Engineering Mathematics Bhaskar Rao

MATHEMATICS -I

SECOND EDITION

P.B. Bhaskar Rao

M.Sc., Ph.D.

Retd. Professor, Former Chairman, Board of Studies, Department of Mathematics

Osmania University Hyderabad

S.K.V.S.Sriramachary

M. Bhujanga Rao

M.Sc., Ph.D. M.A., M.Phil., B.Ed.

Professor & Head (Retd.) Department of Mathematics University College of Engineering (Autonomous)

Osmania University Hyderabad

Professor, Dept. of Mathematics University College of Engineering (Autonomous) Director of Centre for Distance Education Osmania University Hyderabad

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CHAPTER -1

Ordinary Differential Equations of

First Order and First Degree ... 1

CHAPTER -2

Linear Differential Equations with

Constant Coefficients and Laplace Transforms ... 69

CHAPTER-3

Mean Value Theorems and

Functions of Several Variables ... 111

CHAPTER-4

Curvature and Curve Tracing ... 213

CHAPTER-5

Application of Integration to

Areas, Lengths, Volumes and Surface areas ... 313

CHAPTER-6

Sequences of Series ... 385 _

CHAPTER-7

Vector Differentiation ... 475

CHAPTER-8

Laplace Transforms ... 623

1

Ordinary Differential Equations of

First Order and First Degree

1.1

Introduction

Differential euqtions play an important role in many applications in the field of science and engineering, such as (i) problems relating to motion of particles (ii) problems involving bending of beams (iii) stability of electric system, etc. For example, Newton's law of cooling states that the rate of change of temperature of a body varies as the excess temperature of the body to that of its surroundings. If

8(t) is the temperature of the body at time 't' and 8₀ is the temperature of the room de

in which the body is kept, then dt gives the rate of change of temperature with time.

de

dt

=

K(8 - 8₀) ; K is constant

Similarly Newton's second law of motion for a particle of mass m moving in a straight line can be written as

d

2

x

m dt2

=

F

Where m is the mass, x is the distance of the particle at time 't' measured from a fixed origin and F the external impressed force.

A differential equation is an equation involving an unknown function and its derivatives. Ifthere is only one independent variable and one dependent variable the equation is called (Ill ordinary differential equation.

If there are more than one independent variable the equation is called a partial differential equation as this involves partial derivatives.

For example: 4d3y +3xdy _ y=e( dx3 dx .... (a)

(

- -

d

3 J4 ( 2 J8

y + - -

d Y (dY)

+ -

12

+ Y -x 6 _ 8 dx;3 dx2 _dx .... (b) .... (c) .... (d) .... (e) .... (t) .... (g)

The first four equations (a), (b), (c) and (d) are ordinary differential equations and the remaining three are partial differential equations.

Order

0/

a differential equation: The order of a differential equation is the order of the highest ordered derivative appearing in the equation.

Degree

0/

a differential equation: The degree of a differential equation is the power to which the highest ordered derivative appears in the equation after clearing the radicals if any.

In the above examples:

Example: 1.1(a) is a differential equation of order 3 and degree I.

Example: 1.1(b) is of third order and fourth degree differential equation.

Example: 1.1(c) is a second order, first degree differential equation.

Example: 1.1(d) is a second order, second degree ditferential equation.

1.2

Example

Formation of an ordinary D.E :

The differential equations ar~ formed by eliminating all the arbitrary constants

th~t are involved in the functional relationship between the dependent and independent variables.

For example:

y = cx2 + c2 where c is an arbitrary constant. .... (I) To eliminate 'c': (only one constant)

dv _ 0 = c.2x+ 0

dx

From(l) I (~V c= -2x dx Substitution of c in (I) gives

y = _I dy x

2

+

_1_(d

y

₎₂

2x d\: 4x2 _d); ~ + 2x3

2 -

4x2_{y = 0} ( d

)2

d dx dx

is the required D.E and y = cx2

+

c2 is called the solution of the D.E.

Note:

Depending on the number of constants in the given equation differentiate it as many number oftimes successively. Then the elimination of the arbitrary constants from the resulting equations and the given equation gives the required differential equation whose order is equal to the number of constants.

1.3

Example

Eliminate the arbitrary constants a, b from xy

+

x2

=

aeX

+

be-X _{and form the} differential equation.

Solution:

The given equation is

xy

+

.x2

= ae-'"

+

be-x _{... ( I)} The number of arbitrary constants is two. Differentiating (I) w.r., to 'x' two times successively. dy _ x dx

+

Y

+

2x = ae-' - be-x ... (2) d2y ely dy X ---2 + - + - + 2.1 = aex _{+ be-}x d-c dx dx

From (I), (2) and (3) el imination of a, b gives the D.E. from (I) and (3) we get

dl)' 2dy . .

x - - ? + - - + 2

=

xy + x2 IS the requIred D_E. dx- dx

.... (3)

1.4

Example

Form the differential equation by eliminating the constants a and b from

a.x2

+

by

= 1 Solution: Differentiating ax2 +

by

= I w.r.t 'x' dy 2ax+ 2by- = 0 d-c Again differentiating wr.t., 'x' d2y dy dy 2a+2by - ? +2b-.- =0 dx- dx dx

Elimination of a, b from (I), (2) and (3) gives

?

i

-I

x-x _{yYl 0 =0}

(Y.h + yl2 0 Expanding the determinant we get

x d

2

y +x(dy

)2 _

y(dY)=o

dx2 dx dx

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

1.5

Example

Form the differential equation by eliminating the constants from y = a secx + b tan x Solution Given equation is y= asecx

+

btanx Differentiating w.r. to 'x' dy

= asecx tanx

+

bsec2x dx

dy

= secx[a tan x

+

b sec xl dx

Further differentiation gives

d

2

y

- , 2 = a sec x tan2x

+

a sec3 x

+

h2sec2x tan x

(X

I.e., d

2

Y 2 1 2 3

- - 1 = asecx tan x

+

bsec-xtanx + bsec xtanx + asec x

dx-.... (I) .... (2) .... (3) I.e.,

d

2 y

dx₂

=

secx t'lnx(atanx

+

bsecx)

+

sec2x(btanx

+

asecx) ... (4) Substituting

and

in (3) we get

i.e.,

asecx

+

btanx

=

y from (I)

(~)

atanx

+

bsecx

= - -

from (2) secx

[

d

(dY)

2

---f)

= secx tanx

~

+

sec2x(y)

dx secx

d2y dy

- - - tanx - - ysec2x = 0

1.6

Example

Form the differential equation of all circles passing through the origin and having <

their centres on the x -axis.

Solution

Take any tangent to the circle as y-axis, the centre lies on x-axis. Let 'a' be the radius of the circle.

