Applied Maths i u v Solution

7/31/2019 Applied Maths i u v Solution

1/36

APPLIED MATHS I

Sol u t i on :CSVTU Ex a m i n a t i onPapers

Depa r tm en t of Ma th em atics

DIMAT

Ordinary Differential Equations & Applications


2/36

UNIT V (I semester)

Depar tm ent o f M athemat ics, DIM AT Page 2

APPLIED MATHS ITime Allowed : Three hours

Maximum Marks : 80

Minimum Pass Marks : 28

Note : Solve any two parts from each question. All questions carry equal marks.

UNIT V

Ordinary Differential Equations & Applications

SOLUTION (Nov-Dec-2005)

(a)Solve the differential equation: 2 22 ( 1) 0xydy x y dx .Ans: 2 22 ( 1) 0xydy x y dx

Which is of the form 0Mdx Ndy

Where 2 2( 1), 2M x y N xy

2 , 2M N

y yy x

M N

y x

, So differential equation is not exact.

Now,2 2 4 2

( )2 2

M N

y y yy xf x

N xy xy x

So, its I.F. =2

( ) 2ln

2

1dxf x dx xxe e ex

By multiplying I.F. to both side of equation (1)2

2 2

11 2 0

y ydx dy

x x x

----------------- (2)

Which is now exact differential equation.

Where2

2 2

11 , 2

y yM N

x x x

So, Solution is given by cNdyMdxtermsxcontainingNottconsy

tan

2

2 2

tan

11 0

y cons t Not containing x terms

ydx dy c

x x

2 1yx c

x x

2 2x y x cx (Ans)


3/36

UNIT V (I semester)


(b) Solve 2 22 cotdy dyy x ydx dx

.

Ans:

2

22 cotdy dy

y x ydx dx

2 22 cotp yp x y 2 2 2 2 2 22 cot cot cotp yp x y x y y x

2 2 2( cot ) cosp y x y ec x

( cot ) cosp y x y ecx

cot cos , cot cosp y x y ecx p y x y ecx

cot cos , cot cosdy dy

y x y ecx y x y ecxdx dx


x ecx dx x ecx dxy y


x ecx dx x ecx dxy y

ln ln(cos cot ) ln sin ln , ln ln(cos cot ) ln sin lny ecx x x c y ecx x x c

(cos cot )ln ln . , ln ln

sin (cos cot )sin

ecx x cy c y

x ecx x x

(cos cot ). ,

sin (cos cot )sin

ecx x cy c y

x ecx x x

2

(1 cos ). ,

sin (1 cos )sin

x cy c y

x x x

2

(1 cos ). 0, 0

sin (1 cos )sin

x cy c y

x x x

2

(1 cos ). 0

sin (1 cos )sin

x cy c y

x x x

(Ans)

(c)A pipe 20 cm in diameter contains steam of 0200 C. It is covered by a layer ofinsulations 6 cm thick and thermal conductivity 0.0003. It the temperature of

the outer surface is 030 C, find the heat loss per hour from two metre length

of the pipe.

Ans:


4/36

UNIT V (I semester)


Let us consider a cylinder of radius x cm and length 1 cm.

Surface area = 2A x sq cm.Let Q be quantity of heat flowing across the surface.

So, .2 .dT dT

Q KA K xdx dx

2

Q dxdT

K x

ln2

QT x c

K

At 10, 200x T

200 ln102

Qc

K ---------(1)

Again 30, 16T x

So, 30 ln162

Qc

K -----------(2)

(1) (2) we get 170 ln16 ln102

Q

K

170 ln1.62

Q

K

170 2 340 0.0003

ln1.6 ln1.6

KQ

Heat loss per hour through 2 m long pipe 3600 200 Q

340 0.00033600 200

ln1.6

490917.3 Cal. (Ans)

**************


5/36

UNIT V (I semester)


SOLUTION(Apr-May-2006)

(a)Find the integrating factor and solve: 3 2 21 (1 ) 03 2 4

y xy dx y xdy

.

Ans:

3 221

(1 ) 03 2 4

y xy dx y xdy


Where3 2

21, (1 )3 2 4

y xM y N y x

2 211 , (1 )4

M Ny y

y x

M N

y x


Now,

2 2 2

2 2

1 31 (1 ) (1 )34 4 ( )

1 1(1 ) (1 )

4 4

M N y y yy x

f xN x

y x y x

So, its I.F. =3

( ) 3ln 3dxf x dx xxe e e x

By multiplying I.F. to both side of equation (1)

3 3 53 2 41 (1 ) 0

3 2 4

x y xx y dx y x dy

-----------------(2)


Where3 3 5

3 2 41, (1 )3 2 4

x y xM x y N y x


tan

3 3 53

tan

03 2


x y xx y dx dy c

4 4 3 6

4 12 12

x y x y xc (Ans)

(b)Solve the differential equation 4 2y xp x p .Ans: 4 2y px x p

Differentiate with respect to x we get

3 2 44 2dy dp dp

p x x p pxdx dx dx

3 2 42 4 2

dp dpp x x p px

dx dx


6/36

UNIT V (I semester)


4 3 22 4 2 0

dp dppx x p p x

dx dx

32 2 2 0dp dp

px x p x pdx dx

32 2 1 0dp

x p pxdx

2 0dp

x pdx

2dp

dxp x

2dpdx

p x

ln 2ln lnp x c

2

cp

x

Now, putting the value of p in 4 2y px x p we get

2cy cx

2xy x c x (Ans)

(c) In a LR series circuit the current attains one-third of its final steady(maximum) value in 5 seconds. What is the time-constant

L

R

of the circuit?

