Differential Equations. 1.Solve; = (1 + x) + y(1 + x) = (1 + x)(1 + y) Integrating Which is the required solution

Differential Equations

PREPARED BY:R.RAJENDRAN. M.A., M. Sc., M. Ed.,K.C.SANKARALINGA NADAR HR. SEC. SCHOOL,CHENNAI-21

1.Solve;

= (1 + x) + y(1 + x)

= (1 + x)(1 + y)

xyyxdxdy 1

xyyxdxdy 1

dxxy

dy)1(

1

Integrating

dxxy

dy)1(

1

cx

xy 2

)1log(2

Which is the required solution.

2.Solve;

Put x + y = z

22)( adxdy

yx 22)( a

dxdy

yx

dxdz

dxdy 1

The given eqn becomes

22 1 adxdz

z

1dxdz

dxdy

2

2

1za

dxdz

2

22

2

2

1z

zaza

dxdz

dxdzaz

z 22

2

dxdzaz

aaz

22

222

dxdzaz

a

22

2

1

Integrating

dxdzaz

a22

2

1

dxdzaz

adz 222 1

cxa

z

aaz

12 tan

1

cxa

yxayx

1tan

ca

yxay

1tan

3. Solve 3ex tan y dx + (1 + ex) sec 2y dy = 0

Given equation is

3ex tan y dx + (1 + ex) sec2y dy = 0

3ex tan y dx = – (1 + ex) sec2y dy

dyyy

dxe

ex

x

tansec

13 2

Integrating

dyyy

dxe

ex

x

tansec

13 2

3 log(1+ex) = – log(tan y) + log c

log(1+ex)3 + log(tan y) = log c

log(1+ex)3(tan y) = log c

(1+ex)3(tan y) = c

Which is the required solution

4. Solve (x2 – y)dx + (y2 – x)dy = 0, if it passes through the origin.

The given equation is

(x2 – y)dx + (y2 – x)dy = 0

x2 dx – ydx + y2 dy – xdy = 0

x2 dx + y2 dy = ydx + xdy

Integrating,

xdyydxdyydxx 22

)(22 xyddyydxx

cxyyx 33

33

Since it passes through the origin,

0 + 0 = 0 + c

c = 0

The required solution is

xyyx 33

33

xyyx 333

5. Solve: xxydxdy

cos2cot

xxydxdy

cos2cot

It is a linear diff eqn with

P = cot x, Q = 2cosx

dxP

eIF

dxx

e cot

dxx

x

e

sin

cos

xe sinlog

xsin

The solution is

cdxQy I.F I.F

cdxxxy sincos2sin x

cdxxy 2sinsin x

cx

y 2

cossin x

cx

y 2

cossin x

6. Solve: 22 12)1( xxxydxdy

x

It is a linear diff eqn with

dxP

eIF

dx

x

x

e

1

22

The solution is

cdxQy I.F I.F

cdxxx

xx

y 1

1

111

222

22 12)1( xxxydxdy

x

2

2

2 11

12

xxx

yxx

dxdy

221

,1

2

x

xQ

xx

P

dxx

x

e

1

22

)1log( 2xe 211x

cdxxx

xy

)1(11

2/322

cdxxx

xy

)1(11

2/322

cdx

xx

xy

)1(2

21

11

2/322

cx

xy

123

)1(2

11

1

12

32

2

cx

xy

21

)1(2

11

1

2

12

2

cxx

y

22

1

11

1

7. Solve: (1 + y2)dx = (tan–1 y – x )dy

The given equation

(1+y2)dx = (tan–1y – x)dy can be written as

2

1

1tan

yxy

dydx

22

1

11tan

yx

yy

dydx

2

1

2 1tan

1 yy

yx

dydx

It is a linear diff eqn in x with

2

1

2 1tan

,1

1y

yQ

yP

The solution is

cdyQ ) (I.F (I.F)x

cdyy

yx yy e

1

tane

1-1- tan2

1tan

dyP

eIF

dy

ye

1

12

ye1tan

)1.........( e1

tane

1-1- tan2

1tan

cdyy

yx yy

Put u = tan – 1 y

211ydy

du

dyy

du 211

Eqn (1) becomes

cdux uy uee -1tan

cduuex uuy ee -1tan

ceuex uuy -1tane

ceyex yyy 11-1 tantan1tan tane

cyex yy

)1(tane 1tantan 1-1

It is the required soln.

