10MAT11: ENGINEERING MATHEMATICS-I

Unit 6

Dr. S S Benchalli

Associate Professor and Head

Department of Mathematics

(Govt. Aided Institution)

Basaveshwar Engineering College

Bagalkot-587102

Email: sbenchalli@gmail.com

#######################################################################

Text Book:

� Higher Engineering Mathematics B S Grewal

Reference:

� Higher Engineering Mathematics John Bird

� Advanced Engineering Mathematics. Zill and Cullion

� Internet

Unit –VI:

Goal:

� One of our goal in this course is to solve or find solutions of differential equations.

� Sketch the family of curves given by the Differential equation

� A particular curve of a family may be determined when a point on the curve is specified

Notation:

Leibnitz notation

Prime notation:

Newton’s dot notation:

Learning outcomes

Upon successful completion of ordinary differential equation, it is expected that a student will be able to

do the following.

� Identify an ordinary differential equation & its order

� Verify whether a given function is a solution of a given ordinary differential equation

� Classify ordinary differential equations as variable separable, homogenous, linear, exact and

reducible to these forms.

� Find solutions of separable differential equations

� To be able to derive an ordinary differential equations as the mathematical model for a physical

phenomenon.

� Solve first order homogeneous differential equations.

3

3

2

2

..dx

yd

dx

yd

dx

dy

nyyyy ........,, ′′′′′′

32s becomes 32..

2

2

−=−=dt

sd

Introduction to Differential Equations:

� In Mathematics, a differential equation is an equation in which the derivative of a function

appear as variables.

� Many of the fundamental laws of Physics, Chemistry, Biology and economics can be formulated

as differential equations

� Definition: The derivative of a function y = Φ(x) is itself another function Φ’(x)

found by an appropriate rule.

For example: The function is differentiable on the interval(-∞, ∞),

and its derivative is

If we replace by the symbol y, we obtain

Definition:

A differential equation is an equation which involves differential coefficients or the

differentials.

Definition:

A differential equation is simply an equation involving an unknown function and its

derivatives

Definition ODE:

A differential equation involving derivatives with respect to a single independent variable is

called an ordinary differential equation.

Partial Differential Equations:

A differential equation involving partial derivatives is called a partial differential equation.

Order of a Differential Equation:

dxdy

21.0 xey= 20.1xe x 2.0=dxdy

21.0 xe xy2.0=dxdy

dy

dx

x

xndt

xd

dyedxeyx

+=

=+

=+

dy

dxxy 3

0 2

0 1

2

2

2

54 2

4 2

3/222

2

2 2 2

2 2 2

4.

5. 1

6. 0

td y d x dxe

dt dt dt

d y dyk

dx dx

u u u

x y z

+ + =

= +

∂ ∂ ∂+ + =

∂ ∂ ∂

The order of a differential equation is the order of the highest order derivative occurring in it.

Degree of a Differential Equation:

The degree of a differential equation is the degree of the highest order derivative occurring in it,

when the derivatives are free from radicals and fractions.

Examples 1, 2, 3, 4 and 6 are of first degree Ex: 5 is of second degree because it can be written as

Family of Curves: Integrating both sides of the derivative gives

Where c is an arbitrary constant represents a family of curves, each of the curves in the

family depending on the value of c. Examples include

and these are shown in the following figure.

A particular curve of a family may be determined when a point on the curve is specified. Thus, if

3y x c= + passes through the point (1, 2) , from which C = -1. The equation of the curve

passing through (1,2) is therefore y = 3 x -1

Example: Sketch the family of curves given by the equation 4d y

xd x

= and determine the

equation of one of these curves which passes through the point (2,3).

Solution: Integrating both sides of 4

d yx

d x=

with respect to x gives

24 . 2d y

d x x d x i e y x cd x

= = +∫ ∫

2

22

32

1

=+dx

ydkdxdy

3dy

dx=

3 ,y x c= +

8-3xy and 3,33,83 ==+=+= xyxyxy

3 ,y x c= +

To determine the equation of the curve passing through the point ( 2, 3), from which c = -5

Hence equation of the Curve passing through the point (2,3) is y = 2x2-5.