Ans: In LR circuit

0 1R

tL

EI e

R

-------------- (1)

Final maximum value of current =0

E

R

Given that after 5 second current attains one-third of final maximum value.5

0 0113

R

LE E

I eR R

51

13

R

Le

5

2

3

R

Le

5 2

ln 0.405463

R

L

0.08109R

L

12.332L

R

So, time-constantL

R

of the circuit is 12.332.

**************


7/36

UNIT V (I semester)



(a)Solve the differential equation:( sin cos ) ( sin cos ) 0xy xy xy ydy xy xy xy xdx .

Ans: ( sin cos ) ( sin cos ) 0xy xy xy ydy xy xy xy xdx

Which is of the form 1 2( ). ( ). 0f xy ydx f xy xdy

Then its1 2

1. .

. ( ). . ( ).I F

x f xy y y f xy x

1 1

( sin cos ) ( sin cos ) 2 cosxy xy xy xy xy xy xy xy xy xy

Now multiplying I.F. both side of equation (1) we get

tan 1 tan 10

2 2 2 2

y xy x xydy dx

x y

----------- (2)


Where tan 1 tan 1,2 2 2 2

y xy x xyM Nx y

So equation (2) is an exact differential equation.

So, its solution is given by cNdyMdxtermsxcontainingNottconsy

tan

tan

tan 1 1

2 2 2y cons t Not containing x terms

y xydx dy c

x y

1 1 1ln sec ln ln

2 2 2xy x y c

2sec cx xy ey

(Ans)

(b)Solve 39( log ) (2 3log )y xp p p p .Ans: 39( log ) (2 3log )y xp p p p -------------- (1)

Differentiating with respect to x we get,

2 29 log log 3 (2 3log ) 3dy dp dp dp dp

p p x p x p p pdx dx dx dx dx

2 29 9 log 9 log 9 9 9 log 0dp dp dp

p p p x p x p p pdx dx dx

2 2

log log log 0

dp dp dp

p p p x p x p p pdx dx dx

2(1 log ) (1 log ) (1 log ) 0dp dp

p p x p p pdx dx

2(1 log ) ( ) 0dp

p p x pdx

2( ) 0

dpp x p

dx


8/36

UNIT V (I semester)


2( )dx

p x pdp

1dxx p

dp p Which is linear differential equation of first order.

Its

1

ln

. .

dppp

I F e e p

Hence solution is . .x p p pdp c

3

.3

px p c -------------(2)

Equation (1) and equation (2) combined is the solution.

(c)A stream pipe 20 cm in diameter contains steam of 0180 C. It is covered by alayer of insulations 16 cm thick. If the temperature of the outer surface is

030 C, find the temperature half way through the covering under steady state

condition.

Ans:

.2 .dT

Q K xdx

2

Q dxdT

K x

ln2

QT x c

K

At 180, 10T x

180 ln102

Qc

K ---------------(1)

Again at 40, 15T x

30 ln 262

Q

cK ---------------

(2)

By subtracting (2) from (1) we get

150 ln 2.62

Q

K --------------(3)

When 18x

ln182

QT c

K ---------(4)


9/36

UNIT V (I semester)


Subtracting (1) from (4) we get

180 ln 1.82

QT

K --------(5)

Dividing (5) by (3) we get

180 ln1.8

150 ln 2.6

T

0ln1.8 150 180 87.73ln2.6

T C (Ans)

**************

SOLUTION (May-June-2007)

(a)Find the integrating factor and solve the differential equations:(i) 2 2 3 2 3 2( 2 ) ( ) 0xy x y dx x y x y dy .Ans: 2 2 3 2 3 2( 2 ) ( ) 0xy x y dx x y x y dy ---------------- (1)

2 2( ) (2 ) 0xy ydx xdy x y ydx xdy

Which is of the form ' '( ) ( ' ' )a b a bx y mydx nxdy x y m ydx n xdy

Where 1, ' ' 2, 1, ' 2, ' 1a b a b m n m n

Its . . h kI F x y where1 1 ' 1 ' 1

,' '

a h b k a h b k

m n m n

2 2 3 3,

1 1 2 1

h k h k

, ( 3) 2( 3)h k h k

3h k

So. 3 33 3

1. . h kI F x y x y

x y

By multiplying I.F. to (1) we get

2 2

1 2 1 10dx dy

x y x xy y

which is exact differential equation.

