engineering analysis and numerical...

بسم هللا الرحمن الرحيم

Engineering Analysis and Numerical Methods

Stage: Third

Civil Engineering Department

Tikrit University -Civil Engineering Department-Third Stage Eng. Anal.& Num. Meth. Dr. Adnan Jayed Zedan

Text Book Advanced Engineering

Mathematics By

C.R. Wylie


References

- Advanced Engineering Mathematics by Kreyszig - Differential Equation by Iyengar - Advanced Mathematics by Agarwal , et .al - Integral Calculus and Differential Equations by Chatterjee


Syllabus

Ordinary Differential Equations of First order (16 hrs)

Linear Differential Equations with Constant Coefficient (12 hrs)

Simultaneous Linear Differential Equations (12 hrs)

Numerical Solutions of Ordinary Differential Equations (8 hrs)

Finite Differences (4 hrs)

Interpolation (4 hrs)

Numerical Differentiation (8 hrs)

Numerical Integration and Computer Application (4 hrs)

Fourier Series (16 hrs)

Partial Differential Equation and Boundary Value problems (12 hrs)

Numerical Solution for Partial Differential Equations (8 hrs)

Matrices and its Applications (16 hrs)

Tikrit University -Civil Engineering

Department-Third Stage Eng. Anal.& Num. Meth. Dr. Adnan Jayed Zedan

Ordinary Differential Equations of First Order **Definitions: Differential equation (DE) : An equation involves one or more derivatives or differentials. **Type: Ordinary or Partial: Ordinary derivatives occur when the dependent variable "y" is a function of one independent variable "x"; Partial derivatives occur when the dependent variable "y" is a function of two or more independent variables ; i.e. **Order : (highest derivative) **Degree: (power of highest derivative)


Example (1) Ordinary, Order 2, Degree 1 Ordinary, Order 3, Degree 2


Example (2) Partial, Order 4, Degree 1 Partial, Order 2, Degree 1



Linear and non-Linear differential equations

If a differential equation is of first degree in the

dependent variable y and its derivatives (consequently ,

there cannot be any term involving the product of y and

its derivatives) then it is called a linear differential

equation otherwise it is non -linear .


OR A linear differential equation (of y = f(x) ) is of the form : (1)

Or

A non linear D.E. ,cannot be put in form (1) .


Examples: linear non linear because ( non linear because (


A solution of D.E.; is a relation between the dependent and

independent variables, and it satisfies the equation identically:

is a general solution of:

The general solution of D.E. of nth order, is one contains n essential

constants (parameters). By essential we mean that the n constants

cannot be replaced by a smaller number.


For example, contains 3 constants and can be reduced Where d = a + b and e = b - c


However, there are equations such that; (which has only the single solution y=0) (which has no solutions at all)

Also, there are differential equations which posses solutions,

containing more essential parameters than the order of the

equation.


C1=1

C1=0.5

C1=0.25

C1=-1

C1=-.5

C1=-.25

C1=1

C1=0.5

C1=0.25

C1=-1

C1=-.5

C1=-.25


Both can be pieced together to give a D.E.

x

yxcy

x

yc

xofvaluesallforx

yy

22,

2

121

x

yxcy

x

yc

22, 222

22cxyfor

x

yyAlso


Any solution found from the general solution by assigning particular

values to the constants is called a particular solution.

general solution

particular solution

Solution which cannot be obtained from any general solution by

assigning specific values to the constants are called singular solution.

xbxay sincos

01 banda

xy cos


If a general solution has the property that every solution of the

differential equation can be obtained from it by assigning

suitable values to its arbitrary constants, it is said to be a

complete solution.

Example: Show is a solution of

for all values of a and b.

Solution:

xx beaey 2

02 yyy

)(2

)2()4(2

2

22

xx

xxxx

beae

beaebeaeyyy


beeeaeee xxxxxx )224()2( 222

000 ba

o.k.