Concept of a Solution of an ODE:

Any function Φ , defined on an interval I and possessing at n derivatives that are continuous on I

which when substituted into an nth

order ordinary differential equation reduces the equation to an

identity, is said to be a solution of the equation on the interval

How do you Solve a Differential Equation for Unknown Function:

The problem is loosely equivalent to the familiar reverse problem of differential calculus: Given a

derivative, find an antiderivative

For example: In our initial discussion we have already seen that is a solution of

on the interval

Solution Curve Or Integral Curve: A solution curve is the graph of a differentiable function

Solution of a Differential Equation:

A solution or Integral of a differential equation is a relation between the independent and the

dependent variables which satisfies the given differential equation.

General Solution or Complete Integral: A solution containing the number of arbitrary constants equal to the order of the equation is called

the general solution or complete integral.

( )?.y x= Φ

20.1xy e=

0.2dy

x ydx

= ( ),− ∞ ∞

Φ

Particular Solution:

Any solution obtained from the general solution by giving specific values to one or more of the arbitrary constants is called a particular solution.

For example: ( )2y a x b= + is the general solution of the second order

differential equation 22

20

d y d yy

d x d x

+ =

, as it contains two arbitrary constants. If we put a =

1, b = 0 then 2

y x= particular solution

The Solution of Equations of The Form :

A differential equation of the form is solved by direct integration, i.e.

Example: Determine the general solution of 32 4dy

x xdx

= −

Solution: Rearranging 32 4d y

x xd x

= − gives:

3 322 4 2 4 2

4d y x x

xd x x x x x

−= = − = −

Integrating both sides gives: 342log

3

xy x c= − +

Problem: Find the particular solution of the differential equation 5 2 3 ,

d yx

d x+ =

given

the boundary conditions 2

15

y = when x = 2.

Solution: 5 2 3,d y

xd x

+ = then 3 2 3 2

5 5 5

dy x x

dx

−= = −

On integration we get 2

3

5 5

x xy c= − +

( )dy

f xdx

=

( )dy

f xdx

=

∫ += cdxxfy )(

Substituting the boundary conditions 2

1 25

y and x= = to evaluate c, gives c Hence the

particular solution is 23

15 5

x xy = − +

Example: The bending moment M of the beam is given by ( ) ,dM

w l xdx

= − − where w and l

are constants.

Determine M in terms of x given: 21

2M w t=

when x =0.

Solution: ( ) :d M

w l x w l w xd x

= − − = − +

Integrating with respect to x gives 2

2

w xM w l x c= − + + When

21, 0

2M wl x= =

Thus ( )( )

2

201

02 2

ww l w l c= − + + From which, 21

2c w l=

Hence the particular solution is ( )2

21

2 2

w xM w lx w l= − + +

i.e ( )2 212

2M w l lx x= − +

The Solution of Equations of the Form ( ) :dy

f ydx

=

Or

( )d y

d xf y

= and then the solution is obtained by direct integration i.e

( )d y

d xf y

=∫ ∫

Example: Find the general solution of 3 2

d yy

d x= +

Solution:, ,3 2

d yd x

y=

+

Integrating both sides gives

3 2

d yd x

y=

+∫ ∫

( )1

log 3 2 ,2

x y c= + + General Solution

Example: Determine the particular solution of ( ) ydx

dyy 312 =−

Given that 1y = when 12 .

6x =

Solution: Rearranging

2 1

3

1

3 3

yd x d y

y

yd y

y

−=

= −

Integrating gives: 1

3 3

yd x d y

y

= −

∫ ∫

i.e. 2 1

lo g ,6 3

yx y c= − +

which is the General Solution.

When 1

1, 2 ,6

y x= = thus 1 1 1

2 log1 ,6 6 3

c= − + from which, C=2.

Hence the particular solution is 2 1

log 2.6 3

yx y= − +

Example: The variation of resistance, R ohms, of an aluminium conductor with temperature cθ o is

given by ,

d RR

dα

θ=

where α is the temperature coefficient of resistance of aluminium.

If R R=o

when 0 .cθ = o Solve the equation for .R

Solution: d RR

dα

θ=

of the form ( ) ,

dyf y

dx=

Rearranging gives: dRd

Rθ

α=

Integrating both sides gives:

1. . log

dRd

R

i e R c

θα

θα

=

= +

∫ ∫

Which is the general solution.