Where2 2

1 2 1 1,M N

x y x xy y

So, Solution is given by cNdyMdx

termsxcontainingNottconsy

tan

2

tan

1 2 1


dx dy cx y x y

12 ln ln

2x y c

xy

(Ans)

(ii) 3 2 2 4( ) 2( ) 0xy y dx x y x y dy .


10/36

UNIT V (I semester)


Ans: 3 2 2 4( ) 2( ) 0xy y dx x y x y dy


Where 3 2 2 4( ), 2( )M xy y N x y x y

2 23 1, 4 2M N

xy xyy x

M N

y x


Now,2 2 2

3 2

4 2 3 1 1 1( )

( ) ( 1)

N M

xy xy xyx yf y

M xy y y xy y

So, its I.F. =

1( ) ln

dyf y dy yye e e y

By multiplying I.F. to both side of equation (1)4 2 2 3 5( ) 2( ) 0xy y dx x y xy y dy -----------------(2)


Where 4 2 2 3 5( ), 2( )M xy y N x y xy y


tan

4 2 5

tan

( ) 2y cons t Not containing x terms

xy y dx y dy c

2 4 62

2 3

x y yxy c (Ans)

(b)Solve the following differential equations:(i) 3xp a bp where dyp

dx

Ans: 3xp a bp 3

a bpx

p

Differentiating with respect to y we get

6 5

6

. 6dp dp

p b a bp pdx dy dy

dy p

61

dp dpbp a bp

dy dy

p p

6 7 1dp

a bpdy

6 7dy a bp dp 27

62

bpy ap c ---------- (2)

Solution is the elimination of p from (1) and (2).

(ii) 2 32y xp y p .


11/36

UNIT V (I semester)


Ans: 2 32y xp y p ---------- (1)2 3

2

y y px

p


3 2 2 2 3

2

1 2 31

2

dp dp

p yp y p y y pdy dydx

dy p

4 2 3 2 3

2

2 31 1

2

dp dp dpp yp y p y y p

dy dy dy

p p

4 2 32 2 2dp dp

p p yp y p ydy dy

4 2 32 2 0dp dp

p yp y p ydy dy

3 31 2 1 2 0dpp yp y ypdy

31 2 0dp

yp p ydy

0dp

p ydy

dp dy

p y

ln ln lnp y c

py c

Putting the value of p in (1) we get

3

2c c

y xy y

2 32y cx c (Ans).

(c)A coil having a resistance of 15 ohm and an inductance of 10 henry isconnected to a 90 volt supply. Determine the value of the current:

(i) After 0.67 sec(ii)After 2 sec. (e = 2.718)

Ans: Here R = 15 ohm, L = 12 Henry, E = 90 volt.

It is RL-circuit.Equation of RL-circuit is

diRi L E

dt

15 12 90di

idt

5(6 )

4

dii

dt


12/36

UNIT V (I semester)


5

6 4

didt

i

5ln( 6) ln

4i t c

5

46t

i ce

5

46t

i ce

-----------(1)At t = 0, i = 0, so equation (1) becomes 0 6 6c c

Putting the value of c in (1) we get5

46 6t

i e

-----------(2)(i) Current after 0.67 sec

50.67

0.837546 6 6 6 6 6 0.4328 6 0.5612 3.3672e e

(Ans)

(ii)Current after 2 sec5

20.546 6 6 6 6 6 0.6065 6 0.3965 2.379e e

(Ans).

**************


(a)Explain necessary and sufficient condition for a differential equation0Mdx Ndy to be exact.

Ans: Necessary condition for a differential equation 0Mdx Ndy to be exact is

M N

y x

.

(b)Solve the differential equation: 32 1/ 2( ) 0xxy e dx x ydy .Ans: Which is of the form 0Mdx Ndy

Where32 1/ 2,

xM xy e N x ydy

2 , 2M N

xy xyy x

M N

y x


Now,2 2

2 2 4 4( )

M N

xy xy xyy xf x

N x y x y x

So, its I.F. =4

( ) 4ln

4

1dxf x dx xxe e ex



13/36

UNIT V (I semester)


32 1/

3 4 20

xy e ydx dy

x x x

-----------------(2)


Where

32 1/

3 4 2,

xy e y

M Nx x x


tan

32 1/

3 4

tan

0x


y edx dy c

x x

32

1/

2

1

2 3

xye c

x (Ans)

(c)Solve 4 2y px x p .Ans: 4 2y px x p

Differentiate with respect to x we get

3 2 44 2

dy dp dpp x x p px

dx dx dx

3 2 42 4 2dp dp

p x x p pxdx dx

4 3 22 4 2 0

dp dppx x p p x

dx dx

32 2 2 0dp dp

px x p x pdx dx

32 2 1 0dp

x p px

dx

2 0dp

x pdx

2dpdx

p x

2dpdx

p x

ln 2ln lnp x c

2

cp

x


2cy c

x

2xy x c x (Ans)


14/36

UNIT V (I semester)


(d)The equation of electromotive force in terms of current i for an electricalcircuit having resistance R and condenser capacity c, in series is

iE Ri dt

C . Find the current i at any time t, when 0 sinE E wt .