Note: Satisfies

xx beyoraey 2

21

02 yyy

02

111 yyy

02

222 yyyTikrit University -Civil Engineering


But; xx beaeyyy 2

21

Is not solution of

02 yyy

02 yyy

02 yyy

Since

Is not linear, while;

Is linear


Example: Given (1)

Find second-order differential equation.

Solution:

If the given function has n constants, differentiate nth

times and then eliminate the constants.

(2)

(3)

By adding and subtract equations (1) and (3)

xbaey x cos

xbaey x sin

xbaey x cos


xaeyy 2xe

yya

2

xbyy cos2x

yyb

cos2

Substitute a and b into Eq. (2)

xx

yye

e

yyy x

xsin

cos22

xyxyyyy tantan2

0)tan1(2)tan1( yxyyx


0cos yxbaex

0sin yxbaex

0cos yxbaex

(1) (2) (3)

0

cos1

sin1

cos1

yx

yx

yx

0)sincos(

)coscos()cossin(

xyxy

xyxyxyxy


0)sin(coscos2)sin(cos xxyxyxxy

Divide by cos x getting:

0tan2tan xyyyxyy

0)tan1(2)tan1( yxyyx


This is only second-order D.E., but there are other D.E.

Differentiate Eq. (3) twice more,

and since it is a 4th order D.E. we expect the

solution to contain 4 constants and we can

show that:

Satisfied for all values of a, b, c and d.

xbaey x cos

xbaey xiv cos

yyiv

xdcexbaey xx sincos

yyiv Tikrit University -Civil Engineering


Separable First-Order D.E.

Often a first order D.E. can be reduce to

(1)

And such an equation is said to be separable;

The general solution is:

(2)

Other forms:

(3)

(4)

dyygdxxf )()(

Cdyygdxxf )()(

dyygxFdxyGxf )()()()(

)()( yNxMdx

dy


Solution of Eq. (3) is: And the solution of Eq. (4) is:

CdyyG

ygdx

xF

xf )(

)(

)(

)(

CdxxMyN

dy )(

)(


Example: Solve Solution:

1 yyx

xy

y 1

1

x

dx

y

dy

1

)ln()ln(

)ln()1ln(

ax

cxy

xay 1Tikrit University -Civil Engineering


Particular solution curve means one-member

For all values of a straight lines pass through (0,-1)

b.) Find the orthogonal trajectories curves.

Slope of straight line

Or

Hence, the slope of the orthogonal trajectories

aydx

dy

x

yy

1

1

y

xy


1

y

xy

xdxdyy )1(

cx

yy

22

22

cyyx 2222

12)12( 22 cyyx

222 )1( Ryx


x

y

x

yR

P(x,y)

Circlex + y = R2 2 2


x

y

x

y

-

x + y = R2 2 2 -

hk

- -

(x-k) +(y-h) = R2 2 2


x

y

y+1=ax


Example: Solve

)3(2)2( ydyxdxdydxxy

Solution:

0)2()62( dyyxydxxxy

0)2()3(2 dyxydxyx

032

2

dy

y

ydx

x

x


0)3

31()

2

42(

dy

ydx

x

cyyxx )3ln(3)2ln(42

yxcyxc eeeyx 2234 )3()2(

yxkeyx 234 )3()2(

yxcyx 2)3()2(ln 34


Example: Solve

Solution:

0)1()4( 22 dyxdxy

041 22

y

dy

x

dx

1

11

2tan

2

1tan c

yx

1

11 22

tantan2 cy

x


In simpler form:

1

11 2tan)2

tantan2tan( cy

x

21

1

tan2tan2

1

2tan2tan

c

xy

yx

C

x

xy

y

x

x

2

2

1

2

21

21

2

Cxyx

xyx

2)1(2

)1(42

2


Example: Solve

Solution: Best first step:

ydydxyxydydx 2

dyxydxy )1()1( 2

dyy

y

x

dx211

cyx )1ln(2

1)1ln( 2

Cyx )1ln()1ln( 22

Cy

x

)1(

)1(ln

2

2


2

2

2

)1(

)1(ke

y

x C

2keC

is necessarily +ve

)1()1( 222 ykx

)1()1( 2

2

2

yk

x

2k

)1()1( 2

2

yx

0


The general solution defines the family of conics:

1)1( 2

2

yx

(a) general solution

If =1 , 1)1( 22 yx , the solution is a circle

If >0 , the solutions are ellipse

If <0 , the solutions are hyperbolas


Particular solution curve which passes through point )5

13,

5

7(

1)5

13(

)15

7(

2

2

=-1

Hence, particular solution is:

(b) 22 )1(1 xy


x

y


The upper branch of any curve of Eq. (a) for x>0, can be associated

with the upper branch of curve Eq. (b) for x≤1. In the D.E. we divided

by (1-x) and (1-y2), hence x = 1, y =±1 were implicitly ruled out.

Had we desired the particular solution through (1,yo), (xo,1) or (xo,-1),

we could not have found it from the general solution, even if it is

existed.

So we return to the original D.E. and search for the solution by some

methods other than separating the variables.


x=1 can be calculated from

by putting =0, So y = 1, y = -1 are singular solutions. General, Particular,

D.E.,

)1()1( 22 yx

1)1( 2

2

yx

0

,

22 )1(1 xy

1x)1()1( 2y

ydy

x

dx

1y

,


x

y

y=1

y=-1

x=1

(x ,1)o

(x ,-1)o

(1, y )o


Homogeneous First-Order D.E.

If all terms in M(x,y) and N(x,y) in:

are all of the same degree in x and y then either of the substitution of will reduce the D.E. to a separable equation. But, generally if the substitution of

dyyxNdxyxM ),(),(

yyandxx

vyxoruxy


will convert and

, then M(x,y) and N(x,y)

are called homogeneous function of n

degree.

The D.E. is said

homogeneous when M(x,y) and N(x,y) are

homogeneous functions of the same degree.

),(int),( yxMoyxM n

),(int),( yxNoyxN n

dyyxNdxyxM ),(),(


Example: Solution: Substitute

?hom)ln(ln),( 22 ogeneousyeyyxxyxFIs y

x

yyandxx

y

x

yeyyxxyxF

)ln(ln),( 2222

y

x

yeyyxx )]ln(lnln[(ln 22


])ln(ln[ 22 y

x

yeyyxx

),( yxF

Hence, homogeneous, degree (1) OR Subst. either or

vxy )(),( vFxyxF

vyx )(),( vFyyxF


Example: Solve

Solution:

Is not separable

Is homogeneous of degree (2)

M and N both are of degree (2)

Substitute

02)3( 22 xydydxyx

vxy xdvvdxdy

0)(2)3( 222 xdvvdxvxxdxxvx

0)(2)31( 222 xdvvdxvxdxvxTikrit University -Civil Engineering


0223 22 vxdvdxvdxvdx

02)1( 2 vxdvdxv

01

212

dvv

vdx

x

cvx )1ln(ln 2

cv

x

1ln

2Tikrit University -Civil Engineering


cev

x

12

k

x

y

x

1

2

k

x

xy

x

2

22

kxy

x

22

3

)( 223 xykx Tikrit University -Civil Engineering Department-Third Stage Eng. Anal.& Num. Meth. Dr. Adnan Jayed Zedan

Example: Solve

Solution: Not separable, but homogeneous. Substitute

Easier rather than

0)( 222 dxtxtxdtx

vxt xdvvdxdt

vtx

0)()( 22222 dxxvvxxxdvvdxx

02 dxvvdxdxxdvvdxTikrit University -Civil Engineering


0)1( 2 dxvxdv

01 2

x

dx

v

dv

cxv lntan 1

vcx 1tanln

vcex1tan

vceex1tan x

t

ekx1tan



Example: Solve

Solution: Homogeneous. Substitute

x

y

xeydx

dyx

vxy xdvvdxdy

dxxevxxdvvdxx x

vx

)()(

0 dxevdxxdvvdx v

0 dxexdv v


0ve

dv

x

dx

0 dvex

dx v

cex v ln

cex x

y

lnTikrit University -Civil Engineering Department-Third Stage Eng. Anal.& Num. Meth. Dr. Adnan Jayed Zedan

