Ordinary Differential Equations - University of Kashmiregov.uok.edu.in/eLearning/tutorials/MM14301CR_16... · 2017-06-05 · Lecture notes on Ordinary Differentia l Equations -(Unit

Lecture notes on Ordinary Differential Equations-(Unit-I) Khanday M. A.

M.A/M.Sc. Mathematics 3rd

Semester University of Kashmir, Srinagar 1

Ordinary Differential Equations

Introduction:

Ordinary differential equations-ODE‟s serve as a tool in mathematical models for many

exciting “real-world” problems. It has applications not only in science, but it also plays a

key role in many areas in engineering, social sciences, biomedical engineering and other

areas such as economics, political sciences etc. We believe that many problems of future

technologies will also be described by means of differential equations. Physical problems

have motivated the development of much of mathematics and this is especially true of

differential equations. Due to differential equations mathematics and physical problems

come closer and closer. An understanding of the required mathematics will aid in the

solution of physical problems, and an understanding of the physical model often makes

the mathematics easier. Additional applications of differential equations are (heat,

radioactive decay, orthogonal trajectories, chemical reactions, economics, chemical

diffusion, and vibrating strings).The increasing availability of technology (including

graphing and programmable calculators, computer algebra systems, and powerful

personal computers) has caused many to question the existing syllabuses in university

courses in differential equations. However, we believe that the importance of applications

will continue to motivate the study of differential equations.

In the first two chapters of this course, we deal with the introduction of ordinary

differential equations. Singular solutions, p- discriminant and c- discriminant of the

differential equations, Initial value problems of the first order ordinary differential

equations (ODE), Integration in Series, Simultaneous and Total differential equations.

In many branches of science and engineering we come across equations, which contain

an independent variable and the derivative of the dependent variable with respect to the

independent variable is called a differential equation.

Examples:-

(i) xdx

dylog

(ii) xdx

dy

dx

yd34

3

2

2

(iii) ∂z/∂x + ∂z/∂y = k, where k is a constant.




Differential equations involving only one independent variable are called ordinary

differential equations.

Differential equations, which involve partial differential coefficients with respect to more

than one independent variables, are called Partial differential equations.

ORDER AND DEGREE OF A DIFFERENTIAL EQUATION

The order of a differential equation is the order of the highest derivative involved in the

differential Equation.

The degree of the differential Equation is the power of the highest order derivative

involved in the equation, when the equation has been made rational and integral as far as

the derivatives are concerned.

Examples:-In the given below differential equation

2

5

2

2

2

1

dx

dyK

dx

yd

The order is two and the degree is also two.

Linear and Non- Linear Differential Equations

A differential equation is said to be a linear differential equation if it can be expressed in

the form.

QyPdx

dyP

dx

ydP

dx

ydP nnn

n

n

n

11

1

10 ....

where QandPPPPP nn ,,...,,, 1210 are either constants or functions of independent

variable x.

Thus, if a differential equation when expressed in the form of a polynomial involves the

derivatives and dependent variable in the first power and there are no product of these,

and also the coefficient of the various terms are either constants or the functions of the

dependent variable, then it is said to be a linear differential equation, otherwise it is a

non- linear differential equation.

It follows from the above definition that a differential equation will be non- linear if,

(i) Its degree is more than one.

(ii) Any of the differential exponent has exponent more than one.

(iii) Exponent of the dependent variable is more than one.




(iv) Products containing dependent variable and its differential coefficients are

present.

Solution of a Differential Equation

Any relation between the dependent and independent variables which, when substituted

in the differential equation reduce it an identity is called a solution of the differential

equation. A solution of the differential equation does not contain the derivative of the

dependent variable with respect to the independent variable or variables.

Complete Primitive or the General Solution

The solution of an ordinary differential equation of nth order containing n arbitrary

constants is called the complete primitive or the General solution.

Particular Integral:-Any solution which obtained from the complete primitive by giving

particular values to the arbitrary constants is called a Particular Integral.

Exercises:Solve the following differential Equations.

(i) dx

dyPwherexyPyPxP 02 .

(ii) dx

dDwherezDyDandzDy ,0)1()1(0 .

Solutions:(i) We have dx

dyPwherexyPyPxP 02

We can write (P + x) (P+y) = 0

Implies either (P + x) = 0 or (P+y) = 0

i.e., dy/dx = -x or dy/dx = - y

or dy = - x dx or dy/y = - dx

Integrating both sides we get

y = - x2/2 + C1 or log y = - x + C2

Thus (y + x2/2 + C1 )(log y + x –C2) is the required solution of the differential equation.

(ii) Dy – z = 0 -------- (i)

(D-1) y – (D+ 1) z= 0 ---- (ii)

Operating equation (i) by (D +1), we get on subtraction,

(D2 + D –D + 1) y = 0

or (D2 + 1) y = 0

Now the auxiliary equation is m2 + 1 = 0




Implies m = ± ί

Therefore y = A Cos x + B Sin x is the complementary function.

Also from (i) Z = Dy

or Z = D( A Cos x + B Sin x)

= -A Sin x + B Cos x,

This is the required complete solution of the given differential Equation.

Home Assignments

Solve the following Differential Equations

1. ( D + 1) y = Z + ex and ( D + 1) Z = y + e

x

2. P2 + 2 xp – 3x

2 = 0, where p = dy/dx

3. x + yp2

= P ( 1 + xy)

4. p2 – x

5 = 0.

5. y = p Sin p + Cos p.

6. p2 – 2p Cosh x + 1 = 0.

7. p3 – p ( x

2 + xy + y

2) = 0

8. t dx/dt + y = 0 and t dy/dx + x = 0.

Hint: Take t = ep,

so that dp/dt = 1/t and dx/dt = (dx/dp).(dp/dt) = 1/t. dx/dp

Double Point:A point on a curve is called a double point if two branches of the curve

passes through it, It is called a triple point if three branches of the curve passes through it.

In general a point is called a multiple point of the kth

order if k branches of the curve pass

through the point.

Classification of Double Points: Since two branches of the curve passes through a double

point, there must be two tangents to the curve at the point, one to each of the two

branches.

If the two tangents are real and different, the double point is called a Node.

If the two tangents are real and co-incident, it is called a Cusp.

And if the tangents are imaginary so that there are no real points on the curve of the

double point.

Envelope:The envelope of the family is the locus of ultimate intersection of consecutive

curves of the family. i.e., If f ( x, y, α ) = 0 is a family of curves corresponding to




different values of α. Then the ultimate intersection of curves f (x, y,α ) = 0 and f ( x, y, α

+ ∂α ) = 0 is called the envelope of the family.

e.g. If x Cos α + y Sin α = p is the equation of the straight line, then

x Cox (α + ∂α ) + y Sin (α + ∂α ) = p

Solving these equations

x = p Cos (α + ½ ∂α ) / Cos (½ ∂α) , y = pSin (α + ½ ∂α )/ Cos ( ½ ∂α )

When ∂α → 0, we have the ultimate point of intersection

x = p Cos α , y = p Sin α

Eliminating α between these equations, we have

x2 + y

2 = p

2as the equation of the circle.

Singular Solution:-If an infinite system of curves which all touch to a fixed curve, which

we call their envelope represent a Complete solution of certain differential equation of

the first order, then the envelope represents a solution of the differential equation. For

every point of the envelope x, y and p have the same value for the envelope and the

curves of the family that touches it there. Such a solution is called a Singular solution.