Then centre is (a, 0)

:. Equation of the circle is (x - a)2

+

.r

= a2

y x

,

y'

Fig. 1.1 Differentiating (I) w.r. to 'x' dy 2(x - a)

+

2y - = 0 dx dy x- a = y -dx dy a=x+ y -dx From (1) and (2)

[x-(x+

y:lJ'

+

y'

=[x+

y:J

.r

(:r

+.r

=

x2

+.r

(:r

+

2XY( : ) 2X_{Y :}

+

x2 -

.r

=

0 is the required D.E

.... (1)

x

1.7

Example

Form the differential equation of all central conics whose axes coincide with the axes of coordinates.

Solution

The equation of all central conics whose axes coincide with the axes is ax2

+

bl

= 1 Differentiating (1) w.r.t., x dy 2ax+ 2by - = 0 dx Differentiating (2) w.r.t., 'x' again d2y

(d

y

)2

a+by - +b - =0 dx2 dx Eliminating a, b from (1), (2)(3) x2 y2 X y -dy

o

=0 dx y~+ ~ _d2 _(d

r

dx2 dx 0 .... (1) .... (2) .... (3)

=>

Xyd2; +x(dy)2 _y(dY)=O is the required D.E.

dx· dx dx

Exercise - 1 (a)

1. Eliminate the arbitrary constants from the following and find the corresponding differential equation :

(i) y

=

mx + c (m, c arbitrary constants)

(ii) y = a e2x + b e-2x (a, b arbitrary. constants)

d2

[ Ans : ~ - 4y - 0] dx2

(iii) y = ax

+

bx2 (a, b arb. constants)

d2 _dy

[ADS: x2

----? -

2x -d

+

2y = 0]

dx x

(iv) (x - h)2 + (y - kf = a2, (h, k a, b arb. constants)

(v) y = (a

+

bx)e-X

, (a, b arb. constants)

(vi) y = a sin x + b cosx (a, b arb. constants)

d2y 2dy [ Ans: - + - + Y = OJ dx2 dx

d

2y [ADS: - l 2

+

Y = 0] (X

2. Find the differential equation of all circles with centre on the line y

=

x and having radius' I '.

3. Form the differential equation of all the circles with centre on the line y = -x and passing through the origin.

[ADS. : (xl

+

y) (:

-1)

=

2(y -

x) (x

+

y:)]

4. Find the differentiall equation of all the parabolas with vertex at the origin and foci on the x -axis.

dy

[ADS:

y

-2xy dx = 0]

5. Find the differential equation of all parabolas with the origin as focus and axis along x-axis.

dy dy

-( ?

Methods to Solve

1.8

The differential equations of the first order and of the First

Degree:

1.8.1 Separation of Variables

Sometimes the differential equation

/

dy

dx = q(x, y) can be written as

f(x) dx

+

g(y)dy

=

0 if the variables can be separated.

Integration of (1) gives the solution of the equation.

i.e., Jf(x)dx+ Jg(y)dy=c

where c is an arbitrary constant.

1.8.2 Example

Solve et"tany dx

+

(1-~sec2ydy = 0

Solution

The given equation

e-'"tany dx

+

(1-e-'")sec2ydy =0 can be rearranged as e( sec2 _y - - d x + - - dy=O I-eX tany Integrating J eX Jsec2 y --dx+ - - dy=c I_eX tany i.e.,

-Iog( e-'" -I )

+

log t~ll1Y = c tany

-X-I = c or tan y = c(eX

-I)

e

1.8.3 Example

dy

Solve -

=

1

+

xl +

Y

+

xly dx

Solution

The given differential equation can be written as dy = (1 +xl) (1

+

y) dx dy x3 -1 - 2

=

(1+xl)dx

=>

tan-I y

=

x + -

+

C +y 3

1.8.4 Example

dy

Solve dx - 2xy = x, where YCO)

=

1

Solution

dy

- =x(1+2y) dx dy I+2y =xdx On integration I x2

"2

Iog(1+2y)

=

2

+c

Giveny= 1 when x

=

0 Substituting in (1) 1

"2

Iog(3)

=

0

+

C C

=

~

log3

=

log(v'J) Hence the solution is

1 x2 (1+2Y)

"2

Iog(I +2y)=

2

+ Iogv'J (i.e.,) log - 3 - =xl

1.8.5 Example

dy Solve - =(4x+y+ If dx Solution dy - = (4x

+

y

+

1)2 dx Substituting 4x + y + I = t in (I) i.e., Integrating i.e., dy dt 4 + - = -dx dx dy = dt -4 dx dx dt - -4

=

t2 dx I t -tan-I-=x+c

2

2

tan-I (4X+;+ I) = 2(x + c)

The solution can also be written as 4x + y + I = 2tan(2x + c) where C is an arbitrary constant.

Exercise -1(b)

1.9

Solve the Following Differential Equations

1. :

=~Y

2. (2 - x)dy - (3

+

y) dx

=

0

.... (I)

[ADS: ~

+

e-Y _{= c ]} [ADS (3

+

y) (2 -x) = c]

dy 6. (x-y)2 dx

=

a2 dy 7. - =(3x+4y+ 1)2 dx 8. dy =(2x+y+

If

dx dy 9. dx

=

tan (x + y) 10 dy

=

2 . dx (x+2y-3) [ ADs: yc

=

(a

+

x) (I - ay) ] ( x- y-a) [ADs: (x - y) + log

=

x

+

c] x-y+a

[ADs: 2(3x+4y+ 1)= .J3tan-12.J3x+c]

1

-I

(2X

+

Y

+

I )

[ ADs:

.J2

tan

.J2

= x +c]

[ADs: log[sin(x + y) + cos(x + y)] = x - y + c]

[ADs: (x + 2y - 3) - 410g (x + 2y + I)

=

x + c ]

1.9.1 Homogeneous Equations

The differential equation of the form dy f(x,y)

dx g(x,y)

where f, g are homogeneous functions of same degree in x, y is called a homogenous differential equation.

Such a differential equation can be written as

Substitutingy

=

vx

dy dv

-

=v+x-dx dx

The D. E (I) becomes

dv \11 (v)

v

+

x dx

=

~(v) dx ~(v)dv

-; +

v~(v)-\lf(v)

= 0

Integration yields the solution

J

-dx +

J

"'()

~(v)dv

( )

=

c where v

= -

y X v'" v - \11 v x \

1.9.2 Example

Solve Solution dy (xl

+

I) -

=xy dx S bU stltuttngy · .

