10MAT11: ENGINEERING MATHEMATICS-I Unit 6 · 10MAT11: ENGINEERING MATHEMATICS-I Unit 6 ... Higher...
Transcript of 10MAT11: ENGINEERING MATHEMATICS-I Unit 6 · 10MAT11: ENGINEERING MATHEMATICS-I Unit 6 ... Higher...
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: [email protected]
#######################################################################
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
+′=
−