Discriminant:-The discriminant of the quadratic equation ax2+bx+c = 0, a(≠0) is b

2 – 4ac.

If Φ (x, y, c) = 0 is the solution of the differential equation f(x, y, p) = 0

The p-discriminant is obtained by eliminating p between f(x, y, p) = 0 and ∂f/∂p = 0. and

the c- discriminant is obtained by eliminating c between Φ (x, y, c) = 0 and ∂ Φ /∂c = 0.

Extraneous Loci:-Let Φ (x, y, c) = 0 be the primitive of the differential equation

f(x,y, p) = 0.

If ψ (x, y) = 0 is its singular solution, then ψ (x, y) = 0 is a factor in both between p-

discriminant and c- discriminant.

Each discriminant have other factors which correspond to other loci associated with the

primitive. In general these loci do not satisfy the differential equation. Therefore these are

sometimes called extraneous loci.

Tac- Locus:-p- discriminant gives equal value of p, but these values may belong to two

curves of the system that are not consecutive, i.e. these curves have different c‟s. Locus

of such points is called the tac- locus.

For example, consider a family of circles, all of equal radii, whose centers lie on a

straight line. the two circles , which are not consecutive, touch at p i.e, have same values




of p but the tangent at p is not the direction of the line of centers which is the locus of

points of contact of the circles. The line of centers is the tac- locus.

Tac locus

Nodal Locus:-c- discriminant gives equal values of c but these values may belong to the

nodes( i.e. double points with distinct tangents) which are also ultimate points of

intersection of the consecutive curves, Locus of such points is called Node- Locus.

Cusp- Locus:-c-discriminant gives equal values of c but these values may belong to the

cusps( i.e. double points with coincident tangents) which are also ultimate points of

intersection of the consecutive curves. Their locus is called the Cusp Locus.

Note:- The p- discriminant equated to zero may include the envelope E, as a factor once,

Cusp locus C, once and the tac- locus T, twice and the c- discriminant equated to zero

may include envelope E, once the cusp locus C thrice and the Node locus N twice.

i.e., p- discriminant ~ ECT2 = 0. and c- discriminant ~ EC

3N

2 =0

The singular solution is obtained as a common factor from the c and p discriminate and it

must satisfy the differential equations.

Examine the equation for Singular Solution 4x p2 = (3x - a)

2

Solution:- From the given equation, we have p = dy/dx

So that we can write p = ± (3x -a)/ 2√x

or we can write

dy = ± (3x -a)/ 2√x dx

Integrating both sides, we get

xaxdy 2/)3( dx

or y = c ± ( x 3/2

- a√x)

or (y –c)2 = x( x

3-a)

2 …(I)

which is the complete solution of the given equation.

Now, we have p- discriminant,

x(3x- a)2 = 0.

Also, (I) can be written as, c2 -2 yc + y

2 – x(x

3-a)

2 = 0




Therefore c-discriminant can be calculated as

4y2 – 4.1. (y

2 –x(x

3-a)

2 = 0,

or 4y2 -4y

2 + 4x(x

3-a)

2 =0, or 4x (x

3-a)

2 = 0

or x (x3-a)

2 = 0

Now comparing p- discriminant and c- discriminant with ECT2 and EC

3N

2, it clearly

follows that x= 0 is common and thus is the Singular Solution of the given equation.

The factor x3-a =0 which occurs in the c-discriminant and does not occur in p-

discriminant is the Node locus.

Also, 3x –a =0 is the Tac- locus.

Examine for Singular Solution xp2 – 2 yp + 4x = 0

Solution.We have the given equation is quadratic in p, therefore

P = dx

dy =

x

xyy

2

1642 22 =

x

xyy 22 4= 4

2

x

y

x

y

This is of homogeneous nature.

Put y = vx , so that dy/dx = v+ x dv/dx

Thus we have,

v + x dv/dx = v ± 42 v

or x

dx

v

dv

42

Integrating both sides, we have

log cxvv

log122

2

or 4

4

2

2

vv = cx

or 2

22

4

4

2 x

xy

x

y = cx

or y + 22 4xy = 22cx

or 2222 4)2( xycxy

or 0122 cyxc is the required primitive of the given differential equation.




Now p discriminant is given by

4y2 – 16x

2 = 0,

or y2 = 4x

2

Also, we have c-discriminant is given by

y2 – 4x

2 =0

or y2 = 4x

2

Thus the Singular Solution is y2 = 4x

2

Examine the equation y2 -2pxy + p

2(x

2 -1) = m

2, for singular solution.

Solution: The given equation can be written as

(x2 -1) p

2 – 2pxy + (y

2 –m

2) =0 …(I)

or (px –y)2 = p

2 + m

2

or 22 mppxy

which is of the Claurits form

Hence the general solution is

22 mccxy

or 222mccxy

or 021 2222 mycxyxc ---- (II)

Hence from (I) and (II) both p- discriminant and c- discriminant are

0))(1(44 22222 myxyx

or 2222 mxmy

This is the required singular solution.

Home Assignments

Examine for Singular Solution ( If any)

(i) 2ppxy

(ii) 0)( 22 axxp

(iii) 094 2 xp

(iv) 1)2(4 2 xp




Exercise: Reduce the equation 0)2( 222 yyxpypx to Claurites form by putting

u = y and v = xy and find its complete primitive and also its singular solution.

Solution: We have yu and xyv

So that ydx

dyx

dx

dvand

dx

dy

dx

du

Now p

yxp

dx

dy

ydx

dyx

dx

dudx

dv

du

dv

or ypxdu

dvp

xp

yp

1

Where du

dvp 1

Putting the value of p in the given equation, we have

0)2()(

2

1

2

2

1

22

yyx

xp

y

xp

yx

or 0)())(2( 2

11

2 xpxpyxx

01

2

1 ypxyp

or 2

11 pypxy

or 2

11 pupv ,

which is of Claurits form of differential equation..

Hence the general solution is

2ccuv

or 02 xyycc

Therefore c-discriminant is

042 xyy

or 0)4( xyy

Also from the given equation p-discriminant is

04)2( 2222 yxyxy

0)4( 22 xyyy or 0)4(. 2 xyyy




Now 0)4( xyy occurs both in c and p discriminate and y = 0 and )4( xy = 0 satisfy

the given differential equation.

Hence y = 0 and 0)4( xy is the Singular Solution.

Exercise: Solve the differential equation for the Singular Solution.

0263214222 xxpxxx …(i)

Solution: Here p-discriminant is 02632122 xxxxx

Also from (i) 214

26322

2

xxx

xxp

or

xxx

xx

xxx

xxp

232

263

212

263

23

22

or dy dxxxx

xx

232

263

23

2

,


21 xxxy + c

or 212

xxxcy

or 212 22 xxxycyc

Now c- discriminant is

02144 22 xxxyy

or 021 xxx

Thus 021 xxx , is a Singular Solution.

Also comparing the two discriminate to ECT2≈ 0and EC

3T

2 ≈ 0 , we have seen that the

tac- locus is 3x2 -6x + 2 = 0 or

3

11

i.e.,

3

11...

3

11 and is a tac- locus of Imaginary points of contact.

Examine for Singular Solution of the following differential equation

022 yxpyp

Solution: Put y2 = v, Differentiating both sides w.r.t. x we get




dx

dvyp2 ( say )

Implies that y

p2

,

Substitute the value of p in the given equation we have

02

22

2

y

yx

yy

or 0

4

2

y

y

x

y

or 044 22 xy Implies 04

22

xy

or4

2 xv ,

This is of Claurits form of differential equation

Therefore, the complete primitive is given by

22 ccxv ( by replacing θ by 2c )

i.e., 22 2 ccxy is the complete solution of the given differential equation.