=

vx, -d dy

=

v

+

x -dv x dx dv xvx v

+

x - = 2 2 2 dx x +v x dv v v+ x - = -dx 1+ v2 dv v x - = - - - v dx 1+ v2 dv _v3 x = -dx 1+ v2

J

- + dx

J

~+ I

JI

-dv=c X v3 v

v-2 logx + - + logv = e

-2

1 logxv- - = e 2V2

(

y)

X2

logx - ~ = e ~ 2ylogy - xl =

2ey

x 2y

i.e., 2ylogy =

2ey

+ x2

1.9.3 Example

Solve Solution i.e., dy y2 x-+-=y dx x dy xl-+y=xy dx Substitutingy

=

vx, dy dv -=v+x-dx dx dv v+x -

=v-vl

dx dv x -

=-vl

dx dx dv

- + -

=0 X v2 Integration yields V-2+1 1 logx+ - - =e ~logx-- =c -2+1 v

x logx - - = c

y

ylogx -x = cy or ylogx = x + cy

1.9.4 Example

Solve [x+ YSin(Yx)]dx= xSin(Yx)dy

Solution

The given differential equation can be written as

dy x+ ysin(X)

= __

--,---.0.-,.---.:-dx

xSin(j~)

dy dy Substitutingy

=

vx ~ -

=

v + x. -dx dx Integrating dv x+ vxsinv v+x - = -dx xsinsin(v) dv l+vsin v+x-

=

-dx sinv dv l+vsinv 1 x - = dx sin v -v=-.-SI11 v dx sin v dv

=

-x -cosv

=

logx + c logx + cos(Yx)

=

c

Exercise - 1 (c)

1.10

Solve the Following Equations

dy y2 l. dx - xy-x2 2. (2 - 2xy)dx = (x2 - 2xy)dy 3. 2xy

+

~

-

r) dy

=

0 dx [ADS : y

=

ceix ] [ ADS : xy(y - x)

=

c ] [ADS:

r

+

y

=

cy ]

[ADS: logx

=

2tan-i (Yx)

+

c]

5. xdy - ydx

=

~

x 2

+

y2 dx

7. xcos(Yx) (ydx

+

xdy)

=

ysin (Yx) (xdy - ydx)

y

[ ADS: cos -

=

logcx ] x

[ADS: sec (Yx)

=

yxc]

8. (ry - x3_{)dy - ~}

₊

_{xy) dx = 0 [ ADS:}

y~

_x 2 _{+ y} 2

₌

_cx.e tan

-I(YI)

Ix _] 9. (r

+

y) dx

=

2xydy

dy

10. X dx

=

y[logy - logx

+

1]

1.10.1 Non-homogeneous Differential Equations

The D.E of the form

dy

=

ax+by+c

dx Ax+By+C

( ADS: (r -

y)

=

xc ]

( ADS : y

=

xecx ]

.... (J) where a, b, c A, B, C are constants, is called a non-homogeneous dif.ferential equation.

Case (i)

If --::;:.-a b

A B

Substituting x

=

X

+

h, Y

=

y

+

K

where h, k are (constants) to be chosen so as to satisfy. ah + bk + c

=

0, Ah + Bk + C

=

0

Solving these equations, values of h, k are obtained. The given D.E then reduces to a homogeneous D.E.

dY aX +bY dX AX +BY which is then solved taking Y = VX and then substitute X = x - hand Y = y - k in the solution.

Case (ii) a b

If

A B

Then the differential equation will be of the fonn dy _ (ax + by)+c

dx - m(ax+by)+C

since Ax

+

By will be constant m times ax

+

by. Now substitute ax

+

by = t, Differentiation gives i.e., a+bdy=dt dx dx dt - - a dy dx = -dx b

D.E (2) then reduces to dt

- - a t+c

~=---b mt+C

1.10.2

Then the solution is obtained by using the method of separation of variables.

Example

Solve dy x+2y-3 dx 2x+ y-3 Solution dy = x+2y-3 dx 2x+ y-3 Substituting x = X + h, y = Y + k

where h, k are chosen to satisfY h + 2k - 3 = 0 and 2h + k - 3 = 0 solving we get

h=I,k=I

i.e., we take x = X + I, Y = Y + I

The D.E (I) reduces to

dY X-2Y = -dX 2X+Y dY dV Substituting Y = VX, dX = V + X dX dV X+2VX V + X = -dX 2X+VX V+XdV =1+2V

~

XdV =1-V2 dX 2+V dX 2+V 2+v dv=dX I-v X dX +(1 +_3_) dv = 0 X v-I

10gX

+

v

+

3 log( v-I) = loge

... - (I)

v + log( v-I )3 + logX = loge v-log(v-I)3 X = loge Y V-x

~

) 3 X +10

-X

X=loge

I.e., 10g(Y - X)3 + XY = Xloge

1.10.3

Example

log(y - \ - x + \)3 + (x - \) (y - I) = (x -I) loge log(y - x)3 = (x - 1) (loge - y + \)

Solve (2x + 3y + \) dx + (2y- 3x + 5)dy = 0

Solution dy 2x+3y+ 1 dx 2y-3x+5 Substituting x = X + h, y = Y = k, : = : Choosing h, k so that 2h + 3k + 1

=

0, 3h - 3k - 5

=

0 we get h = 1, k

=

-1

The given differential equation reduces to

dV 2X+3Y Substituting dX 3X-2Y Y=YX dV dV - =Y+X-dX dX dV 2X+3YX Y + X dX

=

3X - 2YX Y+X dV =2+3Y dX 3-2Y X dV

=

2(1 + y2 ) dX 3-2Y

(3-2V)dV= 2dX _{1+ V2 X} Integrating

f

dv

f

2v

fdX

3 - - - --dv=2 - + c 1+ v2 1+ v2 X 3tan-1(V) - log( I

+

v2)

=

210gX

+

c

3tan

1

(~ )-IO~

X';/')

~

210g)(

+

c

B~ X=x-h=x-I Y=y-k=y+1

3tan-

1

(~::

)-IOg[

_(x

-Il~: ~

₊

I)']

~

_210g(x-

I)

₊

_c

1.10.4, Example

Solve Solution dy =x-y+l dx 2x-2y dy (x- y)+1

- =

--'---'-'--dx 2(x-y) Substituting x - y

=

V dY=I_ dV dx dx l_dV=V+I dx 2V dV_ I V+l_V-1 - -

-dx 2V 2V 2V - dV=dx V-I

1.10.5

Integrating or

Example

Solve Solution Substituting 2 JV-I+I V-I dV=x+c 2[V + 10g(V-l)] = x + c 2[x - y + log(x - y -1)1 = x + c 210g(x - y - I) = 2y - x + c (x - y -1)2 = e2j~x C dy 4x+6y+1 dx 2x+3y -5 (~V 2(2x+3y)+ I = -d'( (2x+3y)-5 2x

+

3y = V dy

(N

2 + 3 = -dx dx dy

=~[dV

-2]

dx 3 £Ix

The differential equation reduces to

~(dV

-2)=- -2V + I 3 dx V-5

dV

6V+3 IO-2V+6V+3 - =2+ - - = -dx 5-V 5-v 5-V 4V+13 dV=dx 13+4V 5-V

Integrating

fl

1 - 33 ) dv = _ 4 Idx + e

J~

4V + 13 33 v- -log(4V+ 13)=-4x+e 4 33

(2x + 3y) - - log[4(2x + 3y) + 13] = - 4xm + C

4

i.e., 4(6x+3y-e)=33Iog[4(2x+3y)+ 13]

