

cot ( ),y h y a function of y alone. Hence

( )h y dye

cotsin

ydye y


Multiplying by sin y, the given DE becomes

6/24/2018 Part I 78

sin ( cos 2 ) 0x xe ydx e y y dy


0M dx N dy

Note that now

sin ; cos 2x xM e y N e y y

Integrating, we easily see that the solution is

c an arbitrary constant. 2sin ,xe y y c

Example 5 Find an I.F. for the following DE and hence solve it. 2 3( 2 ) 0ydx x x y dy

Here

31 1 4M N

xyy x

2 3; 2M y N x x y


6/24/2018 Part I 79

M N

y x

Ny Mx

Now

3

2 4

1 (1 4 )

( 2 )

xy

xy x y xy

2 2( ),g z

xy z

a function of z =x y alone. Hence

( )g z dze

2

2 2 2

1 1dzze

z x y


6/24/2018 Part I 80

2 2

1 1( 2 ) 0d x y d y

x y xy

Multiplying by 2 2

1,

x y the given DE becomes


0M dx N dy

Integrating, we easily see that the solution is

c an arbitrary constant. 21,y c

xy

6/24/2018 Part I 81


Bernoulli differential equations

A Bernoulli equation is a differential equation of the form:

This is solved by:

(a) Divide both sides by yn to give:

(b) Let z = y1−n so that:

ndyPy Qy

dx

1n ndyy Py Q

dx

(1 ) ndz dyn y

dx dx

6/24/2018 Part I 82


Bernoulli differential equations

Substitution yields:

then:

becomes:

Which can be solved using the integrating factor method.

1(1 ) (1 )n ndyn y Py n Q

dx

(1 ) ndz dyn y

dx dx

1 1

dzPz Q

dx

Example 1 Solve the following D.E.

Solution

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constants positive are and where,' 2 BAByAyy

(a) Divide both sides by yn to give:

(b) Let z = y1−n so that:

ByyAyy 22 '

BAzz

BAz

BAy

ByAyyyyz

x

yynz

x

z n

'

)(''

)1('

1

222

Example 1 Solve the following D.E.

Solution

6/24/2018 Part I 84

constants positive are and where,' 2 BAByAyy

Ax

AxAxAxAx

AxAxAx

CeA

Bxzxy

CeA

BCee

A

Be

CexBeexz

AxxAhBxrAxp

1)()(

)(

,)( ,)(

1

engineering analysis i - philadelphia university i.pdf2 2 2 2 3 4 3 0 is an equation of the 1st...

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