Now p-discriminant of the given differential equation is

022 yxpyp is 022 yx

and c- discriminant is given

022 yx

Hence 022 yx is the required Singular Solution of the given differential equation.

Examine for Singular Solution (if any) by using a suitable substitution.

0263 2 yxypxp

Solution: We have 0263 2 yxypxp ------(i)

Put vyx 3 and ux 2 ,

Differentiating both sides, we get

dvdydx 3 and duxdx 2

Thereforedu

dv

xdx

dydx

2

3

i.e., xdx

dy

xdx

dx

2

3

2 or

x

p

x 2

3

2

1




or xp 231 or 3

21 xp

Now from (i)

023

216

3

213

2

yx

xy

xx

023

216

9

4413

22

yx

xy

xxx

022123

441 22

yxxy

xxx

031244 233 xxyxxx

or 031 22 yxx

Implies that using the substitution,

We can write, 01 2 vu

or

1

uv

This is of Claurites form of differential equation

Therefore the complete primitive of the differential equation is

ccuv

1

i.e., c

cxyx1

3 2 or yxcxc 3122

or 01322 cyxcx

Now c- discriminant is

043 22 xyx

or 0496 222 xyxyx ; 023 22 xxyy

Also p- discriminant is

023 22 xxyy

Hence 023 22 xxyy is the required Singular Solution of the given differential

equation.




Solve by using the suitable substitution or otherwise for Singular Solution of the

differential equation

084 23 yxypp

Solution: Put 2vy ,

Differentiating both sides w.r.t. x we get

dx

dvv

dx

dy2 or p = 2 v θ where θ =

dx

dv

Therefore from the given differential equation, we have

08242 23 yvxyv

or on simplification, we have

3 xv

It is of Claurits form of differential equation.

Therefore, the complete primitive is

3kkxv or 222 kxky or 2cxcy

Now p- discriminant can be calculated by taking partial derivative of the function

fyxypp 23 84 (say)

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

,0

p

f or 043 2 xyp or xyp

3

42

Now 0843

4 2

yxypxyp

or 083

8 2 ypxy or ypx 3

or 222 9yxp or 22 9

3

4yxyx

or yx 274 3

Also in a similar way, the c- discriminant of the solution is

yx 274 3

Therefore, the required singular solution of the given differential equation is

0274 3 yx




Home Assignments

Examine for singular Solution (if any) of the following differential equation.

1. 22 32 xxpp

2. 042 232 yxpxp

3. 22 yppxyx

4. 02 22222 ybxyppxa

5. 022 axxp

6. 122 pyxp

7. 022 yxpyp

8. apxpy 2

9. 222 bpapxy

10. 022 yxpp

11. ypxp log

12. pCospxSinySinpxCosy

Find the differential equation if the complete primitive is given. Also find the

Singular Solution (if any).

012 22 xcyc

Solution: Differentiating w.r.t. x we get

022 xcp or p

xc

Therefore, from the given differential equation

012 2

2

2

xyp

x

p

x

or 021 222 xxypxp

Now p- discriminant is

014 222 xyx

Also, c- discriminant is




014 22 xy

Thus, the singular solution is a circle

122 yx

and 0x is a tac-locus.

Home Assignment

Find the differential equations if the complete primitive is given by the following.

1. 0cos2 2222 ccxyx

2. 012 xycyxc

3. 012 222 ccxyyx

4. 02 34222 xxcycxy

Exercise: Reduce 01222 xypyxxyp to Clairaut’s form by the

substitution X= x2 ; Y= y

2. Hence,show that the equation represents a family of

conics touching the four sides of a square.

Solution:Let X = x2 ; Y= y

2

So that we can write on differentiation.

2 x dx = dX and 2 y dy = dY

or

px

y

dX

dY or

y

xp where

dX

dY

Therefore 012

2

2

xy

y

xYX

y

xxy

or 112

2

22

YX

y

xy

or 112

YX

Y

XY

or YXYX 2

or 11 YX

or

1XY




This is of Clarautis form of differential equation.

Therefore its complete primitive is

C

CCXY

1 or CXCCCY 11

012 YCYXXC

Thus c- discriminant is

0412

XYYX

or 042221 22 XYXYYXYX

or 02221 22 XYYXYX

or we can write

01222 222244 yxyxyx

This represents the family of conics touching the four sides of a square.

Exercise: Find the Singular Solution of the differential equation.

02 22222 ybxyppxa

Solution:- We can write the differential equation as

02 222222 ybxyppxpa

or 2222bpayxp

or 222 bpapxy

It is of Calirauits form of differential equation

Thus the complete primitive is given by

222 bcacxy

Now p- discriminant is

044 222222 ybxayx

or 222222 baxbya

orwe can write this as

12

2

2

2

b

y

a

x

Similarly the c-discriminant is




12

2

2

2

b

y

a

x

Hence the singular solution of the given differential equation is 12

2

2

2

b

y

a

x

This represents the equation of the ellipse.

Miscellaneous Methods for Equations of the Second and Higher Orders:

Here we reduce equations of the second order to the equations of the first order. We

shall show that the order can always be so reduced if the equation

(i) does not contain y explicitly

(ii) does not contain x explicitly

(iii) is homogeneous

Case I. If y is absent; i.e., If y does not occur explicitly in an equation of the second

order, we obtain an equation containing only dx

dp , p and x, so of the first order.

Case II. If x absent, In this case we write p for dx

dy and for 2y we write

dy

dpp

Since 2ydx

dp

dy

dp

dx

dy

dy

dpp

Examples for Solution

1. 12

2 xCosy

Solution:- Here y is absent, we can simply write py 1 and dx

dpy 2

Therefore we have

12 xCosdx

dp or xSec

dx

dp 2

or xdxSecdp 2


axp tan Where „a‟ is a constant.

Implies dxaxdy )(tan

Integrating again, we have

kaxSecxbaxCosxy loglog




This is the required solution of the given differential equation.

2. 2

12 1 yyy

Solution:- Here x is absent, thus we can write 1ydx

dyp

and dy

dppy 2

So that 21 pdy

dppy

or 12

p

dy

dppy or

y

dydp

p

p

21

2

2

1

or ayp log1log 2 where „a‟ is a constant.

or 222 1 yap

or we can write this as in terms of dy as

221 yadx

dy

Integrating we get by putting ay = Cos θ, So that ady = -Sinθdθ

dxya

dy

221 = bx

a

Sin

a

1

Sin

d 1

or θ = -a ( x + b)

Implies that

Cos-1

ay = -a (x + b) or ay = Cos ( ax + c)

Solve 2

2

2

131 2yyyy by using above exercise

Solution: Here both x and y are absent

py 1 , dy

dppy 2 and

dy

dp

dx

dp

dy

dp

dx

dpy

3

or dy

dp

dx

dy

dy

dp

dxdy

pdpy

2

3 =

2

2

22

22

dy

dp

dy

pdp

dy

dpP

dydy

pd

dx

dyp

Now from the given equation




2

2

2

131 2yyyy , we have

2

2

2

2

22 2

dy

dppp

dy

dpp

dy

pdpp

or

22

2

2

21

dy

dp

dy

dp

dy

pdp

or

2

2

2

1

dy

dp

dy

pdp

This gives

2

12 1 ppp

which is similar to above example.