Exercise -1(d)

1.11 Solve the Following Equations

dy y-x+5

\. - +

-=---dx y+x+3

y+4 1 y+4

( ) [ ( )2

1

[Ans:tan-J _{x-I +"2 log x-I +1 =log(x-I)+loge]}

dy x+2y-3 2. dx - 2x+3y-5

[Ans: (2+Ji) IOg[y-1 -

~]

- (2-Ji) log [y-I +

~]+ ~

logx= loge]

x-I ,,3 x-I ,,3 ,,3

3 _dy = _2y'---_x_-_4 . dx y-3x+3

[Ans: (x - 2)2 - 5(x - 2) (y - 3) + (y - 3)2 = e {2(y-3)-[5+h1{X-2)]}] 2{y -3)-(5 -h1(x-2» 4. (2x+5y+ l)dx-(5x+2y- l)dy=O

dy y-x+ 1

5.

dx y+x+5

( y-2) I

[ADS : tan~l x _ 3

+

2

log [(y - 2)2

+

(x - 3)2] = loge] dy _ 3y+2x+4

6. dx 4x+6y+5

[ADS: .7'j(2x+3y)-

:9

(14x+2ly+22)=x+c] 7. (4x-6y-l)dx+(3y-2x-2)dy=O

dy x-y+3

8. dx

=

2x - 2y + 5

1.12 Linear Differential Equations

3

[ADS: x - y

+

41og(8x - 12y - 5) = c ]

x

[ ADS: (x - y)

+

log[2

+

x - y ] = -

+

C ]

2

A differential equation of the from : + py = q where p, q are functions of 'x', alone or constants is said to be a linear differential equation offirst order. Multiplying both sides of the equation by el pta [called the integrating factor (I.F)] we get,

el

ptlr

dy + el

ptlr

py

=

q .e

I ptlr

dx

The left hand side is the differential coefficient of y. e

l

pcb:

Integrating

(i) can be written as d(y.e

1ptlr

)

= qe

1ptlr

which gives the desired solution.

Note:

1.12.1

In some cases a differential equation can be reduced to the linear form by taking 'y' as independent variable and x as the dependent variable.

The D.E is written as

p" q I are functions of y or constants

J

Pldy Now the I.F = e

Solution is

Example

Solve Solution JPldy

f

Jpl"Y • x.e = ql·e +c dy (I

+

x2_{) -}

₊

_{2yx - 6x}2 _{= 0} dx

Rearranging the given differential equation to the form dy dx

+

py

=

q We have Here Solution is given by dy 2x 6x2 + y = -dx I +X2 1 +X2 2x 6x2 p= I+X2 ,q= I+X2

J

~dx

J

pdx 2 I.F= e =e I+x log(I+X2 ) I.F = e = (1

+

x2 ) y(1. F)

=

f

p(I.F)dx + c

1.12.2

Example

Solve Solution Here Solution is

1.12.3

Example

Solve Solution xlogx -dy +y=2Iogx dx dy y 2 + = -dx xlogx x I 2 p= - - andq=-xlogx x ylogx =

J~

logxdx + c (IogX)2 ylogx=2 +c 2 dy ( . ) xcosx

-+

Y xsmx+cosx

=

I dx dy xsin+ cosx + y . -dx xcosx xcosx xsin x + cosx I p= , q = -xcosx xcosx

Solution is

1.12.5

Example

Solve Solution

J

~Slll X~.COS~ dx I.F = e XCDSX = e(log(xsecx) I. F = e1og(x sec x) = xsecx y(xsec x) = f _ l - x x sec x dx + c xcosx xysec x = Jsec2 xdx E c xysecx = tanx+c

dy + 2ytanx = sinx given that y = 0 where x =

~,

~ 3 dy + 2ytanx

=

sinx dx p = 2tanx q = sinx IF = e

f

2 tan xd!: ~ = e210gsecx ylF= JqxIF.dx+c ysec2_x

₌

_Jsecx.sec2 _{xdx + c}

1.12.6

ysec2_{x =}

J

_{sec x tan xlir}

₊

_c

')

ysec-x = secx + ('

Given that y = 0 when x =

;7j

1C

o

= sec -

+

c => C = -2 3

Substituting c = -2 in (I)

ysec2 _{x = secx - 2} is the required solution

Example dy Solve (x

+

2.v) - , = y (X Solution X +

2.v

= ydy dx dx x - - - =

2.v

dy y is in the form of -1 P = - q = 2,1 I ' 1 Y Y y .... (I)

1

f

2 1 x- = 2y x-dy+c y y x

- =y+C

Y 1.13.7 Example Solve Solution dy (x

+

y+ I) - = 1 dx dy dx- X_=y+l PI=-lq\,=y+1 IF = e

f-

1dy = e-Y Solution is given by x(IF)

=

fql(/F)dy+c xe-Y _{= f(y}

₊

_I)e-Y _dy

₊

_c

i.e., or

Exercise 1(e)

x

+

y

+

2 = ceY

1.13 Solve the Following Differential Equations

I. (I

+

y)dx = (t~n-Iy - x)dx

dy

2. cos2x -

+

Y

=

tanx

dx

[ADS: xetan- I y

=

tan-Iyetan-Iy --etan- I y

+

c]

dy 3. x - +2y-x21og=0 dx c I .r2 (ADS: y = - ? +- x210gx - - I x- 4 16 dy n/

4. dx

+

ycot x

=

4x cosec.x, if y = 0, when x

=

~2

5. yeYdx = (y

+

2x&)dy 6. (x

+

31) dy = y dx dy 7. (xy -I ) 3

+

y3 = 0 dx+ y dy 8. -

=

x3 - 2xy if y

=

2 when x = I dx dy 10. dx

=

x+ ycosx I +sinx n2 (ADS: ysinx = 2x2 -2

( ADS : xy~2

=

c - e~Y

I

3 ( ADS : x =

21

+

cy ] I . I

I

ADS: x = ceY + - +

II

Y [ADS: 2y - x2 _{+ 1 = 4} el~~ _]

x

2 [ADS: y(l+sinx) = c - - ] 2

1.13.1 Non-linear Differential Equation of First Order

Ber noulli j. equation:

The differential equation of the form ely

dx

+

PY = llyn .... (I)

where p, q are functions ofx alone is said to be a Bernoulli's differential equation. Dividing (I) throughout by yn

dy y-n_

+

py-n+l

=

q dx Substitutingynt-l(coefficient ofp) "" v 1 dy dv (1-n) = -y" dx dx (2) reduces to I dv - - - +pv=q

(I-n)dl'

dv - + (1-n)pv = (I-n)q dx which is linear in v. .... (2)

The avove D.E can be solved by using the method given in 1.12.1 example.