Substituting the boundary conditions 0R R when θ= =o

gives:

10 lo g R c

α= +

o

for which 1logc R

α= −

o

Hence the particular solution is

Hence Re R R e

R

α θ α θ= ⇒ =o

o

The Solution of Equation of The Form ( ) ( ). :dy

f x f ydx

=

A differential equation of the form ( ) ( ). :

d yf x f y

d x=

where ( )f x is a function of

x only and ( )f y is a function of y only, may be rearranged as

( )( ) ,

dyf x dx

f y=

( )

1 1l o g l o g

1l o g l o g

1l o g

l o g .

R R

R R

R

R

R

R

θα α

α

θα

α θ

= −

= −

=

=

o

o

o

o

And then the solution is obtained by direct integration i.e.

( )( ) .

d yf x d x

f y=∫ ∫

Example: Solve the equation 24 1d y

xy yd x

= −

Solution: Separating the variables gives 2

4 1

1

yd y d x

y x

=

−

Integrating both sides gives: 2

4 1

1

yd y d x

y x

=

− ∫ ∫

On integration, the general solution is ( )22log 1 logy x c− = +

or ( )2

2log 1 logy x c− − =

or ( )2

2 1lo g

yc

x

− =

or ( )

22 1

log .cy

ex

−=

Example: Determine the particular solution of 3 22 ,

tde

d t

θθ −= given that

0 0.t when θ= =

Solution: 3 22 tde

d t

θθ −=

( )( )3 22 te e

θ−=

Separating the variables gives:

3

22

tde d t

eθ

θ−

=

i.e.

2 32 ,te d e dt

θ θ =

Integrating both sides gives 2 32 ,t

e d e dtθ θ =∫ ∫

Thus the general solution is

232

2 3

tee c

θ

= +

When 0 , 0 ,t θ= = thus: 1 2

2 3e e c

θ θ= + from which

1.

6c = −

Hence the particular solution is : 2 31 2 1

2 3 6

te e

θ = −

Or 2 33 4 1.t

e eθ = −

Example: Find the curve which satisfies the equation ( )21dy

xy xdx

= + and passes through the

point ( )0,1 .

Solution: Separating the variables gives:

( )21

x d ydx

yx=

+

Integrating both sides gives ( )21log 1 log

2x y c+ = +

when 10, 1 log1 log1

2x y thus c= = = +

from which 0c = .Hence, the particular solution is ( )21log 1 log

2x y+ = i.e.

( )1

2 2log 1 logx y+ = from which ( )1

2 21y x= +

Hence the equation of the curve is 21 xy −=

Example: When a cake is removed from an oven, its temperature is measured at 300 .Fo

Three

minutes later its temperature is 200 .Fo How long will it take for the cake to cool off to a room

temperature of 70 ?.Fo

Solution: Given 7 0 ,m

T =

By Newton’s Law of cooling, we have ( ) ( ), 0 300m

dTk T T T

dt= − =

i.e. ( ) ( )70 1dT

k Tdt

= − −−− and determine the value of k so that ( )3 200,T =

separating the variables gives:

7 0

d Tk d t

T=

−

Integrating both sides gives

( ) 1log 70T kt c− = +

( )270 2kt

T c e⇒ = + −−−−−

Using initial conditions:

when t = 0, T = 300

( ) .0

20 70 kT c e= +

2 2300 70 230,c c= + ∴ =

equation (2) becomes 70 230 ktT e= + ------- (3)

At t = 3: T(3)=200, equation (3) becomes

3

3

3

3

70 230

200 70 230

130 230

130 1 13log 0.19018

230 3 23

k

k

k

k

T e

e

e

e or k

= +

= +

⇒ =

⇒ = = = −

( ) 0.1901870 230

tthus T t e

−= + ---------(4) We expect the cake to reach the room temperature after a reasonably long period of time

T(t) t(time) 750 20.1 740 21.3 730 22.8 720 24.9 710 28.6 70.50 32.3

Example:

Solve the initial value problem ( ), 4 3dy x

ydx y

= − = −.