Therefore we have from above the solution in terms of p is

cayCospa

or cayCosa

p 1

or

dxcayCos

dya

Implies dxdycaySeca

Þ a logSec ay+ c( ) + tan ay+ c( )

a= x + d

or dcaycaySecx tanlog

is the required solution of the given differential equation.

Solve the following

(i) xyxy 1223

(ii) x

nn eyy 12

(iii) Integrate and interpret geometrically

ky

y

2

2

32

11

Solution: (i) Multiplying each term by 2x , we have

3

2

2

3

3 12xyxyx




Let zex , xdx

dzxz

1log

dz

dy

xdx

dz

dz

dy

dx

dy 1

dz

dy

xy

11 ,

Let dz

dD

Therefore Dyxy 1

Also, dx

dz

dz

yd

xdz

dy

xdxdz

yd

xdz

dy

xdz

dy

xdx

dy

2

2

2

2

22

11111

or DyyDx

y 2

22

1 or yDDyx 12

2

Similarly, yDDDyx 213

3

Therefore from the given differential equation we have

zeyDDyDDD 312121

or zeyDDD 312121

or zeDD 32121

Now the auxiliary equation is 012mm

i.e., m = 0, 1, 1

Therefore the Complementary function of the equation is

xxCBAeCzBA z log

Also the particular integral is

333

2

3

2 2.3

112

1

12xeee

DD

zzz

Therefore the complete solution of the equation is

3log xxxCBAy

(ii) x

nn eyy 12

Solution:- We know as from above that

Dyy 1 , yDy 2

2 , . . . , yDy n

n




Therefore from the given equation, we have

xnn eyDD 12

Now the Auxiliary equation is

022 11 mmmm nnn

i.e.,

,0,...0,0,0,2

1 timesn

mm

Therefore the Complementary function is

2220222 ...... nxxnxe hxdxcxbaeehxdxcxbae

Also the particular Integral is

xx

nn

x

nneee

DD

11121

1

2

1

Hence the complete solution of the integral is

y = 222 ... nxx hxdxcxbaee

(iii)

ky

y

2

2

32

11; where k is constant.

Solution:- We know that py 1 , dx

dpy 2

Therefore from the given equation, we have

1+ p2( )3

2

dpdx

= kÞ kdp

dx= 1+ p2( )

3/2

or we can write

dxp

dpk

2/321

, Integrating both sides we get

dxp

dpk

2/321 ; Put p = tanθ ; So that dp = Sec

2θdθ

Therefore dxdSec

Seck

3

2

p 21 p

Implies that k

axSin

, or

21 p

p

k

ax

1




Implies that

2

2

2

2

1 k

ax

p

p

or

22

2

axk

axp

or we have

22 axk

saxp

Let 222 taxk

So that tdtax

Then we have

dx

axk

axdy

22

2

i.e., tbt

tdty

or baxky 22

or 22 axkby

222axkby

or we can write this as

222kbyax

which is the required equation of circle with centre (a,b) and radius k given by

k

y

y

2

2

32

11

Home Assignments

1. xyxy 412

2. 2

12 yyy

Homogeneous Equations

If x and y are regarded as of dimension 1.

y1 is of dimension zero

y2 is of dimension -1

y3 is of dimension -2




we define a homogeneous equation, the equation in which all the terms are of same

degree, e.g., 012

2 yxyyx is called Cauchy‟s Homogeneous equation.

Solve the following homogeneous differential equations.

(i). 1

2

12 3yyxyxyy .

(ii) 012

2 yxyyx .

(iii) .0512

2 yxyyx

(iv) 2

1

22

2

22 yxyyyx .

(v) 1

2

1

22

2

2 242 xyyyxyyyx [Hint, using substitution y= z2]

Solution:- (i) We know that ,1

1dt

dy

xdx

dt

dt

dyy

2

2

22

2

2

2

2

12

11

11

,11

dt

yd

xdt

dy

x

dt

yd

dx

dt

xdt

dy

x

dt

dy

dx

d

xdt

dy

xdx

dyy

Substitution in 1

2

12 3yyxyxyy and multiplying by x, we get

dt

dyy

dt

dy

dt

dy

dt

ydy 3

2

2

2

i.e., dt

dyy

dt

dy

dt

ydy 4

2

2

2

This is an equation where t is absent

By putting qdt

dy , we can obtain easily

,2 2 byyq

Giving byct 2log4

1

Hence 442 aeeby ct

Which is the required solution of the equation.

(ii) 012

2 yxyyx




Solution:- Put tex so that t = log x

So that we have ,1

1dt

dy

xdx

dt

dt

dyy

and

2

2

22

2

2

2

2

12

11

11

,11

dt

yd

xdt

dy

x

dt

yd

dx

dt

xdt

dy

x

dt

dy

dx

d

xdt

dy

xdx

dyy

or we can write

dt

dy

dt

ydyx

2

2

2

2

Thus the question becomes,

02

2

y

dt

dy

dt

dy

dt

yd

or 022

2

ydt

dy

dt

yd

Let Ddx

d and

dt

dy

Therefore, we have ydx

dyxxDy

Because, dx

dyx

dx

dy

dt

dx

dt

dyy

i.e., xDy y

Similarly, yyDx 122

Therefore from the given equation we have

01 yyy

or 012

y

Now the auxiliary equation is

012m or 1,1m

Also the complementary function is given by




xxbaebta t log

Since the particular integral is zero, therefore the complete solution of the given

differential equation is

xbxaxxxbay loglog

(iii) .0512

2 yxyyx ( Try yourself )

(iv) 2

1

22

2

22 yxyyyx

Solution:- Let vxy , So that we have dx

dvxvy 1

And

2

2

2

2

12 22dx

vdx

dx

dv

dx

vdxvy

Therefore from the given equation we have

2

222

2

22 22

dx

dvxvxxv

dx

vdx

dx

dvvxx

or 2

2

2

2222

dx

dvx

dx

vdvx

dx

dvvx

Put tex so that t= log x and xdx

dt 1

dt

dv

xdx

dt

dt

dv

dx

dv.

1.

Also

dt

dv

dt

vd

xdt

dv

xdx

dt

dt

vd

xdx

vd2

2

222

2

2

2 111

Therefore we have

2

2

2

22

dx

dv

dt

dv

dt

vdv

dt

dvv

2

2

2

2

dx

dv

dx

vdv

Now for ;pdx

dv

,

2

2

dt

dv

dv

dp

dt

vd

dt

dp

i.e., ,dv

dpp

dt

dp




Thus, we can write

22 pdv

dpvp or p

dv

dpv 2

or we can write this as v

dv

p

dp2


cvp loglog2

or ,vbdt

dv where cb

Therefore, we have bdtv

dv

or 24 abtv

or 2log4 BxAxy is the required solution of the given differential equation.

(v) 1

2

1

22

2

2 242 xyyyxyyyx

Solution:- Let y =z2, so that ,2z

dz

dy and

dx

dzz

dx

dz

dz

dy

dx

dyy 21

and 2

22

2 22dx

zdz

dx

dyy

Substitute these things in the given differential equation

1

2

1

22

2

2 242 xyyyxyyyx gives

dx

dzzxz

dx

dzzxz

dx

zdz

dx

dzzx 2224222 2

2

24

2

22

22

implies that

dx

dzxz

dx

zdx

2

22

,022 zxDzzDx where dx

dD

,0122 zxDDx …(1)

As before, put tex

Implies that dt

dz

xdx

dz 1 or

dt

dzxDz




Also, we have

dt

dz

dt

zd

xdx

zd2

2

22

2 1

For dt

d and D

dx

d

we have yxdx

dt

dt

dyDy

dx

dyy

11

or yxDy and yyDx 122

Thus, we have from the equation (1),

0122 z or 012

z

Now the Auxiliary equation is 012m , m = 1, 1

Also the complementary function is

xxbaebtaz t log

Hence 22 log xxbazy is the required solution of the given equation.