Ans:i

E Ri dt

C

0 sini

Ri dt E wtC

By Differentiating with respect to t we get

0 cosdi i

R E w wtdt C

0 cosE w wtdi i

dt RC R -------------- (1)

Which is first order linear differential equation.

Its I.F. =1 t

dtRC RCe e

Solution of (1) is0 cos

t t

RC RCE w

Ie e wtdtR

0

2

2 2

1cos sin

1

tt RC

RCE w e

Ie wt w wt cR RC

wR C

2

0

2 2 2

1cos sin

1

t t

RC RCE wC R

Ie e wt w wt cR C w RC

0

2 2 2 cos sin1

t

RCE wC

I wt wRC wt ceR C w

(Ans)

**************

SOLUTION (May-June-2008)

(a)Write the general form of the first order differential equation of nth degree.Ans:

1

0 1( , ) ( , ) ...................... ( , ) 0

n n

n

dy dyf x y f x y f x y

dx dx

.


15/36

UNIT V (I semester)


(b)Solve: (1 ) (1 ) 0y xy dx x xy dy .Ans: (1 ) (1 ) 0y xy dx x xy dy ------------- (1)


Then its2 2

1 2

1 1 1. .

. ( ). . ( ). (1 ) (1 ) 2I F

x f xy y y f xy x xy xy xy xy x y


2 2

1 1 1 10

2 2 2 2dx dy

x y x xy y

----------- (2)


Where2 2

1 1 1 1,

2 2 2 2M N

x y x xy y

2 2 2 2

1 1,

2 2

M N

y x y x x y

M N

y x

. So equation (2) is an exact differential equation.


tan

2

tan

1 1 1


dx dy cx y x y

1 1 1ln ln

2 2 2x y c

xy

ln ln 1 2xy x xy y cxy (Ans)

(c)Solve:2

tan1

pp x

p

.

Ans:2

tan1

pp x

p

1

2tan

1

px p

p

1

2tan

1

px p

p

-------------- (1)


2 2

22 2

1 1 2

1 1

dx dp p p dp

dy p dy dyp

2 2

22

1 1 1

1

p p dp

p dyp

22

1 2

1

dp

p dyp

22

2

1

pdpdy

p

22

2

1

pdpdy

p

2

1

1y c

p


16/36

UNIT V (I semester)


21

1c y

p

11p

c y

------------------ (2)

Equation (1) and (2) are the solution.

(d)The equation of electromotive force in terms of current i for an electricalcircuit having resistance R and condenser capacity c, in series is

iE Ri dtc

. Find the current i at any time t, when 0 sinE E wt .

Ans:i

E Ri dtC

0 sini

Ri dt E wtC


0 cosdi i

R E w wtdt C

0cosE w wtdi i

dt RC R -------------- (1)

Which is first order linear differential equation.

Its I.F. =1 t

dtRC RCe e

Solution of (1) is

0 cost t

RC RCE w

Ie e wtdtR

0

2

2 2

1cos sin

1

tt RC

RCE w e

Ie wt w wt cR RC

wR C

20

2 2 2

1cos sin

1

t tRC RC

E wC RIe e wt w wt c

R C w RC

02 2 2

cos sin1

t

RCE wC

I wt wRC wt ceR C w

(Ans)

**************

SOLUTION (Dec-2008)

(a)Select the correct answer: integrating factor for 32 1/ 2( ) 0xxy e dx x ydy is:I. 4x .

II. 4x .III. 3x .IV. 3x .


17/36

UNIT V (I semester)


Ans: 4x

.

(b)Solve the differential equation: 2 2(3 2 ) ( 2 ) 0xy ay dx x axy dy .Ans: 2 2(3 2 ) ( 2 ) 0xy ay dx x axy dy ----------------(1)

Here 2 2(3 2 ), ( 2 )M xy ay N x axy

Then 3 4 , 2 2

M N

x ay x ayy x

x

N

y

M

So, given differential equation is non-exact.

Now,2

3 4 2 2 1( )

2 ( 2 )

M N

x ay x ay x ayy xf x

N x axy x x ay x

So, its I.F. =1

( ) lndxf x dx xxe e e x


2 2 3 2(3 2 ) ( 2 ) 0x y axy dx x ax y dy -----------------(2)


Where 2 2 3 2(3 2 ), ( 2 )M x y axy N x ax y


tan

2 2

tan

(3 2 ) 0y cons t Not containing x terms

x y axy dx dy c

3 2 2x y ax y c (Ans)

(c)Solve the differential equation: 42 2 . 0dy dyx x ydx dx

.