8. Solve (D2 – 13D + 12)y = e –2x

The given equation is(D2 – 13D + 12)y = e –2x

The characteristic equation is p2 – 13p + 12 = 0(p – 12)(p – 1) = 0 p = 1, 12The complementary function isCF = A ex + B e 12x

Particular integral is

xeDD

PI 22 1213

1

xe 22 12)2(13)2(

1

xe 2

122641

xe 2

421

The general solution is

y = CF + PIxxx eBeAey 212

421

9. Solve (D2 + 6D + 8)y = e –2x

The given equation is(D2 + 6D + 8)y = e –2x

The characteristic equation is p2 + 6p + 8 = 0(p + 2)(p + 4) = 0 p = – 2, – 4The complementary function isCF = A e–2x + B e– 4x


xeDD

PI 22 86

1

xeDD

2

)2)(4(1

xeD

2

)2)(42(1

xx ex

xe 22

221


y = CF + PIxxx e

xBeAey 242

2

10. Solve (D2 – 4D + 13)y = e –3x


(D2 – 4D + 13)y = e –3x

The characteristic equation is

p2 – 4p + 13 = 0

xeDD

PI 32 134

1

xe 32 13)3(4)3(

1


y = CF + PI

xx exBxAey 32

341

)3sin3cos(

12

1314)4()4( 2

p

252164 p

2364

ii

322

64

CF = e2x(Acos3x + B sin3x)

34131291 3

3x

x ee

11. Solve (D2 – 4)y = sin2x

The given equation is(D2 – 4)y = sin2xThe characteristic equation is p2 – 4 = 0(p – 2)(p + 2) = 0 p = –2, 2The complementary function isCF = A e –2x + B e 2x


xD

PI 2sin4

12

x2sin)42(

12

82sin

x


y = CF + PI

xBeAey xx 2sin4122

12. Solve (D2 – 2D – 3)y = sinx cosx

The given equation is(D2 – 2D – 3)y = sinx cosxThe characteristic equation is p2 – 2p – 3 = 0(p – 3)(p + 1) = 0 p = –1, 3The complementary function isCF = A e –x + B e 3x


32cossin

2

DDxx

PI

322sin

21

2

DDx

)322(22sin

2

Dx

)72(22sin

D

x

)72)(72(22sin)72(

DD

xD

)494(22sin)72(

2 D

xD

xDD

2sin)494(2

)72(2

)49)2(4(22sin7)2(sin2

2 xxD

)4916(22sin72cos4

xx

)2sin72cos4(130

1xx


y = CF + PI

)2sin72cos4(130

13 xxBeAey xx

13. Solve (D2 + 14D + 49)y = e–7x + 4

The given equation is(D2 + 14D + 49)y = e–7x + 4The characteristic equation is p2 + 14p + 49 = 0(p + 7)(p + 7) = 0 p = – 7, – 7 CF = (Ax + B)e–7x


49144

2

7

DDe

PIx

2

0

2

7

)7(4

)7(

De

De xx

2

072

)70(4

2

xx eex

494

2

72

xex


y = CF + PI

494

2)(

727

xx ex

eBAxy

14. Solve (D2 – 13D + 12)y = e–2x + 5ex

The given equation is(D2 – 13D + 12)y = e–2x + 5ex

The characteristic equation is p2 – 13p + 12 = 0(p – 12)(p – 1) = 0 p = 12, 1 CF = A e12x + B ex