Home Assignments

1. xyxyyx log212

2

2. 12 1

2

2

3

3

4 xyyxyxyx

3. 2

12

2 )1(202 xyxyyx

4. 2

22

1

113

xyxDDx

5.xeyxyyx 24 12

2

6. Sinxxy 2

2

7. yySecy tan2

2

An equation of the type yfy 2

This form of differential equations occurs frequently in Dynamics especially in problems

on motion under a force directed to a fixed point and of magnitude depending solely on

the distance from that fixed point.

Here we multiply each side by 12y as it acts as an integrating factor for yfy 2 .

Thus we get




yfyyy 121 22

Integrating, we get

dyyfdxdx

dyyfy 22

2

1

Which represents the equation of energy

Applying the method to the equation of the SHM, we get from xpdt

xd 2

2

2

,

dt

dxxp

dt

xd

dt

dx 2

2

2

2

Integrating with respect to t, we get

,tan 22222

2

xaptConsxpdt

dx

Hence,

,11

22 xapdx

dt

On integration we get

tConsa

xSin

pt tan

1 1

or ptaSinx

Solve the following

1. yyy 3

2 , given that ,01 y when y = 1

2. ,2

2

yey given that y = 0 and ,11 y when x =0

3. ,tan2

2 ySecy given that y = 0 and ,11 y when x = 0

4. 2

2

2

2

x

ga

dt

xd , given that x = h and 0

dt

dx when t = 0.

5. 222

2

uh

pu

d

ud

, in the two cases(i) P = μu

2; (ii) P = μu

3 given that 0

d

du,

when ,1

cu where μ, h and c are constants.

Solution:- 1. Multiplying both sides by an integrating factor 12y , we get,

1

3

121 22 yyyyyy




or we can write this as 1

32

1 2 yyyydx

d


dyyyy 32

1 2

= cyy

cyy

2

424

2242

or cyy

y 24

2

12

Now using the initial conditions, we have, for ,01 y when y = 1

c 12

10 or c = 1/2

Therefore we get

2

1

2

12

2

1

2

22242

42

1

yyyy

yy

Therefore we have

dxy

dyyy

1

2

2

12

2

1


dxy

dy

12

2

Implies that,. cxyCoth 12

or

2

1 cxyCoth

or

2

cxCothy

2. ,2

2

yey given that y = 0 and ,11 y when x =0

Solution:- Multiplying both sides by the integrating factor 2 y1 , we get

yeyyy 2

121 22

Now integrating both sides, we get

cedyey yy 222

1 22

i.e., cey y 22

1

Now using the initial conditions, we get




y = 0 and ,11 y when x =0

ce 01 Or c = 0

Therefore yey 22

1 or yey 1

or dxdye y ,

Integrating both sides we get,

axe y

Now again using final condition, we have from y = 0 when x = 0,

Therefore a = -1

Thus, we have xe y 1

or 1log xy

Hence, 1log xy

3. ,tan2

2 ySecy given that y = 0 and ,11 y when x = 0

Solution:- Multiplying both sides by the integrating factor 2 y1 , we get

,tan22 1

2

21 yySecyy


cydyyySecy 2

1

22

1 tantan2

i.e., cyy 22

1 tan

Using initial conditions, we get,

c 0tan1 2 or c = 1

or ySecyy 222

1 tan1

or Secyy 1

thus dxSecy

dy , on integration we get,

or dxdyCosy

axSiny or axSiny 1

Again using final conditions y = 0 , x = 0 , we get

aSin 00 1 or 0a




Therefore, we get

xSiny 1 is the required solution of the given differential equation.

4. 2

2

2

2

x

ga

dt

xd , given that x = h and 0

dt

dx when t = 0.

Solution:- Multiplying both sides by the integrating factor 2dt

dx, we get

dt

dx

x

ga

dt

xd

dt

dx2

2

2

2

22

Integrating both sides, we get,

bx

gadxx

gadt

dx

12

12 2

2

2

2

bx

ga

dt

dx

22

2

Using initial conditions x = h and 0dt

dx, we get,

02 2

bh

ga or

h

gab

22

Therefore,

hx

xhga

h

ga

x

ga

dt

dx 2222

222

Implies that 22,2 gawherekxh

xhk

hx

xhag

dt

dx

.

i.e., ,

xh

xhk

dt

dx

Integrating both sides we get,

,

dtkdxxh

hx

,ctkxh

dxxh

Put h – x = z2

So that – dx = 2zdz




Therefore ,22

ctkdzzz

zhh

or we have ,2 2 ctkdzzhh

Put

Sinhdz

Coshz

Therefore, ,2 22

ctkdSinhCoshh

,2 2 ctkdSinhh

,2

212 ctkd

Coshh

,22

3

ctkdCosdh

,2

22

3

ctkSin

h

,2

3

ctkCosSinh

,12

12

3

ctkh

z

h

z

h

zCosh

,212

3

ctkzhh

z

h

zCosh

,11

2

3

ctkxxhhh

xhCosh

Using final conditions, we get

,01

012

3

ch

Cosh

or 0c

Therefore, we have,

txxhhh

xhCos

k

hh

11

or

21

2

1xhx

h

xhhCos

g

h

at




5. 222

2

uh

pu

d

ud

, in the two cases

(i) P = μu2 ; (ii) P = μu

3 , given that 0

d

du, when ,

1

cu

Solution:- we hav 222

2

uh

pu

d

ud

Case I :- let P = μu2

Therefore 222

2

2

2

huh

uu

d

ud

The above equation can be written as

2

2

huuD

Where

d

dD

or 2

2 1h

uD

Now the auxiliary equation of the differential equation is

012 m or m

Therefore the complementary function of this is given by

bSinaCosu

Also the particular integral of the equation is given by

2

0

2

2

0

2

120

22

...1

11

1.

hu

hD

uh

DuhD

IP

Therefore the Complete primitive is given by

2hbSinaCosu

Case II let P = μu3,

Therefore, 222

2

uh

pu

d

ud

gives us

222

3

2

2

h

u

uh

uu

d

ud




or uhd

ud

1

22

2

It is of the form ufu 2

Therefore, multiplying both sides by an integrating factor, d

du2 we get

uhd

du

d

ud

d

du

122

22

2

On integration we have

au

h

uduhd

du

212

122

2

2

2

2

i.e., auhd

du

2

2

2

1

Now using the initial conditions, 0

d

du, when ,

1

cu

Therefore we can have,

ach

22

110

or

22

11

cha

Therefore

2

22

22

2

2

2

11

111

uch

chu

hd

du

or

c

uc

hd

du 22

2

11

Implies that on integration,

d

huc

duc

2221

1

Put cu = Cost

Implies c du = -Sin t dt

Therefore




bh

dttCosc

Sc

)(1

1

int22

or bh

dt 21

or bh

t 2

1

bh

cuCos

2

1 1

Using boundary conditions, we have

bCos 11

or 0b

Therefore, 2

1 1h

cuCos

i.e.,

21

hCosCu

or

1

2hCoshCu

according as 2hor

Factorization of the Operator:

Method of Operational factors, If It is possible to factorize the operation, the method of

Operational factors may be applied.