Solution: By rewriting the equation as 2 2

12 2

y d y x d x w e g e t

y xy d y x d x a n d c

= −

= − = − +∫ ∫

2 2 2 2

1, 2x y c c c⇒ + = =

This solution of the differential equation represents a family of concentric circles centered at the

origin.When x = 4, y = -3 so that c2 = 25. Thus the initial value problem determines the circle x2 +

y2 = 25. with radius 5. Particular solution is 22 5 , 5 5y x x= − − − < < The

solution curve is the lower semicircle, shown in green color, that contains the point (4,-3)

Example:Weight Reduction Model:

� Weight of person = x kg

� Tries to reduce weight

� Weight loss per month= 10% of weight

� Starting weight = 100kg

xdt

dx1.0−=

Initial conditions x =100 at t = 0.Determine x(t) as a function of t.

Analytical Solution of Simple model:

Recall the model: ( ) 1000t x1.0 ==−= xdt

dx

Cross multiplying, 1.0 dtx

dx−=

Integrating both sides from o to t dt1.0 ∫∫ −=x

dx

C+log x(t) = -0.1t. Using initial conditionsC= - log 100. Thus the final solution is

( )( ) tet

tx 1.0100tor x 1.0100

log −=−=

Example: Solve the differential equation ( )

2 2d yx y a

d x+ =

2

(1)

( 2 )

x y d x

a d y

x yL e t t

a

+ ⇒ = − − − − −

+= − − − − −

Differentiating equation (2) with respect to y we get 11

d x d t

a d y d y

+ =

1d x d t

ad y d y

⇒ = −

( )

2

2

2

1

1

var1

dtt a

dy

dtt a

dy

dy dtiable separable

a t

⇒ = −

⇒ + =

⇒ =+

on integration

( )1

1

ta n

ta n .

yt c

a

x ya y c

a

−

−

= +

+ ′⇒ = +

Homogeneous Equation:

Definition: A function is said to be homogeneous of the nth

degree in x and y if it can be put in the

form ( )/ .nx f y x

These equations can be put in the form ( )( )

1

2

,

,

f x yd y

d x f x y=

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

where 1f and

2f are expressions homogeneous of the same degree in x and y .

The solution is obtained as follows

Put y v x= in (1) we get

1

2

( , )

( , )

f x yd y

d x f x y=

from (1)

( ) ,F v= a function of v

vvFdx

dvx −= )(

(v.s.) )( x

dx

vvF

dv=

−By direct integration we get the solution of the given equation.

Example:

Solve 2 2( ) 2 0 .x y d x xy d y+ − =

The given equation can be put in the form

2 2

2

d y x y

d x x y

+= -------(1).

This is homogeneous in &x y

Solution: Put y v x= in equation (1) we get

2

2

2

1

2

1

2

2( . . )

1

d v vv x

d x v

d v vx

d x v

v d v d xv s

v x

++ =

−⇒ =

=−

on integration we get 2

2

2

log(1 ) log log

log(1 ) log log

log(1 )

v x c

v x c

v x K

− − − =

⇒ − + = −

′− =

)( vFdx

dvxv =+

2

2

2

2 2

(1 ) .

(1 ) .

v x C

yx C

x

x y x C

⇒ − =

⇒ − =

⇒ − =

Example: Solve 2 2( ) 0y d x x y x d y+ + = -----(1)

Equation (1) can be put in the form

2

2,

dy y

dx xy x= −

+

Which is homogeneous in x and y

Solution: Put y = vx and we obtain 2

1

d v vv x

d x v+ = −

+

22.

1

( 1 )0 ( 3 )

( 2 1 )

d v v vx

d x v

v d x

v v x

− +⇒ =

+

+⇒ + = − − − − −

+

Now ( 1) 1(4)

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

v v

v v v v v v

+= + − − − − − −

+ + +

Again, 12)12(

1

++=

+ v

B

v

A

vv

BvvA ++= )12(1

A =1, B=-2

12

21

)12(

1

+−=

+ vvvv

Equation (4) becomes

1 1 1 2

( 2 1 ) ( 2 1 ) 2 1

1 1

2 1

v

v v v v v

v v

+= + −

+ + +

= −+

Equation (3) becomes 1 10

2 1

d x

v v x− + =

+

on integration we get 1lo g ( 2 1) lo g

2v lp g v x c− + + =

cxvv =++− log)12log(log 21

( )c

v

vx=

+ 21

12log

( )k

v

vx=

+ 21

12

Non-homogeneous equation of the first degree in x and y:

These equations of the form

(1)dy ax by c

dx a x b y c

+ += −−−−−−

′ ′ ′+ +

put ,x x h and y y k′ ′= + = + where h and k are some constants

to be determined.