Ans:

4

22 . 0dy dyx x y

dx dx

4

2 2 .dy dy

y x xdx dx

2 4 2y x p xp

Now differentiating with respect to x we get

4 2 32 4 2 2dy dp dp

xp x p p xdx dx dx

4 2 32 4 2 2dp

p xp x p x pdx

4 2 30 2 4 2dp

xp x p x p

dx

3 32 2 1 2 1 0dp

x xp p xp

dx

32 1 2 0dp

xp x pdx

Now 2 0dp

x pdx

22

dp dp dxx p

dx p x

2

dp dx

p x


18/36

UNIT V (I semester)


1ln ln ln

2

cp x c p

x

So, solution is 2 4 2y x p xp 4

2

22

c cy x x

x x

42y c c x (Ans)

(d)The temperature of a body decreases at a rate k where 0 is the amount thebody is hotter than the surrounding air. The body is heated by a source

which makes the bodys temperature increase at a rate ' 'at where t is time

and a is constant. If this source is applied at t = 0, and the body is then at

the temperature of the surrounding air, show that:1 1 kta t e

k k k

.

Ans:

Given that The temperature of a body decreases at a rate k where 0 is

the amount the body is hotter than the surrounding air. The body is heated by a

source which makes the bodys temperature increase at a rate ' 'at where t istime and a is constant.

So,d

at kdt

dk at

dt

which is first order linear differential equation..

Its . .kdt ktI F e e

So, solution is . kt kt e a te dt

. ( )kt kt kt d

e a t e dt e dt t dt

dt

.kt kt

kt e ee a t dt k k

2.

kt kt kt e ee a t C

k k

-------------(1)

Given that at t = 0, 0 .So, equation (1) becomes

2

aC

k

So, putting the value of C in (1) we get

2 2.

kt kt kt e e ae a t

k k k

2 2

1.

kt kt kt e ee a t

k k k

2 2

1 ktt ea

k k k

1 1 kta t ek k k

(Proved).

**************


19/36

UNIT V (I semester)


Solution (Apr-May-2009)

(a)Fill up the blank:Integrating factor for

3

2 1/ 2( ) 0x

xy e dx x ydy is __________.

Ans: Integrating factor for3

2 1/ 2( ) 0xxy e dx x ydy is 4x . (Ans)

(b)Solve the differential equation: 3 2 2 3( ) ( ) 0x y x dy x y y dx .Ans: 3 2 2 3( ) ( ) 0x y x dy x y y dx ---------------- (1)

2 2( ) ( ) 0ydx xdy x y xdy ydx

Which is of the form ' '( ) ( ' ' )a b a bx y mydx nxdy x y m ydx n xdy

Where 0, ' ' 2, 1, 1, ' 1, ' 1a b a b m n m n

Its . . h kI F x y where1 1 ' 1 ' 1

,' '

a h b k a h b k

m n m n

1 1 3 3,

1 1 1 1

h k h k

1 1,3 3h k h k

1, 1h k

So. 1 11

. . h kI F x y x yxy

By multiplying I.F. to (1) we get

2 21 1 0x y dy xy dyy x

which is exact differential equation.

Where 2 21 1,N xy M x yx y


tan

2

tan

1 1


x y dx dy cy y

2 2

log log2

x yx y c (Ans)

(c)Solve the differential equation: 2tan ,1 p dyp x pp dx .Ans:

2tan

1

pp x

p

1

2tan

1

px p

p


20/36

UNIT V (I semester)


1

2tan

1

px p

p

-------------- (1)


2 2

22 2

1 1 2

1 1

dx dp p p dp

dy p dy dyp

2 2

22

1 1 1

1

p p dp

p dyp

22

1 2

1

dp

p dyp

22

2

1

pdpdy

p

22

2

1

pdpdy

p

2

1

1y c

p

21

1c y

p

11p

c y

------------------ (2)


(d)A pipe 20 cm in diameter contains steam at 0150 C and is protected with acovering 5 cm thick for which K = 0.0025. If the temperature of the outersurface of the covering is 040 C, find the temperature half way through the

covering under steady state condition.

Ans:

.2 .dT

Q K xdx

2

Q dxdT

K x

ln2

QT x c

K

At 150, 10T x

150 ln102

Qc

K ---------------(1)

Again at 40, 15T x

40 ln152

Qc

K ---------------(2)



21/36

UNIT V (I semester)


110 ln1.52

Q

K --------------(3)

When 12.5x

ln12.52

QT c

K ---------(4)

Subtracting (1) from (4) we get150 ln1.25

2

QT

K --------(5)


150 ln1.25

110 ln1.5

T

0ln1.25110 150 89.5

ln1.5T C (Ans)

**************


(a) Explain necessary condition for a differential equation 0 NdyMdx to beexact.

Ans: - The necessary condition for differential equation 0 NdyMdx to be exact is

x

N

y

M

(b) Solve: 3 2 2 4( ) 2( ) 0xy y dx x y x y dy .Ans: - 3 2 2 4( ) 2( ) 0xy y dx x y x y dy ----------------(1)

Here 3 2 2 4, 2 2 2M xy y N x y x y

Then 2 23 1, 4 2M N

xy xyy x

x

N

y

M

So, given differential equation is non-exact.

Now,2 2 2

3 2

4 2 3 1 1 1( )

( 1)

N M

xy xy xyx yf y

M xy y y xy y

So, its I.F. =

1( ) ln

dyf y dy yye e e y

By multiplying I.F. to both side of equation (1)4 2 2 3 5( ) 2( ) 0xy y dx x y xy y dy -----------------(2)


Where 4 2 2 3 5( ), 2( )M xy y N x y xy y


tan

4 2 5

tan

( ) 2y cons t Not containing x terms

xy y dx y dy c


22/36

UNIT V (I semester)


2 4 62

2 3

x y yxy c (Ans)

(c) Solve:

21tan

p

pxp .