Let the linear equation of second order

RQydx

dyp

dx

yd

2

2

be written as f (D) y =R

or RyDfDf 21 .

Which is such that Df 2 operates on y and Df1 on the result and we get the same

result as if .Df operates upon y.

Examples for solution:-

1. 0211 12 yyxyx

Solution:- We can write the given differential equation as

0211 2 yDyxyDx




or 0211 2 yDxDx

or, we have 02211 2 yDxDx

02211 2 yDDxDx

01211 yDDDx

0121 yDDx …(i)

Let zyD 1 …(ii)

Therefore we have

021 zDx or 02 zDxD

Implies that zdx

dz

dx

dzx 2

or 1

2

x

dx

z

dz

Now integrating both sides we get,

axz log1log2log

or 21loglog xaz

or 21 xaz

Substitute in (ii)

211 xayD

21 xay

dx

dy

This is a linear differential equation,

Now integrating factor is xdx

ee

Therefore, axeyedx

dye xxx 2

1

or xexayedx

d 22 1


dxexaye xx

2

1

dxexexa xx 1212




dxeexexa xxx 1212

beexexa xxx 21212

Implies that baeexaexaye xxxx 21212

xbeaxaxay 21212

or xbexay 12


2. 01 12 yyxxy

Solution:- We have 0112 yDxxD

012 yDxDxD

0111 yDDxD

011 yDxD …(i)

Let zyD 1 …(ii)

Therefore 01 zxD or zdx

dzx

or x

dx

z

dz

Integrating both sides of the equation, we get

axz loglog Or axz

Therefore from (ii), we have

axyD 1

axydx

dy

which is a linear differential equation,

Therefore the integrating factor is xdx

ee

Thus we have, xx axeyedx

d


dxexeaye xxx




baeaxe xx

or xbeaaxy

Therefore xbexay 1 is the complete primitive of the differential equation.

Home Assignments

1. 2

12 1 xyyxxy

2. ,221 1

2

2 xxyyxxy given that y = 2 and ,01 y when x = 0.

3. xeyxxyxxyx 22

1

2

2

2 5645341 given that y = 1 and

21 y when x = 0.

4. xeydx

dyx

dx

ydx 1

2

2

5. 3

2

2

22 xydx

dyx

dx

ydx

6 xexyDxDx22 32723

ONE INTEGRAL BELONGING TO THE COMPLEMENTRY FUNCTION

When the integral belonging to the equation

y2 +Py1 +Qy = 0 ...(I)

is known, say zy , then the more general equation of the second order

y2 +Py1 +Qy = R ...(2)

where P, Q , R are functions of x, can be reduced to one of the first order by the

substitution y= vz.

Differentiating, we get, 111 vzzvy

21122 2 vzzvzvy .

Hence (2 ) becomes

,2 12112 RQzPzzvPzzvzv

i.e., zdv1

dx+ v1 2z1 + pz( ) = R ...(3)

Since by hypothesis, .012 QzPzz

Equation is linear equation of the first order in v1.




Similarly a linear equation of nth order can be reduced to one of the (n-1)th if one

integral to the complementary function is known.

Examples

1. Show that 012 QyPyy is satisfied by xey if 01 QP and by xy if

0QxP

Solution:- The given differential equation is y2 +Py1 +Qy = 0 ...(1)

So for xey , we have xeyy 21

Substituting these in (1), we have

01

,0,01

QPtherefore

ebutQPe xx

Thus it follows that xey is the solution of (1), if 01 QP .

Also, we have for y = x,

0,1 21 yandy

Therefore (1) becomes

0QxP

Hence, xy is the solution of (1), if 0QxP .

2. 3

12

2 8xyxyyx … (1)

Solution:- Dividing each term by 2x , we have

xx

y

x

yy 8

2

12 ,

Now comparing the equation with 012 QyPyy .

We have 2

11

xQand

xP

Thus it is easy to see that 0QxP , therefore it follows that

,solutiontheofpartaisxy

Let vxy be the solution of (1), then we can write

2

2

21 2dx

vdx

dx

dvyand

dx

dvxvy

Substituting these in the given differential equation we have




3

2

22 82 xvx

dx

dvxvx

dx

vdx

dx

dvx

xdx

dv

dx

vdx 83

2

2

Put Pdx

dv , So that

dx

dP

dx

vd

2

2

Thus our equation becomes

83

83 Pxdx

dPorxP

dx

dPx

Which is a linear differential equation,

Now we can easily have

dxx

axdvor

x

axP

3322

Now integrating both sides, we have

bx

axv

2

2

2

bxx

axytherefore

x

yvBut

2

3


3. xexyxyxxyx 3

1

2

2

2 22 …(1)

Solution:- Dividing throughout by 2x , we have

xxeyx

xy

x

xxy

212

2

2

22

Now comparing this equation with 012 QyPyy , we have

2

2121

xxQand

xP

Clearly we observe that 0QxP

Therefore it follows that xy is a part of the solution

Let vxy be the solution of (1),





2

2

21 2dx

vdx

dx

dvyand

dx

dvxvy

Substituting in (1) we get

xexvxxdx

dvxvxx

dx

vdx

dx

dvx 32

2

22 222

i.e., we can easily found

xedx

dv

dx

vd

2

2

Put dx

dP

dx

vdthatsoP

dx

dv

2

2

We have, xePdx

dP

Now this is a linear differential equation,

Therefore xx aexeP

But dx

dvP

Thus we can write dxaexedv xx


beaexev xxx

bxxeaxeexythereforevxySince xxx 2,,

4. xexyxyxxy 2

12 2212 …………………(1)

Solution:- Dividing throughout by x , we have

xex

xy

x

xy

x

xy 2

12

2212

Comparing this differential equation with 012 QyPyy ,

We have, x

Qandx

P2

12

2

Clearly we can have 01 QP ,

Therefore it follows thatxey , is a part of a solution.

Let xvey be the solution of (1),





2

2

21 2dx

vde

dx

dveveyand

dx

dvevey xxxxx

Substituting in (1) we get

xxxxxxx exvexdx

dvevex

dx

vde

dx

dvevex 2

2

2

22122

or we have

xex

x

dx

dv

xdx

vd 222

2

Let dx

dP

dx

vdandP

dx

dv

2

2

Thus we observe that

xex

xP

xdx

dP 22

Now multiplying both sides by an integrating factor 2

1

x, we have

ax

e

x

P x

22

or 2axeP x

or since we have dxaxedvthereforePdx

dv x 2

Integrating again we have bx

aev x 3

3

Hence

b

xaexy x

3

3

is the complete solution of the given differential equation.

Home Assignment

1. xexyxyyx 2

12

2 29 , given that 3xy is a solution.

2. 0222 1

2

2 CosxxSinxySinxyxSinxxCosxxy , given that 2xy is a

solution.

3. 2

2

2

)1()1( xydx

dyx

dx

ydx




4. xeyxdx

dy

dx

ydx 2232)12(

2

2

5. .1 2

2

2

xSinyCotxdx

dyCotx

dx

yd

6. 32

2

22 )1(3121)( xyxx

dx

dyx

dx

ydxx

Picard’s Existence and Uniqueness theorem

Existence Theorem:

Statement:- The initial Value problem

,, yxfdx

dy 00 yxy

....(1)

has at least one solution xy provided the function yxf , is continuous and bounded

for all values of x, say

Myxf , …(2)

and satisfies the Lipschitz Condition

,,, 1212 yyMyxfyxf ...(3)

for all values of arguments.