1.13.2 Example

Solve dy - ytanx = ysecx dx Solution Substituting dy dx - ytanx = ysecx dy 1

y-2 -d - - tan x = sec x

x Y

- - =v _{y ,}

1 dy dv

+ =

-/ dx dx (1) reduces to dv dx

+

vtan x = sec x is linear in v.

Here p = tanx, q = secx

IF = e

f

pd<

flanXd< = e = elogsec x

= secx

SolutIon of th [) E (I) IS therefore

v(lF)

=

Jq(IF}dx + c v. secx = Jsecx.secxdx + C vsecx.= J sec 2 xdx + c Substituting 1

v=--Y 1 - - sec x = tan x

+

c y

i.e., y(tan x

+

c)

+

sec x = 0

1.13.3 Example

Solve (3xy

+

.0)

dx - 3i2dy = 0

Solution

dy

3x2 - -3xy =

.0

dx

is in Bernoulli's fonn Substituting 1 dy _

"7

dx - xy - 3x2 -1 y =v D.E reduces to dv I 1 v = -dx xy 3x2 which is linear in v 1 1 Here _{p=~, q= 3x}2 IF = e

Jpdx

= J!d\ e x =x Solution of the D.E (1) is

v(lF)

=

fq(lF)dx+c -..!..x= f_l_. xdx

+

c y 3x2 x 1 --=-Iogx+c Y 3 i.e.,

y(

110g

x

+

c)

+

x

=

0

1.13.4 Example

dy Solve dx

+

(2xtan-1y -.x3) (1

+

y)

= 0 Solution dy dv 1

+

y2 dx

=

dx .... (1)

The given differential equation becomes dv

-

+

2xv=x3 d-c which is linear in v Here p = 2x, q = x3

J

2xdx x2 IF = e =e Solution of the D.E (1) is

v(lF) =

Jq(IF}dx+

c Writing x2 = t 1.14.5 Example 1 xdx

=

2dt 2

r

I dt

v.e

x

₌

_{{e 2+c}

dy

Solve tany dx

+

tanx

=

cosy cos3x

Solution

dy

tany d'C

+

tanx

=

cosy cos3x Dividing by cosy throughout

dy

secy tany dx

+

secy tanx

=

cos3x Substituting secy = v, we get

dy dv secy tany dx d'C

dv

dx + v . tal1.X

=

cos3x is linear in v

Here p

=

tal1.X, q

=

cos3x

IF =

f

pdt =

flallXdy

= secx

e e

Solution of the D.E (I) is

v(IF)

=

fq(IF)dx + c

v. secx = fcos3 xxsecxdx+c secy secx = fcos2 xdx + c

I

I +cos2x secx secy = 2 +c x sin2x secx secy = - + - - + c

2

4

1.13.6 Example

dy . . cosx

Solve -d = (SII1.X - smy)

-x cosy

Solution

d y . ."

cosy dx

+

smycosx = SII1.XCOSX

Substituting siny = v

dy dv

cosy-=-dx dx

The gIven equation reduces to

dv .

dx

+

vcosx

=

SII1.XCOSX

.... (I)

is linear in v

Here p = COSX, q = SIl1 X cos X ~

f

pdr

feo')

'(tl\ ~1Il X

IF = e = e = e Solution of the D.E (I) is

J .

m"' I

v.e SlIlt = SII1 xcosx.e . (X + c

write sinx

=

t=> cosx dx

=

£II in the RHS

siny e;lIlt

=

[tet - el] + c

sinyeSlI1

\" = eSItlX[sinx-l] + c

Exercise - 4(f)

1.14 Solve the Following Differential Equations

I. (ylogx -I )ydx = x(~y

1

(Ans: - = 1 + logx + ex ) y

dy .

r.::::::::

2. - cosx + YSlnx = ...; ysecx dx

dy tany

4. - - - - = (I+x) ~secy

£Ix l+x

(Ans :2y y 12 -Jsecx = tanx

+

2c )

( Ans :

~

= - sin2x -sinx

-~

+

ce2SIIl \" )

y- 2

dy 5. x -

+

ylogy = xyeX dx dy 2 x3 6. 3-+--y=-) dx x+ I

y

7.

dy + ytanx = isecx dx 8. dy dx X 2 1 Y +xy

I

ADs : xlogy = (x - I)e"

+

c J

x6 _2x5 X4

I

ADs: (x +

I i i

= - + - + - + C

6 5 4

I

ADs: cos2x = y(c

+

2sin x) J

1.14.1 Exact Differential Equations

Let us consider the differential equation Mdx

+

Ndy = 0 where M, N are functions ofx,y.

If this equation is to be exact, then it must have been derived by directly differentiating some function ofx,y.

Hence Mdx

+

Ndy

=

du, say But from differential calculus

au

au

du = -dr:+-dy

ax ay

From (I) and (2) we get

au

N =

au

M= ax' ay

aM

a

2

u

Now =

-ay ayax and -ax- -

-ax-ay-.... (I)

aM oN. I d·· ..

- = - IS t le con ItlOn lor exactness.

oy

AX

:. The differential equation Mdx + N~v = 0

. 'f aM

aN

IS exact I =

-oy

ax

Then the solution is expressed in the form

+

(treatingyas (integrate w.r.t y those

constant integrate w.r.t x) terms that are independent of x) Note:

IfN has no term independent ofx then the solution is fMdx = e

1.14.2 Example

Solve (x + 2y - 3) dy - (2x - Y + \ )dx = 0 Solution (x+2y - 3)dy - (2x - Y + I) dx = 0 M = -(2x - Y + \) N = (x

+

2y - 3) aM - = \

oy

aN

- = \

ax

The given differential equation is exact The solution is

- f(2x - y + \ )dx +

fe

x + 2y - 3) =

(ry

=

e

_2X2 2i

- - 2 - + yx - x +

2 -

3y = e

1.14.3 Example

Solve (x2

₊

_y.)dr:

₊

_{2xy dy = 0}

Solution (x2

₊

_.V)

_dx

₊

_{2xy dy = 0} M=x2+y.

aM

- = 2 y

av

N = 2xy

aM

- =2y

ax

The given differential equation is exact Solution is f(x2

+

y2

}h

+

f2xydy

=

c

1.14.4 Example

Solve (I

+

e

-,:~,

)dx

+

/Y

[I.'... ;

1

dy = 0 Solution x x' ( ) (I

+

e' Y ) dx

+

e Y J - ; ; , dy = 0 t x' [

x)