12135

2

2

DDee

PIxx

)1)(12(5

12132

2

DDe

DDe xx

)1)(121(

5

12264

2

D

ee xx

11

5

42

2 xx xee


y = CF + PI

11

5

42

212

xxxx xee

BeAey

11

5

42

2

xx xee

15. Solve (D2 + 4D + 13)y = cos3x


(D2 + 4D + 13)y = cos 3x


p2 + 4p + 13 = 0x

DDPI 3cos

1341

2

CF = e–2x(Acos3x + Bsin3x)Particular integral is

12131444 2

p

252164 p

2364 p

ii

p 322

64

xD

3cos1343

12

xD

3cos44

1

xDD

D3cos

)1)(1(4)1(

xDD

3cos)1(4

)1(2

)13(43cos)3(cos

2 xxD

403cos3sin3

xx

)3cos3sin3(401

xx


y = CF + PI

)3cos3sin3(401

)3sin3cos(2 xxxBxAey x

16. Solve: (D2 – 1)y = cos 2x – 2sin 2x


(D2 – 1)y = cos 2x – 2sin 2x


p2 – 1 = 0

(p – 1)(p + 1) = 0

p = 1, – 1

CF = A ex + B e–x

The particular integral is

)2sin22(cos1

12 xx

DPI

12sin

21

2cos22

D

xD

x

122sin

212

2cos22

xx

52sin

252cos

xx

xx 2sin52

2cos51


y = CF + PI

)2cos2sin2(51

xxBeAey xx

17. Solve (D2 – 4D + 1)y = x2


(D2 – 4D + 1)y = x2


p2 – 4p + 1 = 0

22 14

1x

DDPI


12

114)4()4( 2

p

24164

2124

322

324

xx BeAeCF )32()32(

= {1 + (D2 – 4D)}–1 x2 = {1 – (D2 – 4D) + (D2 – 4D)2 – } x2

= {1 – D2 + 4D + 16D2} x2

= x2 – 2 + 4 (2x) + 16(2)= x2 + 30 + 8xThe general solution is y = CF + PI

3082)32()32( xxBeAey xx

18. Solve (D2 + 3D – 4)y = x2


(D2 + 3D – 4)y = x2


p2 + 3p – 4 = 0

(p + 4)(p – 1) = 0

p = 1, – 4

CF = Aex + Be–4x

The particular integral is

22 43

1x

DDPI

2

2

43

14

1x

DDPI

2

12

43

14

1x

DD

2

222

..4

34

31

41

xDDDD

222

..16

94

31

41

xDDD

22

1612

1613

14

1x

DD

)2(

1612

)2(1613

41 2 xx

xx

23

813

41 2

813

23

41 2 xx


y = CF + PI

813

23

41 24 xxBeAey xx

19. In a chemical 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, 60grams remain and at the end of 4 hours 21 grams. How many grams of the first substance was there initially?

Let x grams of the substance remain after t hours.

xdtdx xk

dtdx

Where k is constant of proportionality

dtkx

dx Integrating

dtkx

dx

log x = – k t + c1

x = c e– k t …….(1)

When t = 1, x = 60

60 = c e–k ……….(2)

1cktex

x = c e– k t …….(1)

When t = 4, x = 21

21 = c e–4k ……….(3)

When t = 0, x = c

From (2) e – k = 60/c

Sub in (3) we get

3

446060

21cc

c

43

2160c

taking log

3log c = 4log60 – log21

= 4(1.7782) – 1.3222

= 7.1128 – 1.3222

= 5.7906

log c = 5.7906/3 = 1.9302

c = antilog 1.9302

= 85.15 85

Hence initially there was 85gms of the substance.