Proof:- Consider the iterative sequence

yn = y0 + f t, yn-1 t( )éë ùûx0

x

ò dt, n =1,2,3,... …(4)

with 00 yty , for the initial value problem (1). In order that the initial value problem

(1) may have a solution, it is necessary that the sequence xyn of functions converge

to a limiting function y (x) which is a solution of (1) or of the equivalent integral equation

,,)(

0

10 dttytfyxy

x

x

n …(5)

To ensure the existence of a limiting function

)(lim)( xyxy nn

…(6)

we use the fact that yn may be written as a sum of successive differences:




1

0

10

n

i

iin yyyy … (7)

this shows that the sequence xyn will converge if the infinite series ii yy 1

converges.

It is easy to see that the successive differences ii yy 1 satisfy the relation

,)(,,)()(

0

11 dttytftytfxyxy

x

x

iiii … (8)

the equation (8) is true for all integers 0,...3,2,1,0 ifori , we have

,,)(

0

001 dtytfyxy

x

x

… (9)

The condition ensures that the existence of integral (8) and (9).

Consider (9) , we have

,,)(

0

001 dtytfyxy

x

x

,

0

dtM

x

x

by (2)

,0xxM …(10)

Again making use of Lipschitz Condition (3), we get from (8),

,,,)()(

0

0112

x

x

dtytftytfxyxy

,)(

0

01

x

x

dtytyM

,.

0

0

x

x

dtxtMM by (10)

.2

2

0xxM

…(11)

We may now proceed by mathematical induction.

Therefore (10) and (11) show that we shall have




,!

)()(0

11i

xxMxyxy

i

i

ii

…(12)

we must show that the inequality (12) holds when i is replaced by i + 1. For this

purpose, we again make use of (8) and (3). We have

,)(,,)()(

0

11

x

x

iiii dttytftytfxyxy

,)()(

0

1

x

x

ii dttytyM

,

!.

0

0

x

x

i

i dti

xtMM

.)!1(

1

01

i

xxM

i

i …(13)

the relation (13) establishes the validity of (12) for all values of t.

From (13), we see that the absolute values of terms in the series (7) are smaller than the

corresponding terms in the Taylor‟s series for the function

.exp 0xxM .

Now the Taylor series for this function converges for all values of ( x – x0), and so the

function yn(x) converges uniformly to a function y(x) for all values x, in any finite

interval.

Proceeding to the limits as n , we get from (4)

dttytfyxy

x

x

nn

nn

0

)(,lim)(lim 10

or dttytfyxy

x

x

nn

0

)(,lim)( 10 , …(14)

Since yxf , is continuous function of both x and y is the range of values considered

and since yn(x) converges to y(x) uniformly over the interval. Therefore the following

interchanges of limiting operations are justified.

dttytf

x

x

nn

0

)(,lim dttytf

x

x

nn

0

)(,lim




dttytf

x

x

nn

0

)(lim,

dttytf

x

x

0

)(, , …(15)

Thus from (14) and (15), we get

,,)(

0

0 dttytfyxy

x

x

this shows that the iterative sequence (4) converges to a solution of the differential

equation problem (1) for all values of x under the given conditions.

Thus the theorem is completely established.

2: Uniqueness theorem

Statement:- The initial Value problem

,, yxfdx

dy 00 yxy .... (1)

has a unique solution xy provided the function yxf , is continuous and bounded for

all values of x, say

Myxf ,

and satisfies the Lipschitz Condition

,,, 1212 yyMyxfyxf

for all values of arguments.

Proof:- Suppose if possible the initial value problem (1) has two distinct solutions y(x)

and u(x). Then using equation (5) of the above theorem we see that the difference y(x)-

u(x) satisfies the relations

dttutftytfxuxy

x

x

0

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

Since f (x,y) is bounded and satisfies Lipschitz condition, we have

Myxf ),( …(3)

and ,,, 1212 yyMyxfyxf …(4)

using (2) and (3) , we have




02

|)(,||,|

)(,,)()(

0 0

0 0

0

xxM

MdtMdt

dttutfdttytf

dttutftytfxuxy

x

x

x

x

x

x

x

x

x

x

i.e., 02)()( xxMxuxy …(5)

Again using (2) and (4), we have

x

x

dttutyMxuxy

0

)()()()( …(6)

Combining (5) and (6), we obtain

!22)(2.)()(

2

02

0

0

xxMdtxtMMxuxy

x

x

…(7)

Employing inequality (7) on the right hand side of (6), we have

!32

!22.)()(

3

03

2

0

0

xxMdt

xtMMxuxy

x

x

…(8)

Continuing in this way, we shall obtain

,...3,2,1,!

2)()(0

nn

xxMxuxy

n

n …(9)

Now the right hand side of (9) tends to zero as n tends to infinity for all values of x, this

shows the solution is unique.

Method of Variation of Parameters

Here we shall explain the method of finding the complete primitive of a linear equation

whose complementary function is known.

Let xBxAy be the complementary function of the linear equation of

second order

RQydx

dyp

dx

yd

2

2

…(i)




Where A and B are constants and xx , are functions of x.

Therefore xBxAy

Satisfies the equation

RQydx

dyp

dx

yd

2

2

or 0'''''' xBxAQxBxAPxBxA

or 0'' '''' xQxPxBxQxPxA

therefore xQxPx ''' =0 …(ii)

and xQxPx ''' =0 …(iii)

Now let us assume that

xBxAy …(iv)

is the complete primitive of (i) where A and B are functions of x, so chosen that (i) will

be satisfied.

Therefore .'' xdx

dBx

dx

dAxBxA

dx

dy

Let A and B satisfy the equation

0 xdx

dBx

dx

dA

…(v)

Therefore xBxAdx

dy ''

and xdx

dBx

dx

dAxBxA

dx

yd

''''

2

2

Since the coefficients of A and B are zero by (ii) and (iii)

Therefore xdx

dBx

dx

dA '' =R …(vi)

From (v) and (vi), we have

xRxxxxdx

dA ''

Therefore,

xxxx

xR

dx

dA''

Integrating we get,




1''

cdxxxxx

xRA

Similarly B can be determined from (v) and (vi).

Substituting these values of A and B in (iv) we get the complete primitive of (i).

Apply the method of variation of parameters to solve the differential equation

1. xCoydx

ydsec

2

2

…(1)

Solution: The auxiliary equation of the given differential equation is

012 m , im

Therefore the complementary functions is

y = a Cos x + b Sin x …(2)

Let y = A Cos x + B Sin x be the complete primitive of the given differential equation.

Where A and B are functions of x.

Differentiating (2) with respect to x, we get,

BCosxASinxSinxBCosxAy 111

where A1and B1 are chosen such that

011 SinxBCosxA …(3)

Therefore BCosxASinxy 1

Differentiating again both sides, we get

BSinxACosxCosxBSinxAy 112

Substituting these in (1) we have

xCoBSinxACosxBSinxACosxCosxBSinxA sec11

Implies that

0sec11 xCoCosxBSinxA …(4)

From (3) and (4), we have

0011 SinxBCosxA

0sec11 xCoCosxBSinxA

xSinxCosxCosxCo

B

xSinxCo

A22

11 1

secsec




1

1

1

11 Cotx

BA

Implies that

,11 A CotxB 1

Now cxdxdxAA 1

Also, we have bSinxdxCotxdxBB log1

Therefore the complete primitive is

xSinxbxxay sinlogcos .