Ans: - 2tan 1

pp x p

1

2 tan1

px pp

1

2tan

1

px p

p

-------------- (1)


2 2

22 2

1 1 2

1 1

dx dp p p dp

dy p dy dyp

2 2

22

1 1 1

1

p p dp

p dyp

22

1 2

1

dp

p dyp

22

2

1

pdpdy

p

22

2

1

pdpdy

p

2

1

1y c

p

21

1c y

p

11p

c y

------------------ (2)

Putting (2) in (1) we get the solution as

1

11

1tan 1

11 1

c yx

c y

c y

1

1

1tan

1

c y

c yc yx

c y

c y

1 1tan (1 )( )c y

x c y c yc y

(Ans).

(d) Show that the differential equation for the current i in an electrical circuitcontaining an inductance L and a resistance R in series and acted upon by an

electromotive force wtEsin satisfies the equation wtERidt

diL sin . Find

the values of the current at any time t, if initially there is no current in the

circuit.Ans: For LR circuit

Voltage drops across R = Ri .

Voltage drops across L =di

Ldt

.

Given electromotive force = wtEsin

So, by Kirchhoffs Law, sum of voltage drops across R and L = Electromotiveforce.


23/36

UNIT V (I semester)


So, wtERidt

diL sin (Proved).-------------- (1)

wtL

Ei

L

R

dt

disin , Which is first order linear differential equation.

So, itst

L

Rdt

L

R

eeFI

..

So, solution is dttweL

Eie

tL

Rt

L

R

.sin

CwtwwtL

R

wL

R

e

L

Eie

tL

R

tL

R

cossin

2

2

CwtLwwtRwLR

Eeie

tL

R

tL

R

cossin222

t

L

R

CewtLwwtRwLR

Ei

cossin222

Given that at 00 it .

2222220

wLR

ELwCC

wLR

ELw

So, solution t

L

R

ewLR

ELwwtLwwtR

wLR

Ei

222222cossin (Ans).

**************

Solution (May-June-2010)

(a) What is the I.F. of 0)( 2/12 3 ydyxdxexy x .Ans: - 0)(

2/12 3 ydyxdxexy x Which is of the form 0Mdx Ndy

Where32 1/ 2,xM xy e N x ydy 2 , 2

M Nxy xy

y x

M N

y x

, So differential equation is not exact. Now,

2 2

2 2 4 4( )

M N

xy xy xyy xf x

N x y x y x

So, its I.F. =

4( ) 4ln

4

1dxf x dx xxe e ex

.


24/36

UNIT V (I semester)


(b) Solve 0)(cos34 xyecyxdx

dyx .

Ans: - Let

x

t

dx

dt

xy

dx

dt

xdx

dy

dx

dyxy

dx

dtxyt

11

Then 0)(cos34 xyecyxdx

dyx

0)(cos0)(cos1 22334

tectxtx

dx

dtxtec

x

tx

x

t

dx

dt

xx

0)(cos3 tecdx

dtx

3sin

x

dxtdt On integrating

2

2cos2

22

1cos

xctC

xt

2)cos(2 xcxy

Cxxy 2)cos(2 (Ans).

(c) Solve ]1[ 2ppxy Ans: - ]1[ 2ppxy

2

2

1 ppx

y

22

2

2

12 pppx

y

x

y

x

xyyp

x

xy

x

y

x

yp 22

2

22

2

2

212

x

xy

dx

dyy

22

2

02)( 22 xydydxyx ---------------- (1) which is of the form

0 NdyMdx

Where xyNyxM 2),( 22

Now, yx

Ny

y

M2,2

Here

x

N

y

M

So, differential equation is not exact.

)(2

2

22xf

xxy

yy

N

x

N

y

M

So, 2ln22

..

xeeFI xdx

x

By multiplying I.F. both side to equation (1) we get


25/36

UNIT V (I semester)


0212

2

dy

x

ydx

x

y, which is exact differential equation.

So, its solution is cNdyMdxtermsxcontainingNottconsy

tan

Cdydxx

y

012

2

Cx

yx

2

(Ans).

(d) A constant electromotive force E volts is applied to a circuit containing aconstant resistance R ohms in series and a constant inductance L henries. If

the initial current is zero, show that the current builds half its theoretical

maximum in (Llog2)/R seconds.

Ans: -

For LR circuit

Voltage drops across RiR .

Voltage drops acrossdt

diLL .

Given electromotive force = E


So, ERidt

diL

L

Ei

L

R

dt

di , Which is first order linear differential equation.

So, its

tL

Rdt

L

R

eeFI

..