20. For a postmortem report, a doctor requires to know approximately the time of death of the deceased. He records the first temperature at 10.00am to be 93.4F. After 2 hours he finds the temperature to be 91.4F. If the room temperature is 72F, estimate the time of death. (Assume that the normal temperature of a human body to be 98.6F). Given

Let T be the temperature of the body at any time t

The temperature difference is T – 72

By Newton’s law of cooling,

00945.04.216.26

log and 0426.04.214.19

log ee

72 TdtdT

)72( TkdtdT


dtkT

dT

72

Integrating

dtkT

dT

72

log (T – 72) = – k t + c1

T = 72 + c e– k t …….(1)

Initially, t = 0, T = 93.4F

93.4F = 72 + c e0

c = 93.4 – 72 = 21.4F

T = 72 + 21.4 e–k t ……(2)

172 ckteT

When, t = 120, T = 91.4F

91.4F = 72 + 21.4 e– 120k

21.4e– 120k = 91.4 – 72

21.4e– 120k = 19.4

4.214.19120 ke

4.214.19

log120 k

4.214.19

log120

1k

41055.3)0426.0(120

1 k

When T = 98.6F, t = t1

98.6F = 72 + 21.4 e–k t1

21.4 e–k t1 = 98.6 – 72 = 26.6

4.216.26

1 kte

4.216.26

log1 ekt

4.21

6.26log

11 ekt

)0945.0(1055.3

141

t

)0945.0(55.3

104

1

t

min26655.3

9451

t

4hours 26min before the first recorded temperature

The approximate time of death = 10 – 4:26 = 5:34am

21. The number of bacteria in a yeast culture grows at a rate which is proportional to the number present. If the population of a colony of yeast bacteria triples in 1hour. Show that the number of bacteria at the end of five hours will be 35 times of the population at initial time.

Let x be the number of bacteria present in the yeast culture at any time t

xdtdx xk

dtdx


dtkx

dx Integrating

dtkx

dx

log x = k t + c1

x = c e k t …….(1)

When t = 0, x = x0

x0 = c e0

c = x0

1cktex

Sub in eqn (1) we get

x = x0 e k t………(2)

Given x = 3x0, when t = 1

3x0 = x0 ek

e k = 3……..(3)

When t = 5, let x = x1

x1 = x0 e 5k

= x0 (ek)5 = x0 35

= 35 times of the population at initial time

22. The sum of Rs. 1000 is compounded continuously, the nominal rate of interest being four percent per annum. In how many years will the amount be twice the original principal? (loge2 = 0.6931)

Let x be the principal

xdtdx

of %4

xdtdx

04.0

dtx

dx 04.0

Integrating

dtx

dx 04.0

log x = 0.04 t + c1

x = c e 0.04 t …….(1)

When t = 0, x = 1000

1000 = c e0

c = 1000

x = 1000e 0.04t……..(2)

104.0 ctex

x = 1000 e 0.04 t …….(2)

When x = 2000, t = ?

2000 = 1000 e0.04t

e0.04t = 2

0.04t = loge2

0.04t = 0.6931

1732.1704.0

6931.0 t

In 17 years the amount will be twice the original principal

23. Radium disappears at a rate proportional to the amount present. If 5% of the original amount disappears in 50years, how much will remain at the end of 100years. (Take A0 as the initial amount.)

Let A be the amount of radium disappears at any time t.

AdtdA Ak

dtdA


dtkA

dA Integrating

dtkA

dA

log A = –k t + c1

1ckteA ktceA

Initially, t = 0 and A = A0

A0 = c e0 c = A0

kteAA 0

When t = 50years, A = 95% of A0 = 0.95A0

0.95A0 = A0 e –50k

e – 50k = 0.95

When t = 100 years, A = A1

A1 = A0 e – 100k = A0 (e – 50k)2

= A0 (0.95)2 = 0.9025A0

Amount of radium at the end of 100years = 0.9025A0

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Differential Equations. 1.Solve; = (1 + x) + y(1 + x) = (1 + x)(1 + y) Integrating Which is the required solution