Example: Solve

Solution: Homogeneous Substitute

0)3()( dyyxdxyx

vxy xdvvdxdy

0))(3()( xdvvdxvxxdxvxx

0))(3()1( xdvvdxvdxv

0)33()1( 2 vxdvdxvxdvvdxdxv


0)3()31( 2 dvvxxdxvvv

0)3()12( 2 dvvxdxvv

012

32

dv

vv

v

x

dx

0)1(

2

1 2

dv

vv

dv

x

dx

cv

vx

)1(

2)1ln(ln


cv

vx

)1(

2)1(ln

c

x

yx

yx

)1(

2)1(ln

cxy

xxy

2)ln(

xy

xcxy

2)ln(

xy

xc

exy

2

yx

x

kexy

2


Exact D.E. of First-Order

If f(x,y) is differentiable, then there is a total

differential :

Conversely if M(x,y) dx+N(x,y) dy=0 which can be

written in the form;

dyy

fdx

x

fdf

0

dfdy

y

fdx

x

f


With and

Then is a solution.

This D.E. is said to be exact.

But there is a test, in general, for a first

order D.E. when it is exact, although

sometimes it is not difficult to tell by

inspection if the D.E. is exact.

x

fyxM

),(

y

fyxN

),(

kyxf ),(


Theorem

If and are continuous, then

is exact if and only if

y

M

x

N

0),(),( dyyxNdxyxM

x

N

y

M


Proof:

Assume equation is exact, so there is a

function f, such that and

and

order is immaterial for

continuous equations.

x

fM

y

fN

yx

f

y

M

2

xy

f

x

N

2

xy

f

yx

f

22


Then show that there is a function f;

(but and are interchangeable, since

is continuous)

x

fM

x

a

ycdxyxMyxf )(),(),(

x

a

ycdxyxMyy

f)(),(

x

M


x

a

ycdxy

M

y

f)(

x

a

ycdxx

N

y

f)(

x

N

y

MSince

,

)(),(),( ycyaNyxNy

f

),( yxNy

f

)(),( ycyaNIf

y

b

dyyaNycHence ),()(, Tikrit University -Civil Engineering Department-Third Stage Eng. Anal.& Num. Meth. Dr. Adnan Jayed Zedan

functionaIs

dyyaNdxyxMyxf

So

x

a

y

b

),(),(),(

,

dydxy

MNdxyxMyxf

dxy

MNyc

Note

x

a

x

a

][),(),(

)(

:


x

a

y

b

cdyyaNdxyxMThen

exactisdyyxNdxyxMIf

),(),(,

,0),(),(

Corollary:


0)143()232(;

:

dyyxdxyxSolve

Example

exact

x

Nand

y

M

Solution

33

);1(

:


x

a

y

b

cdyyadxyx )143()232(

cyyayxyxxy

b

x

a )23()23( 22

cbbabyyay

ayaaxyxx

)23()23(

)23()23(

22

22

bbabaacyyxyxx 2222 232223

kyyxyxx 22 223

kyxxyyxisSolution 232 22


cyyxyxx

cdyyxyxx

cdyxnowithNxyxx

22

2

2

223

)14(23

)(23

);2(


cyyxyxx

cdyyxyxx

cdydxyxxyxx

22

2

2

223

)14(23

]3)143[(23

);3(


0;

:

22

222

dy

yxy

yxdxyxxSolve

Example

exact

yx

xy

x

Nand

yx

xy

y

M

Solution

22

22

:


)()(3

1

)(3

2)(

2

1

)()(

)(),(),(

2

3

22

2

3

22

2

1

22

ycyx

ycyx

ycdxyxx

ycdxyxMyxf


dxyxxyyxy

yx

dxyx

xy

yxy

yx

dxyx

xy

yxy

yx

dxy

MNyc

2

1

22

22

2

2222

2

2222

2

)(22

1

2

2

1

)(


][

])}({

[

][

1

2)(

2

1)(

22

22222

22

22222

22

22

2

22

22

2

2

1

22

22

2

yxy

yxyxyxy

yxy

yxyxyxy

yxyxy

xy

yxyyxy

yx

yxyyxy

yxyc


,

,

kyyx

ky

yx

ky

yc

yyc

yxy

yyxyyc

32

3

22

3

2

3

22

3

2

22

22

2

)(

3)(

3

1

3)(

)(

][)(


D.E. which is not exact can be made exact by

multiplying by an integrating factor. For example;

is exact, and if it simplified by dividing by (xy2), the

equation :

is not exact and can be restored to its original form

by multiplying it by factor (xy2). Sometimes the

integrating factor can be found by inspection.

032 223 dyyxdxxy

032 xdyydx


.,

1,0)(

:

22

22

solveandfactorais

yxshowydydxxyx

Example

exact

yx

xy

x

Nand

yx

xy

y

M

dyyx

ydx

yx

x

Solution

222

222

2222

)(

2

)(

2

0)1(

:


cyxx

cyxx

yx

ydyxdxdx

dyyx

ydx

yx

xdx

Simpler

22

22

22

2222

ln

)(ln2

1

022

2

1

0)(

;


.,

,0)(: 32

solveandfactorthefind

dyxyxydxExample

0)()(

)(

1

0

2

32

2

2

32

dyxy

yx

xy

xdyydx

xyisfactorThe

dyyxxdyydx

Solution


cy

xy

ydyxy

xyd

orydyxy

xdyydx

2

1

0)(

)(

,0)(

2

2

2


.,int

,)4(: 22

solveandfactoregrationthefind

dyyxydxxdyExample

dy

x

yx

ydxxdy

dyx

y

x

ydxxdy

xisfactorThe

Solution

)4(

)4(

1

2

22

2

2

2

2


2,

x

ydxxdydu

x

yu

cyx

y

cyu

dyu

du

dyu

du

2tan

2

1

2tan

2

1

])2

(1[4

)4(

1

1

2

2


multiplieraasxyTry

xydydxxdycontainsEqIf

multiplieraasy

orx

Try

x

ydydxxdycontainsEqIf

multiplieraasyxTry

yxdydyxdxcontainsEqIf

FactornIntegratiotheonNotes

)(.)3(

11

)(.)2(

)(

)(2

1.)1(

:

22

22

22


22

22

11

22

1

22

1

22

22

22)][ln()7(

)()6(

)(sin)5(

)(tan)4(

22)()3(

)(,)()2(

,)()1(

:

yx

ydyxdxyxd

dyxnydxymxyxd

yxx

ydxxdy

x

yd

yx

ydxxdy

x

yd

ydyxdxyxd

y

xdyydx

y

xd

x

ydxxdy

x

yd

dyy

fdx

x

ffydxxdyxyd

Notes

mnnmnm


Linear first-order equations

)2........()(

)(])([

)1(..............................)()(

;)(

)()()(

ydx

xd

dx

dyxyx

dx

d

xQyxPdx

dy

tscoefficientherenameandxFbyDivide

xHyxGdx

dyxF


.

;)()()(

)3.....()()()()()(

);()1(.

EqseparablesimpleaisThis

xPxdx

xdIf

xQxyxPxdx

dyx

xbyEqMultiply


factoraise

edxxPx

dxxPx

dxxPx

xd

dxxP

dxxP

)(

)(])(exp[)(

)()(ln

)()(

)(


)4......()(

;,)(

;

)(]*[

)(])([

;)1(.