2. xydx

yd2tan44

2

2

… (1)

Solution:- The auxiliary equation of the given differential equation is

042 m im 2

Now the complementary function of the given differential equation is

y = a Cos 2x + b Sin 2x …(2)

Let y = A Cos 2x + B Sin 2x be the complete primitive of the given differential

equation. Where A and B are functions of x.


xBCosxASinxSinBxCosAy 222222 111


022 11 xSinBxCosA …(3)

Therefore xBCosxASiny 22221

Differentiating again both sides, we get

xBSinxACosxCosBxSinAy 24242222 112

Substitute the values of y, y1 and y2 in the given differential equation, we have

xxBSinxACosxBSinxACosxCosBxSinA 2tan422424242222 11

or 02tan42222 11 xxCosBxSinA …(4)

Solving (3) and (4), we get,

0022 11 xSinBxCosA

02tan42222 11 xxCosBxSinA




2

1

2tan242tan24

11 xxCos

B

xxSin

A

xxSinA 2tan221 and xxCosB 2tan221

Thus we have

dx

xCos

xCosdx

xCos

xSinxdxxSindxAA

2

212

2

222tan22

22

1

dxxCosxSec 2222

Therefore, cxSinxxSecA 22tan2log

Also dxCosxdxSindxBB 2221

Hence, the complete primitive of the given differential equation is

hxxCosxxSinxxSecy 2sin22cos22tan2log

4. xe

yy

1

22

Solution:- The given differential equation can be written as

xe

yD

1

212

…(1)

Now the Auxiliary equation is given by

012 ym , 1,1 m

Therefore the complementary function is given by

xx beaey

Let xx BeAey …(2)

be the solution of (1), where A and B are constants.


xxxx eBeABeAey 111


011 xx eBeA …(3)

Therefore, we have

xx BeAey 1

Differentiating again, we have




xxxx eBBeeAAey 112

Substituting these values in the given differential equation, we have;

x

xxxxxx

eBeAeeBBeeAAe

1

211

or we can write, this as

x

xx

eeBeA

1

211

…(4)

Solving (3) and (4) we get

0011 xx eBeA

x

xx

eeBeA

1

211

we get, xx ee

A

1

11 and

x

x

e

eB

1

11

Integrating these, we get

beedx

eedx

eedxAA xx

xxxx1log

1

11

1

11

or bxeeA xx 1log

Also

cee

edxBB x

x

x

1log1

1

or ceB x 1log

Hence the complete primitive of the given differential equation is

xxxxxxx eceebxeeBeAey 1log1log

5. xexyxyyx 2

12

2 …(1)

given that the complementary function is .1 bxax

Solution:- We have the complementary function is

.1 bxay

Let y = Ax+Bx-1

…(2)

be the solution of (1), where A and B are functions.





1

1

2

11

xBBxxAAy

Where A and B are Chosen such that,

A1x+B1x-1 = 0 …(3)

Therefore we have

2

1

BxAy

Differentiating again we have,

3

1

3

12 2 xBBxAy …(4)

Substituting these in (1), we get

xexBxAxxBBxxAAxxBBxAx 211

1

2

1

3

1

3

1

2 2

This gives 2

1

xeA and

2

2

1

xexB

Integrating we get,

ae

dxe

dxAAxx

221

bexeexdxex

dxBB xxxx

22

12

1

2

Hence the complete solution is given by (2) as,

bexeexxa

exy xxx

x21

2

1

2 is the required solution.

6. xeyyyy 2

123 6116

…(1)

Solution:- The given differential equation can be written as

xeyDDD 2)3)(2)(1(

Now the Auxiliary equation of the equation is given by

0)3)(2)(1( mmm implies 3,2,1m ,

Therefore, the complementary function is

xxx cebeaey 32

Let xxx CeBeAey 32 …(2)

is the solution of (1).

Therefore xxxxxx CeeCBeeBeAAey 33

1

22

111 32




where A1, B1 and C1 are chosen such that

A1ex +B1e

2x +C1e3x = 0 …(3)

Differentiating again, we get

xxxxxx eCCeeBBeeAAey 3

1

32

1

2

12 3924

xxx CeBeAe 32 94

where 032 3

1

2

11 xxx eCeBeA …(4)

Differentiating again, we have

y3 = Aex +A1ex +8Be2x + 4B1e

2x +27Ce3x +9C1e3x

…(5)

Substituting these in the given differential equation, we have

.6

3211

39246

92748

232

33

1

22

11

1

32

1

2

1

3

1

32

1

2

1

xxxx

xxxxxx

xxxxx

xxxxxx

eCeBeAe

CeeCBeeBeAAe

eCCeeBBeeAAe

eCCeeBBeeAAe

or xxxx eeCeBeA 23

1

2

11 236 …(6)

Solving (3), (4) and (6), we have

03

1

2

11 xxx eCeBeA

032 3

1

2

11 xxx eCeBeA

0236 23

1

2

11 xxxx eeCeBeA

xxx

xxx

xxx

xxx

xx

xx

xxx

xx

xx

xxx

xx

xx

eee

eee

eee

eee

ee

ee

C

eee

ee

ee

B

eee

ee

ee

A

32

32

32

22

2

2

1

23

3

3

1

232

32

32

1

236

32

1

36

02

0

26

03

0

23

032

0

or xxxxxxx eee

C

ee

B

ee

A632

1

42

1

52

1

2

1

2

Therefore,

21

xeA ; 11 B ;

xeC

212

1

Integrating each term we get,




ce

CandbxBae

Ax

x

2

1;

2

Using these in (2), we have

x

x

xxx

ee

cexbeae

y 32

2

1

2

or xxx ceexbaey 32)( is the required solution.

7. Secnxyndx

yd 2

2

2

Solution:- Here the complementary function is

BSinnxACosnxy

where A and B are constants,

Let BSinnxACosnxy …(1)

Be the complete primitive of the given differential equation, where A and B are

functions of x,

.Sinnxdx

dBCosnx

dx

dABnCosnxAnSinnx

dx

dy

Now choosing A and B such that

0.. Sinnxdx

dBCosnx

dx

dA

…(2)

we have .BnCosnxAnSinnxdx

dy

.22

2

2

Cosnxdx

dBnSinnx

dx

dAnSinnxBnCosnxAn

dx

yd

Substitute in the given differential equation, we have

SecnxCosnxdx

dBnSinnx

dx

dAn

…(3)

Now multiplying (2) by nxSinbyandnxCosn )3( , then subtracting we get,

nxdx

dAn tan

Therefore 12log

1cCosnx

nA




Again multiplying (2) by nxCosbyandnxnSin )3( and adding we get,

1dx

dBn

Therefore 2cn

xB

Substituting the values of A and B in (1), the complete primitive of the given differential

equation is

.log.1

221 Sinnxn

xCosnxCosnx

nSinnxcCosnxcy

Home Assignment

Solve the following by the method of Variation of Parameters.

1. xexy

dx

dyx

dx

ydx 2

2

22

2. 3

2

22 )1(2)1(2 xyx

dx

dyxx

dx

ydx

3. 2

2

2

)1()1( xydx

dyx

dx

ydx

4. xey

dx

dyx

dx

ydx )1(

2

2

Ordinary Differential Equations - University of Kashmiregov.uok.edu.in/eLearning/tutorials/MM14301CR_16... · 2017-06-05 · Lecture notes on Ordinary Differentia l Equations -(Unit

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