So, solution is dteL

Eie

tL

Rt

L

R

CeR

EC

L

R

e

L

Eie

tL

RtL

R

tL

R

CeR

Ei

tL

R


R

ECC

R

E

0

So, solution

t

L

Rt

L

R

eR

Ee

R

E

R

Ei 1 --------------- (1)

Maximum theoretical value ofR

Ei


26/36

UNIT V (I semester)


To become its half i.eR

Ei

2

Equation (1) becomes

t

L

R

eR

E

R

E1

2

22

1

2

11

t

L

Rt

L

Rt

L

R

eee

2log tL

R

R

Lt

2log. (Proved)

**************


(a) Find the integrating factor of the differential equation0)()2( 2222 dyyxxyxdxyxxyy .

(b) Solve the differential equation: 32 1/ 2( ) 0xxy e dx x ydy .Ans: Which is of the form 0Mdx Ndy

Where3

2 1/ 2,xM xy e N x ydy

2 , 2M N

xy xyy x

M Ny x , So differential equation is not exact.

Now,2 2

2 2 4 4( )

M N

xy xy xyy xf x

N x y x y x

So, its I.F. =

4( ) 4ln

4

1dxf x dx xxe e ex


32 1/

3 4 20

xy e ydx dy

x x x

-----------------(2)Which is now exact differential equation.

Where

32 1/

3 4 2,

xy e yM N

x x x


tan


27/36

UNIT V (I semester)


32 1/

3 4

tan

0x


y edx dy c

x x

3

21/

2

1

2 3

xye c

x

(Ans)

(c) Solve 4 2y px x p .Ans:

4 2y px x p Differentiate with respect to x we get

3 2 44 2

dy dp dpp x x p px

dx dx dx

3 2 42 4 2

dp dpp x x p px

dx dx

4 3 22 4 2 0

dp dppx x p p x

dx dx

32 2 2 0dp dppx x p x pdx dx

32 2 1 0dp

x p pxdx

2 0dp

x pdx

2dpdx

p x

2dp

dx

p x

ln 2ln lnp x c

2

cp

x


2cy c

x

2xy x c x (Ans)

(d) The equation of electromotive force in terms of current i for an electrical circuithaving resistance R and condenser capacity c, in series is

iE Ri dt

c . Find

the current i at any time t, when 0 sinE E wt .

Ans

:

iE Ri dt

C


28/36

UNIT V (I semester)


0 sini

Ri dt E wtC


0 cosdi i

R E w wtdt C

0cosE w wtdi i

dt RC R --------------(1)Which is first order linear differential equation.

Its I.F. =

1 tdt

RC RCe e

Solution of (1) is

0 cost t

RC RCE w

Ie e wtdtR

0

2

2 2

1cos sin

1

tt RC

RCE w e

Ie wt w wt cR RC

wR C

2

0

2 2 2

1cos sin

1

t t

RC RCE wC R

Ie e wt w wt cR C w RC

02 2 2 cos sin1

t

RCE wC

I wt wRC wt ceR C w

(Ans)

SOLUTION (Apr-May-2011)

a) Write the necessary and sufficient condition for a differential equation0Mdx Ndy to be exact.

Ans: Necessary condition for a differential equation 0Mdx Ndy to be exact is

M N

y x

.

b) Solve the differential equation : (1 ) (1 ) 0y xy dx x xy dy .Ans: (1 ) (1 ) 0y xy dx x xy dy ------------- (1)


Then its2 2

1 2

1 1 1. .. ( ). . ( ). (1 ) (1 ) 2

I Fx f xy y y f xy x xy xy xy xy x y


2 2

1 1 1 10

2 2 2 2dx dy

x y x xy y

----------- (2)



29/36

UNIT V (I semester)


Where2 2

1 1 1 1,

2 2 2 2M N

x y x xy y

2 2 2 2

1 1,

2 2

M N

y x y x x y

M N

y x



tan

2

tan

1 1 1


dx dy cx y x y

1 1 1

ln ln2 2 2

x y cxy


c) Solve:2

tan1

pp x

p

.

Ans

2tan

1

pp x

p

1

2tan

1

px p

p

1

2

tan1

px p

p

--------------(1)Differentiating with respect to y we get

2 2

22 2

1 1 2

1 1

dx dp p p dp

dy p dy dyp

2 2

22

1 1 1

1

p p dp

p dyp

22

1 2

1

dp

p dyp

22

2

1

pdpdy

p

22

2

1

pdpdy

p

2

1

1y c

p

21

1c y

p

11p

c y

------------------(2)


30/36

UNIT V (I semester)



d) A pipe 20 cm in diameter contains steam at 0180 C and is covered with a material6 cm thick for which K = 0.0025. If the temperature of the outer surface of the

covering is 030 C, find the temperature half way through the covering under

steady state condition.