M = J

+

e:Y , N = e,iy J - Y

aM

-x ,y' - = - e'Y

~

i

oM oN

oy

ox

Solution is

x

X + e Y (y) = c

Exercise -

4(g)

1.15 Solve the Following Differential Equations

1. (el' + 1 )cosxdx + eJ'sinx~v = 0

lADs: (oY + 1 )sinx = c

I

2. (vcosx + siny + y)tb' + (sinx + xcosy + x)dy = 0

I

ADs: ysinx + (siny + y)x = c

I

3. (x2 - ay)dx = (ax - ;l)(~v ADS: x3

+

),3 - 3ll.\y = C

I

4. (ax + hy + g)d'C + (hx + hy + j)dy"= 0 ax2

hi

I

ADs· -_{. 2} + (liy + g)x + ':' + _.!. fy = c

I

5. (x2 +

1-

a2_{)xdx + (x}2

-.v -

_{P)ydy = 0}

I

ADS : x4 + 2x2

1 -

2a2

x

2 -

l -

2b2

;l

= 4c

I

1.15.1 Integrating

factors

If the differential equation Mdx + Ndy = 0 is not exact, it can be made exact by multiplying it with some function of x, y. Such a function is called an integrating jactor.

Rule!t' for fillt/illg the illtegrtltillg factors :

1. Integrating factors found by inspection: Example

Solve x dy- ydx = 0 Solution

Dividing by x2 xdy- ydx ---'-,=-=-- = 0 x2 On integration

Yx

=c

First method /0 find an integrating jac/or :

If the differential equation Mdx

+

Ndy is not exact, but is homogenous and Mx

+

Ny 7:-0, then the integrating factor is 1 . Multiply the differential

Mx+Ny equation by IF. The DE becomes exact.

1.15.2 Example

Solve (x2y - 2xy2)dx - (x3 - 3x2y)dy

=

0

Solution

(x2y - 2xy)dx - (x3 - 3x2y)dy

=

0

The differential equation (I) is homogeneous

M

=

x2y- 2xy

aM

- =x2-4xy

oy

The DE is not exact and Mx

+

Ny

=

x2y 7:-0 ) 1 IF = Mx+ Ny - x2 y2 N = - (x3 - 3x2y)

aN

- =-3x2

+

6xy

ax

Multiplying the DE by the integrating factor

~

• x y (X 2Y_-2Xy2

_)d

_(X2 -3X3Y

)d

= 0 2 2 X 2 2 Y X Y X Y .... (I) .... (2)

write M = - - -1 2 I Y X -x 3 N

= +

-I y2 Y then aMI aNI = =

-ry

ax

y2 i.e., or DE (2) is exact Solution is x - - 210gx

+

3 logy

+

c y x

1.15.3 Second Method to Find the Integrating Factor

If the differential equation Mdx

+

Ndy = 0 is not exact and is of the form j(xy)ydx

+

g(xy)xdy = 0

and Mx- Ny *- 0

1

then Mx _ Ny is an integrating factor

1.15.4 Example

Solve (x2y2

+

xy

+

1 )ydx

+

(x2y2 - xy

+

1 )xdy

=

0

Solution

(x2y2

+

xy

+

1 )ydx

+

(x2y2 - xy

+

1 )xdy = 0 .... (1)

M

=

x2j3

+

xy2

+

y, N

=

x2y2 - x2y

+

x

Mx - Ny

=

2x2y2 *- 0

1 1

-Multiplying (I) by IF

1.15.5 Example

(X2y 2 +xy+

1)+

(X2y2 -xy+

I)

=

0

2x2y2 2x2y2

I{ I 2} I{ II}

- y + - + - dx+- x - - + -1 dy=O 2 x x 2y 2 y xy-DE (2) is exact

Solution of the DE (I) is

~ 'y+~+-!-tx+~

'x_J.-+_I_₁

)d

_Y = c 2

Jl

x x- y

r

2

Jl

y xy

1

[

I] I

- xy + log x - - - -logy = c

2 xy 2

Solve (xysin xy

+

cos xy)ydx

+

(xysin xy - cos xy)xdy

=

0

Solutiou

(xysin xy

+

cos xy)dx

+

(xysinxy - cosxy)xdy = 0 M

=

(xysin xy

+

cos x,v)y, N

=

(xysin xy - cos xy)x Mx - Ny = 2xycosxy =F- 0 I F = -Mx- NY 2xycosxy Multiplyingthe DE by 2xycosxy The DE reduces to

~(ytanXY+~)dX+~(

xtanxy-

~)

dy = c .... (2) .... (1) .... (2)

which is exact (verify) .. Solution is

1 1 1

-Iogsecxy + -Iogx - - log y = c

2 2 2

I.e., logxsecxy

=

2c

+

logy. Taking 2c as log A ~ xsecxy

=

Ay

Exercise - 4(h)

1.16 Solve the Following Differential Equations

I. (x3; + x2y2 + xy + 1 )ydx + (x3; -

xli -

xy + 1 )xdy = 0

1

r

ADS: xy - - - 210gy

=

c

I

2.

(xli

+ xy

+

I )ydx +

(xli -

xy + 1 )xdy = 0

xy

1

r

ADS: xy

+

logx - logy - - = c ] xy

[ ADS: logx2 - logy - _I = c ] xy 4. (x4y4 + x2

i

+ xy)y dx + (x4y4 - x2

i

+ xy)xdy = 0

I

ADS :

~

x2

i -

_I - logx - logy = c ]

2 xy

1.16.1 Third Method to Find the Integrating Factor

aM

aN

If Mdx

+

Ndy = 0 is not exact and

_a-=-~

_ _ ax_ is a function of x alone say fix),

N

J

j (x)dr

1.16.2 Example

Solve (x2

₊

y

₊

_{6x )dx}

₊

_{yxdy = 0} Solution (x2

+

Y

+

6x)dx

+

yxdy = 0 M

=

x2

+

Y

+

6x N

=

yx aM

aN

ay=

3y,

fu

=

y2 aM aN

ay-ili

3y2 - y2 2 N y2x X is a function of x alone

f

2tJx IF

=

e ~

=

x2 Multiplying the given DE by x2

x2(x2

+

y

+

6x)dx

+

yi3dy

=

0 aMI - =3x2

y

ay

DE (2) is exact Solutionis ... (1 ) ... (2) f(x4

+

x2/

+

6x3}tx

+

fy2x2dY = c

1.16.3 Example

Solve (x2

+

y)

dx - 2xydy

=

0 Solution (x2

+

y)dx -

2xydy

=

0 ... (1 )

M

=

.xl

+

.0,

N

= -

2xy

aM

aN

ay

= 2y, ox = -2y

aM

aN

ax ax N is a function of x alone 2y+2y -2 -2xy x

f

-~dx 1 IF = e x =

-x

2 MuItiplyingthe DE by 1 IF= -x2

The given DE reduces to

1+- dx,.--dy=O

[ l)