Then

( )( )

( )

( )

d x d x a n d d y d y

a x h b y k cd y d y

d x d x a x h b y k c

a x b y a h b k c

a x b y a h b k c

′ ′= =

′ ′+ + + +′∴ = =

′ ′ ′ ′ ′ ′+ + + +

′ ′+ + + +=

′ ′ ′ ′ ′ ′ ′+ + + +

If h and k are determined so that 0 & 0ah bk c a h b k c′ ′ ′+ + = + + =

Then d y a x b y

d x a x b y

′ ′ ′+=

′ ′ ′ ′ ′+

, which is homogeneous in x ′ and y ′ , which can be

solved by putting .y vx′ ′=

This method fails when : :a b a b′ ′=

. .a b

i e w h e na b

=′ ′

Because the h and k become indeterminate

Suppose 1a b

a b m= =

′ ′

.

Then Equation (1) can be written as

( )

d y a x b y c

d x m a x b y c

+ +=

′+ + put a x b y v t h e n

d y d va b

d x d x

+ =

+ =

.v c d v

a bm v c d x

++ =

′+

Then by separating the variables and integrating we get the solution.

Example: Solve (2 4 5) ( 2 3) .x y dy x y dx− + = − +

Solution: The given equation can be written as

2 3

2 4 5

( 2 ) 3(1)

2 ( 2 ) 5

d y x y a bH e r e

d x x y a b

x y

x y

− + = = ′ ′− +

− += − − − − − − −

− +

Put v=x-2y

dx

dy

dx

dv21 −=

1

12

d y d v

d x d x

= −

From equation (1) , 3 1

12 5 2

2 61

2 5

v d v

v d x

v d v

v d x

+ = −

+

+⇒ = −

+

2 61

2 5

2 5 2 6

2 5

1

2 5

d v v

d x v

v v

v

v

+= −

+

+ − −=

+

= −+

On integration we get

( ) ( )

2

2

5

. . 2 5 2 .

v v x c

i e x y x y x c

+ = − +

− + − = − +

Example: Solve 2 3

2 3

d y x y a bH ere

d x x y a b

+ − = ≠ ′ ′+ −

Solution: Given equation is non-homogeneous in x and y

Put ,

2 2 3(1)

2 2 3

x x h y y k

dy x y h k

dx x y h k

′ ′= + = +

′ ′+ + + −∴ = − − − − − −

′ ′+ + + −

We find ,h k such that 2 3 0

2 3 0

h ksolve by cross multiplication

h k

+ − =

+ − =

11, 1

6 3 6 3 1 4

h kh k= = ∴ = =

− + − + −

Equation (1) becomes 2

2

d y x y

d x x y

′ ′ ′+=

′ ′ ′+

This is homogeneous in x ′ and y ′

Put

1 2

2

y v x

d y d vv x

d x d x

v

v

′ ′=

′′∴ = +

′ ′

+=

+

On simplification we get

2

2

2(var )

1

2

1

v dxdv iables are separable

v x

v dxdv

v x

′+=

′−

′+=

′−∫ ∫

( )2

log 2(1 )(1 )

2

(1 )(1 ) (1 ) (1 )

1/ 2, 3/ 2

vdv x c

v v

v A Bnow

v v v v

A B

+′= + − − − − −

+ −

+= +

+ − + −

= =

∫

Equation (2) becomes

1 3log

2(1 ) 2(1 )

1 3log(1 ) log(1 ) log ,

2 2

dv x cv v

v v x c

′+ = + + −

′⇒ + − − = +

∫

On simplification we get

2

3

1l o g .

( 1 )

vx c

v

+′=

−