Ans:

.2 .dT

Q K xdx

2

Q dxdT

K x

ln2

QT x c

K

At

180, 10T x

180 ln10

2

Qc

K

---------------(1)

Again at

30, 16T x

30 ln162

Qc

K ---------------(2)


150 ln1.62

Q

K --------------(3)

ln132

QT c

K ---------(4)

Subtracting (1) from (4) we get180 ln1.3

2

QT

K --------(5)


180 ln1.3

150 ln1.6

T

ln1.3

ln1.150 80

61T (Ans)


31/36

UNIT V (I semester)



1. Find the integrating facto of the differential equation0)()2( 2222 dyyxxyxdxyxxyy .Ans:- Rewriting the equation as( + ) + ( 2) = 0 and comparing with

( + ) + ( + ) = 0 We have a=b=1, = = 2 = = 1 I.F=.Where

=

,

=

1 + + 11

=1 + + 1

1,1 + + 1

2=

2 + + 11 = 0 , + 2 + 9 = 0

Solving these ,we get h=k=-3

I.F= ,it becomes

1

+

2

+

1

1

= 0

which is an exact differential equation.

Therefore the solution is + ( ) = 1 1 + 2 = 2 1 =

2. Solve the differential equation: 32 1/ 2( ) 0xxy e dx x ydy .Here M= , = = ( = which is a function of x only .I.F = = =

Multiplying throughout the given differential equation by


32/36

UNIT V (I semester)


1 = 0 For the differential equation (1) ,we have

M=

N= Now

= = .Hence the differential equation (1) is exact .

Now = =

+

(3

)

= + = = 2Since no new term is obtained by integrating N w.r.t y , hence the required solution is

=c3. Solve 4 2y px x p

Ans:-The given equation is

= + .(1)Which is clearly solvable for y .Hence differentiating (1) w.r.t to x we get

= + 4 + 22( 1 2) + . ( 1 2) = 0 (2 + ) ( 1 2) = 0


33/36

UNIT V (I semester)


Neglecting the second factor which does not contain the derivatives of p, we have

(2 + ) = 0

+ 2

= 0

Integrating, we get + 2

=

= = =

Substituting this value of p in eq(1) ,we get = . + . = . +

4. The equation of electromotive force in terms of current i for an electrical circuithaving resistance R and condenser capacity c, in series is

iE Ri dt

c . Find the

current i at any time t, when 0 sinE E wt .

Ans:-Given equation can be written as + = .Differentiating both the sides w.r.t t ,we have + = . + = (1 )Which is a Leibnitzs linear equation .

I.F= = Therefore the solution of the equation (1) is

.

=

.

=

. 1 +

1 + =

( ) +


34/36

UNIT V (I semester)


= i= ( ) +

which gives the current at any time t.

SOLUTION (Apr-May-2012)

(a)Write down exact differential.Ans A differential equation is called an exact differential ,if it can be derived

from its primitive directly by differentiation without any subsequent operation ofelimination or reduction = + dy= (,) + (,)

(b)Solve the differential equation : (1 ) (1 ) 0y xy dx x xy dy .Ans:

(1 ) (1 ) 0y xy dx x xy dy -------------(1)


Then its2 2

1 2

1 1 1. .

. ( ). . ( ). (1 ) (1 ) 2I F

x f xy y y f xy x xy xy xy xy x y


2 2

1 1 1 10

2 2 2 2dx dy

x y x xy y

-----------(2)


Where2 2

1 1 1 1,

2 2 2 2M N

x y x xy y

2 2 2 2

1 1,

2 2

M N

y x y x x y

M N

y x



tan

2

tan

1 1 1


dx dy cx y x y

1 1 1

ln ln2 2 2

x y cxy



35/36

UNIT V (I semester)


(c) Solve the differential equation: 2 32y px y p .Ans

Solving the given equation for x, we obtain2 2

2 2

y y px

p

Differentiating (1) w.r.t y ,we obtain2 2

2

2

2

2 2

1

2 2

1 1 1

2 2

(1 2 ) (1 2 )

dx y dp dpyp y p

dy p x dy dy

dpyp y py

p p p dy

dpyp p y yp

dy

31 2 0dp

p y ypdy

Neglecting the second factor, we obtain

0dp

p ydy

Separating the variables ,we get

0dp dy

y y

Integration gives,

0dp dy

y y

log log logp y c

py c

cp

y

Substituting this value of p in the equation (1) ,we have2 2

2 3

2 2

2

y cx

c

y cx c

s

(d) Solve the differential equation 0 sindiL Ri E wtdt

where L, R , and0E are constants

and discuss the case when t increases indefinitely.Ans: For LR circuit

Voltage drops across R = Ri .


36/36

UNIT V (I semester)

Voltage drops across L =di

Ldt

.

Given electromotive force = wtEsin


So, wtERidt

di

L sin (Proved).-------------- (1)

wtL

Ei

L

R

dt

disin , Which is first order linear differential equation.

So, itst

L

Rdt

L

R

eeFI

..

So, solution isdttwe

L

Eie

tL

Rt

L

R

.sin

CwtwwtL

R

wL

R

e

L

Eie

tL

R

tL

R

cossin

2

2

CwtLwwtRwLR

Eeie

tL

R

tL

R

cossin222

t

L

R

CewtLwwtRwLR

Ei

cossin222


2222220

wLR

ELwCC

wLR

ELw

So, solution tLR

ewLR

ELwwtLwwtRwLR

Ei

222222 cossin (Ans).

Documents

Applied Maths i u v Solution