)()()(

)()(

)()(

)()(

)(

dxxPdxxPdxxP

dxxPdxxP

dxxPdxxP

dxxPdxxP

dxxP

ecdxexQey

finallyandcdxexQye

equationabovetheIntegrate

exQyedx

d

exQyxPdx

dye

efactorthebyEqMultiply


Steps to solve the linear first-order

equations:

1-Compute the integration factor

2-Multiply the given equation by this factor.

3-Integrate both sides of the resulting

equation.

4-Solve the integrated equation for y.

dxxP

e)(


01

2))(1(

;:

2

xwhenywhich

xydxdxdyx

SolveExample

dxx

xydxdy

xbyequationtheDivide

Solution

2

2

1

2

1

:


21

1ln

)1ln(

1

2

2

2

1

1

)(

11

2

1

21

22

2

xee

ex

yx

x

dx

dy

x

xy

dx

dy

xx

dxx

x


cxyx

cdxx

yx

egratingBy

xy

x

x

dx

dy

x

1

2

22

2222

tan1

1

1

1

1

1

;int

1

1

)1(

2

1

1


)1(tan)1(

1

)01(01

01

)1(tan)1(

212

212

xxxy

c

c

xy

xcxxy


0)22(

;:

dyyxxyydx

SolveExample

22

2)2(

22

022

:

xy

y

dy

dx

yxydy

dxy

yxxydy

dxy

ydyxdyxydyydx

Solution


y

yyy

yy

yyydy

y

y

ceyyxy

ceyeey

dyeyxey

yeeey

442

)22(2

2

)(

22

2

22

2ln2

2


The Bernoulli's Equation:

D.E. of the form

is said to be a Bernoulli's equation. (P

and Q are functions of x (or

constants) and do not contain y)

nyQPydx

dy


dx

dy

yn

dx

dvyv

QPydx

dy

y

ybysidesbothdividingBy

yQPydx

dy

n

n

n

n

n

n

1)1(

1

1

1


10.'

1

0

)1()1(

1

1

andnEqsBernoulli

orderfirstseparablenIf

orderfirstlinearnIf

nQPvndx

dv

QPvdx

dv

n


3;: yx

y

dx

dySolveExample

dx

dv

dx

dy

y

dx

dv

dx

dy

yv

yPut

xydx

dy

y

EqBerisyx

y

dx

dy

Solution

2

11

21

111

..

:

3

32

23

3


2

ln

ln22)(

1

)(

22

12

1

2

xe

eeex

vinlinearvxdx

dv

x

v

dx

dv

x

xx

dxdxxP


cxxy

cxx

v

cdxxx

v

cdxex

vdxxP

21

21

21

21

22

2

22

)(

2


,

,

1)2(

)2(1

2

2

2

22

cxyx

cxxy

cxxy


3

1;: ye

x

y

dx

dySolveExample x

dx

dv

dx

dy

y

dx

dv

dx

dy

yv

yPut

eyxdx

dy

y

EqBerisyex

y

dx

dy

Solution

x

x

2

11

21

1

1

11

..1

:

3

32

23

3


2)1ln(

)1ln(212)(

)1(

)(

21

2

12

1

2

xe

eeex

vinlinearevxdx

dv

ex

v

dx

dv

x

xx

dxdxxP

x

x


222

22

22

22

2

22

)1()1(

4

1

42

1

)1()1(

4

1

42

4)1(4)1(2)1(

]2)1(2)1[(2)1(

)1(2

)1(2)1(

x

c

x

e

x

ee

y

x

c

x

e

x

eev

ceexexxv

ceexexxv

cdxex

cdxxexv

xxx

xxx

xxx

xxx

x

x


xeyxydx

dySolveExample

x

sin;: 22

2

dx

dv

dx

dy

y

dx

dv

dx

dy

yv

yPut

xey

xdx

dy

y

EqBerisxeyxydx

dy

Solution

x

x

2

2

22

22

1

11

sin11

..sin

:

2

2


cxdxve

cdxxeeve

eex

vwithEDlinearxexvdx

dv

xexvdx

dv

x

xxx

xxdx

x

x

sin

)sin(

)(

..sin

sin

2

222

2

2

2

2

222

2

2

2


)(cos1

cos1

cos

2

2

2

2

2

2

cxey

cxy

e

cxve

x

x

x


min86.13

100200150

;150

100200

100

200100

100,0

20

20

t

e

lbQWhen

eQ

c

c

lbQtWhen

t

t

o


original

orthogonal

orthogonaloriginal

dx

dydx

dy

dx

dy

dx

dySlope

esTrajectoriOrthogonal

)(

1)(

1)()(

.

4

;:

2

estrajectoriorthogonalFind

xcy

GivenExample


y

xy

orthogonalofSlope

x

yy

x

y

x

ycyy

x

yccxy

Solution

2

:

2

4442

44

:

22

22


cxy

cxy

xdxydy

y

x

dx

dy

y

xy

22

22

2

2

2

2

2


Example: A hemispherical tank of radius; R,

is initially filled with water. At the bottom

of the tank there is a hole or radius; r,

through which water drains under the

influence of gravity. Find the depth of the

water at any time; t, and determine how

long it will be taken the tank to drain

completely.



)(tantan

2

,int

;

:

2

2

headheighteousinsh

gravityonaccelaratig

lawTorricelliorificefromgh

AdtdV

rAanddtwaterof

streamdtervalwithwaterofVolume

dyxdV

waterofvolumetheinDecrease

Solution


y

x

R

y

xdy

r


dtgrdyyRy

dtgyrdyyyR

dtgyrdyx

EqsEquating

dtgyrdV

dyxdV

2)2(

2)2(

2

)2(&)1(.

)2.......(..........2

)1.(....................

22

3

2

1

22

22

2

2


2

5

2

5

2

5

22

5

2

3

15

14

05

2

3

4

,0

25

2

3

4

Rc

cRR

RytWhen

ctgryRy


gr

Rt

Rtgr

tywaterNo

RtgryRy

215

14

15

1420

??,0;

15

142

5

2

3

4

2

2

5

2

5

2

2

5

22

5

2

3


Example: The rate at which a solid substance dissolves

varies directly as the amount of undissolved solid

presented in the solvent and as the difference between

the instantaneous concentration and the saturation

concentration of the substance. Twenty pounds of solute

is dumped into a tank containing 120 lb of solvent, and

at the end of 12 min. the concentration is observed to be

1 part in 30. Find the amount of solute in solution at any

time; t, if the saturation concentration is 1 part of solute

to 3 parts of solvent.



)(

120

)20(

,;

:

ionconcentratsaturationmaterialdundissolvedt

dQ

ionconcentratingcorrespondtheisQ

timethatat

materialdundissolveofamounttheisQ

ttimeat

solutioninmaterialtheofamounttheisQIf

Solution

120 lb

20 lb

Q


,

,

cdtk

dQQQ

dtk

QQ

dQ

QQk

dt

dQ

QQk

dt

dQ

120]

)40(20

1

)20(20

1[

120)40()20(

)40()20(120

)1203

1()20(


ctk

Q

Q

ctk

QQ

cdtk

dQQQ

cdtk

dQQQ

620

40ln

6)40ln()20ln(

620

1]

)40(

1

)20(

1[

20

1

120]

)40(20

1

)20(20

1[


tk

Q

Q

ortk

Q

Q

cck

QtWhen

ctk

Q

Q

6)20(2

40ln

,2ln620

40ln

2ln)0(6020

040ln

0,0

620

40ln


t

t

t

e

eQ

eQ

Q

tQ

Q

kk

QQ

concetAfter

tk

Q

Q

0098.0

0098.0

0098.0

2

)1(40

240

40

0098.0)20(2

40ln

05889.0)12(6)420(2

440ln

412030

1..min12

6)20(2

40ln


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