2y x2 x

aM, _

2y

-a

- - 2 ' y X (2) is exact Solution of (1) is

oN,

2y

=

-y2

2y

J

l+-dx+ J--dy=c x2 x

y2

x- - =c x

·1.16.4 Fourth Method to Find the Integrating Factor

... (2)

If the differential equation Mdx

+

Ndy

=

0 is not exact and ax M By

[

aN

aM

1

is a

Then

1.16.5 Example

IF =

f

t(y)<ly

e

Solve (xy + y) dx + 2(.x2.0 + x + y4)dy = 0

Solution

(xy + y)dx + 2(.x2.0 + x + y4)dy

=

0

M =

.xy3

+ y N = 2(.x2.0 + x + y4) oM - - - = 3xy + 1 8y

oN

- =4xy2+2

ox

-a;

-8;

= (4xy2 + 2)-(3 xy2 + I)

(

oN OM]

M xi+y

is a function ofy alone

f~d)'

F=e Y _=y

y

Multiplying the differential equation by I.F

=

Y (xy4 + .0) dx +

2(.x2Y

+ xy + yS)dy

=

0

oMI .,3 --=4xy +2y 8y oNI .,3 - =4xy +2y

ox

The differential equation (I) is exact

Solution is f{xl +

i

}:Lx

+ 2 f(x2 i + xy + yS) dy = 0

1.16.5 Example

Solve (3x2

y4

+

2xy)dx

+

(2x3

_{_Y' -}

_{x2)(~V = 0} Solution aN _ aM _ (6x2/ -2x)-(12x2y3

+2xL

6x2/ -4x

-_3.

ax ay - 3x2

l

oJ-2x - y{3x2/ + 2x ) - Y is a function of y alone

-

J~ dy I

IF

= e Y =-?

Y

Multiplying the DE by

~,

the DE reduces to y

(

3x y +-y dx+ 2x y - 7 dy= 0

2 2

2X)

(3

x2) :. DE (I) is exact Solution of the DE is x3 2 2 x2 3 - y + - - = c 3 y 2 aNI =6xy- 2x ax

i

Exercise - 4(i)

1.17 Solve the Following Differential Equations

I. (x

+

y3

+

I )dx

+

ydy

=

0

2 3x 2 2 3x 3x

X e x 3x e 3 e

ADs: (ADs: - - - - e +--+(y + l)-=e

I

3 9 9 3 3

2. (x2

+

Y

+ 2x)dx

+

2ydy = 0

3. 2xydy - (x2

+

Y

+

I)dr = 0

5. (y4

+

2y)dx

+ (

xy3

+

2y4 - 4x)dy = 0

1.17.1 Fifth Method for Finding Integrating Factor

If the differential equation Mdx

+

Ntry= 0

is not exact and is of the form

xliyb(mydx

+

nxdy)

+

Xf y(pydx

+

qxdy) = 0 when a, b, r, s, m, n, p, q are constants

Then the IF =

xlII

I

ADs:

y -

x2

+

I = ex

I

2x .1

I

ADs: xy

+ -) +

Y = e ]

y

.... (l)

where h k are constants such that the equation (I) after multiplying with IF becomes exact.

1.17.2 Example

Solve xy\ydx

+

2xdy)

+

3ydx

+

5xdy) = 0

Solution

xy3(ydx

+

2xdy)

+

(3ydx

+

5xdy) = 0 The IF =

x"1

Multiplying (I) by IF =

xhl

it must become exact

(xh+1 1+1

+

3x" 1+1)dx

+

(2xilj.?

+

1+3

+

5xil f-/ I){~Y = 0 is exact if

aM

-

_ay

= (4

+

k)xH1

_.

1+3

+

3x\k

+

1)1

_.

aN

- = 2(h

+

2) xh

+

I

/+

3_+5(h

₊

_1)x"1

ax

For DE (2) to be exact

aM

aN

ax

ax

Comparing the coefficients on both sides

Solving

2h - k = 0 and Sh - 3k = -2

h = 2, k= 4

Substituting h = 2, k = 4 in (I) DE (I) reduces to

(x3y8

+

3x2

yS)

dx

+

(2x4y1

+

5x3y 4)dy = 0 which is exact .. Solution is

... (1 )

1.17.3 Example

Solve (3x

+

2y)ydx

+

2x(2x

+

3y)dy = 0

Solution (3x

+

2y)ydx

+

2x(2x

+

3y)dy = 0 IF =

0/,

MUltiplying (1) by IF .... (1) (0/,+2

+

30+2 /,+I)dx

+

(2xh_{+3/, - 0+}1_{/,+I)dy = 0 .... (2)} oM - = 3(k

+

2)0/,+1

+

2(k

+

1) 0+2_yK

ox

oN

-

=

2(h

+

3) xh_{+2/, ---(h}

₊

_{1 )0/,+ 1}

ox

The DE(1) is to be exact

oM ON

ox

ox

2(k

+

1) = 2(h

+

3)

=>

h = -

%

k =

-/i

Comparing the coefficients

k

+

2

= -

(h

+

J)

and 2(k+ 1)

=

2(h

+

3)

=>

h

=

-,%

k=

-/i

Substituting h, k values in (2)

X /2 y/2 + 2x 12 yl2 dx + 2X72 Y 12 _ X /2 y72 dy

=

0

(

-5/ 3/ I -1/ II) ( II -II -3/

II)

(3) is an exact DE

Solutionis --x 2 -2/ 3/ 13 yl2

+

4X72 y72 II II

=

c 3

Exercise - 4(j)

I Solve the following differential equations I. x(3ydx

+

2xdy)

+

8y4(vdr

+

3xdy = 0

4 2

I

Ans: 3x 3+l_y 3 4 - +1 3

3.

(y2

+

2x2y)dx

+

(2x3 - xy)dy = 0

10 +1 X 1 7 - - - y 3 =c] 10 - - + I 3 I I 2 3 3 ( Ans : 4x 2 y 2 _ _ X 2 Y 2 = ex ] 3

4. (2.~y - 3y4)dx

+

(3x3

+

2xy3)dy = 0

- 36, 24 _10 ' 15

(Ans: 5x 13_{y 13_12x} I ] y I] =c]

1.17.4 Applications to geometry, law of natural growth and Newton's law of cooling:

When some action is applied on a quantity the changes and the action affects all the parts equally. The rate of change depends on the original quantity.

For example the total population 'p' of a country increases with respect to time 't', say, then its rate of change with time is

~

. Under ideal condrtions the rate of change of the population will be proportional to the total population at any given time and is called the law of natural growth.

The growth of the population satisfies the differential equation dp

- =kp

dt

where k > 0 is a constant

Solving the differential equation we have

p

=

cekl

dx

This rate of decrease or decay is found to be proportional to x itself dx

- =-kx

dt (where k> 0 is a constant) is the law of decomposition

Solving the DE x

=

ce-kl

Newton

s

law of cooling:

Newton's law of cooling states that the rate of decrease of the temperature of a body is proportional to the difference of the temperature 9fthe body and that of its surrounding medium.

Let 0 be the temperature of the body at time 't' and 0₀ the temperature of its surrounding medium.

:. Difference between the temperatures = 0 - 8₀ (0 > 00) and the rate of decrease

dO

of the temperature of the body is - dt' since the body is cooling oc.

Using Newton's law ofcooling - -de oc(O - 0 )

dt 0

-dO

or

dt

=

k(O - 0₀) where k> 0 is a constant of proportionality

dO

- = -k (0 - 0 ) dt 0

f~=-k

fdt+c 0-0₀ 10g(0 - 0₀)

= -

kt + c or 0=0

_o

+

Ce-kl 1.17.4 Example

The mass of crystalline deposit increases at a rate which"js proportional to its mass at that time. The deposit has started around a crystal seed of 5 grams. Find an expression of its mass at time 't'. Ifin 30 minutes the mass of the deposit increases by'l' gram, what will be the mass of the deposit after 10 hours.

Solution

Let m be the mass of crystalline at time 't' then by law of growth dm=k'

dt '

dm - = d t

on integration log m = kt + c initially when t = 0, m = 5 (1)

=>

log5 = 0 - c c=-log5 SubstitutilllJ; c value in (1) log m = kt-log5 I.e., _{kt= 10g(}

~)

When t = 30 minutes mass deposit increases by , I ' gram m = 5

+

1 = 6 grams Substituting 1 t =

"2

(hours), m =6 Substituting in (2) log(1Y t= 109;

Now to find the mass after 10 hours (i.e t = 10) from (3) we get

IOg(H

x

IO~

log

7

~

log

(7)

~

IOg(%)"

(

6)20

m = 5"5 grams

.... (I)

.... (2)

1.17.5 Example

The rate at whi~h a certain substance decomposes in a certain solution at any instant is proportional to the amount of it present in the solution at that instant. Initially, there are 27 grams and three hours later, it is found that 8 grams are lett. How much substance will be left after one more hour.

Solution

If m grams is the amount of the substance left in the solution at time 't', then the rate at which it decomposes is dm , which is proportional to m.

dl By law of decay dm

dt

= - km (k> 0)

f~

=-k fdt+c logm = -kt+ c Initially when t = 0, m = 27 From (1) we have log27 = - k.O +c

=>

c= log27

Substitution of'c' in (I) gives log m = - kt + log27

10g(;; )

= -

kt It is given that m = 8 when t = 3 .. From (2) log (28 7 ) = - k, 3 ( 8

)X

-k= log 27 2 -k=log-3 .... (I) .... (2)

Then (2) becomes

10g~

=IOg(%} ... (3) when t = 4 ( ) 4 III 2 log 27

=

log

"3

m=27 x (%r grams 16 m =

3

grams. 1.17.6 Example

The number x of bacteria in a culture grow at a rate proportional to x. The value ofx was initially 50 and increased to 150 in one hour what will be the value ofx after

1 12" hour. Solution dx -=/0: dt dx - =kdt x logx = kt

+

c c is the constant of integration

when t= O,x= 50

..

log 50 = k.O

+

c

or c= log50

(1)

=>

logx = kt

+

log50

x log- =kl 50 x

=

150, when 1= 1 150 .. log 50

=

k.I or k= log3 (2) then gives log (

5

x

O)

= flog3 3 we want to find x when I =

"2

X 3

50 = (3)2

3

X

=

50 (3)2 grams 1.17.7 Example

The rate of cooling of a body is proportional to the difference between the temperature of the body and the surrounding air. If the air temperature is 20°C and the body cools for 20 minutes from 140°C to 80°C, find when the temperature will be 35°C.

Solution

If 8 is the temperature of the body at time '1' then from Newton's law of cooling

-d8 de - a ( 8 - 2 0 ) ~ - =-k(8-20) dl dl

f

~

= - k rldl

+

c 8-20

J'

log(8 -20)

= -

kt

+

c Initially when t = 0, 8 = 140 log(8 - 20)

=

0

+

c, .... (I)

c = log( 120), (I) reduces to

log(140 - 20) = - kt + log 120 or +kt = log( 120) - 10g(S - 20) It is given that q = 80 when t = 20 minutes

k. 20= logI20-log(80-20) I (120) k= 20 log 60 I k= -log2 20 Substituting in (2) (2 1 0 IOg2)t= logI20-log(S-20)

It is required to find t when

0= 35°c t= logI20-log(35 - 20) I -log2 20 20 10g( \250 ) log2 2010g(8) 2010g23 60 10g2 . t= - - = - - = - - =60mmutes

log2 log2 log2

Exercise - 4(k)

.... (2)

I. In a certain reaction, the rate of conversion of a substance at time "t' is proportional to the quantity of the substance still untransformed at that instant. At the end of one hour 60 grams while at the end offour hours 21 grams remain. How many grams of the first substance was there initially?

( ADs: 85 grams approximately I

2. The rate of growth of a bacteria is proportional to the number present. If initially there were 100 bacteria and the amount doubles in '1' hour, how many bacteria will

1

be there after 2"2 hours.

3. Under certain conditions cane sugar in water is converted into dextrose at a rate which is proportional to the amount unconverted at any time. 1f75 grams was there at time t

=

0.0 and 8 grams are converted during the first 30 minutes find the amount

I

converted in 12 hour.

[ADS: 21.5 gms] 4. The rate of cooling of a body is proportional to the difference between the temperature of the body and the surrounding air. If the surrounding air is kept at 30°c and the body cools from 80°c to 60°c in 20 minutes. Find the temperature ofthe body after 40 minutes.

[ADS: 48°c

J

5. If the air is maintained at 30°c and the temperature of the body cools from 80°c to 60°c in 20 minutes. Find the temperature of the body after 40 minutes.

[ ADS: 48°c] 6. The rate at which a heated body cools in air is proportional to the difference between the temperature of the body and that of the surrounding air. A body originally at 80° cools down to 60°c in 20 minutes the temperature of the air being 40°c what will be the temperature of the body after 40 minutes from the original temperature.

[ADS: 50°c

J

7. The rate at which bacteria multiply is proportional to the instantaneous number present. If the original number doubles in 2 hours in how many hours will it triple.

[ ADS: 210g3

J

log2 8. Water at temperature 100°C cools in 10 minutes to 80°C in a room of temperature

25°C. Find the temperature of water after 20 minutes.

[ ADS: 65.5°c] 9. A cup of coffee at temperature 100°C is placed in a room whose temperature is 15°C and it cools to 60°C in 5 minutes, find its temperature after a further interval of 5 minutes.