chapter 2: second order differential equations …chapter 2: second order differential equations...

Chapter 2: Second Order Differential Equations SSE1793

1

2.1 Basic Definition, Classification and Terminologies

Variable coefficients

Coefficients with variable such as , , in it. Normally will be written as , , or

Constant coefficients

Coefficients with constants such as 2, 5, in it.

Will be written as or

Linear DE

Can you tell which one is the independent variable and dependent variable in the given linear differential equation?

The general form of a linear differential equation can be expressed in

the form

where

are called coefficients and

is a function of

Variable / Constant Coefficients


2

Homogenous Equation

When

Non-Homogenous Equation When

Example 2.1.1:

1. 7 10 0y y y

2. 2 23 5 5yy y y x

3. 2 cosx y xy y x

Homogenous / Non-Homogenous Equation

Linear / Non-linear Constant Coefficients / Variable Coefficients Homogenous / Non-homogenous Equation




3

Linearly independent

Linearly dependent At least one of

Example 2.1.2: Determine whether or not the following functions are linearly independent.

a) 1 2sin and 4sin y x y x

b) 2

1 2 and y x y x

Solution:

a) Given that 1 2sin and 4sin y x y x . Equation

can be rewritten as

2

1

4 sinsin

C xx

C ,

gives . Since , this equation is linearly dependent

b) Given that 2

1 2 and y x y x . Equation

has a solution only if .

Since , this equation is linearly independent

Linearly Dependent / Linearly Independent


4

If and are linearly independent, and each of them are the solutions for the DE

Then the general solution for the equation is

If , , are linearly independent, and each of them are the solutions for the DE

Then the general solution for the equation is

Linear Combination of Solutions

Linear Superposition Principle


5

2.2 Solution of Homogenous Equation

How to find the solution of this equation:

2

20.

d y dya b cy

dx dx (i)

1) Assume the solution in the form of mty e . Thus, we have

22; ;

2

dy d ymt mt mty e me m edt dt

(ii)

2) The solution mty e must satisfy (i). Insert (ii) -> (i) gives the

characteristic (auxiliary) equation:

2 0am bm c (iii)

3) Find the roots 2

1 2

4,

2

b b acm m

a

For 2 4 0b ac roots are real and distinct (m1 m2)

For 2 4 0b ac roots are real and equal (m1 = m2)

For 2 4 0b ac roots are complex conjugate numbers

4) The solution of (i) depends on the type of roots.


6

Case 1: Roots are real and distinct, 21

mm ,

General solution: 1 2m x m xy Ae Be .

Example 2.2.1:

Solve '' 5 ' 6 0.y y y

Solution:

1) Use the characteristic equation: 2 5 6 0.m m

2) Find the roots : (m+3) (m+2) = 0 1m = -3; 2m = -2.

3) The general solution is 3 2x xy Ae Be

Example 2.2.2:

Solve 11 0.y y y

Solution:

1) Use the characteristic equation:

2 11 0.m m

2) Find the roots : 2

45

2

1m

1

1 45,

2m

2

1 45.

2m

3) The general solution is 1 2( ) .

m x m xy x Ae Be


7

Case 2: Roots are real and equal, 1 2m m

General solution: xm

eBxAy 1)( Example 2.2.3:

Solve 6 9 0y y y

Solution:

1) Use the characteristic equation:

0962 mm 2) Find the roots : 0)3( 2 m ⇒ 3,3 m

3) The general solution is 3( ) ( ) xy x A Bx e

Case 3: Roots are complex numbers; im

General solution: )sincos()( xBxAexy x

Example 2.2.4:

Solve 10 26 0.y y y

Solution:

1) Use the characteristic equation: 2 10 26 0.m m

2) Find the roots : 2

)26(410010 m

210 4 10 45

2 2

ii

5 ; 1 3) The general solution is

5( ) ( cos sin )xy x e A x B x


8

Example 2.2.5:

Solve the initial value problem: 6 16 0y y y ; (0) 1, (0) 0.y y

Solution:

1)Use the characteristic equation:

2 6 16 0.m m

2) Find the roots : 6 36 4(16) 6 28

2 2m

=2

283

2i = 3 7 .i

3)The general solution is

3( ) cos 7 sin 7ty t e A t B t

4) Using IC: i) y(0) = 1 ; A = 1 ;

ii) (0) 0y

' 3 37 sin 7 7 cos 7 cos 7 sin 7 ( 3 )t ty e A t B t A t B t e

0 7 3B A 37 B 3

.7

B

5) The solution is

3 3( ) cos 7 sin 7

7

ty t e t t

1. 2 7 3 0y y y (Ans: /2 3

1 2( ) x xy x c e c e )

2. 2 0, (0) 5, (0) 3y y y y y (Ans: ( ) 5 2x xy x e xe )

3. 4 5 0y y y (Ans: 2( ) ( cos sin )xy x e A x B x )

Initial Value Problem

Exercise 2.2

These are the

initial conditions


9

2.3 Solution of Non-Homogeneous Equation The solution of

( )ay by cy f x

is given by )()()( xyxyxy

ph

where )(xy

h is the solution of the homogeneous equation and )(xy

p is

called a particular integral.

Two methods to find ( ) :py x

1) Method of Undetermined Coefficient 2) Method of Variation of Parameters

2.3.1 Method of Undetermined Coefficient The form of

py depends on the form of f(x).

Case 1: Polynomial of degree n ;

1

1 1 0...n n

n n nP A x A x A x A

Case 2: Form of mxCe ; where C and m are constants Case 3: Sine/cosine function Case 4: Sums or products of the above functions (Case 1,2,3)

How to find the solution of this

equation


10

Case 1: Polynomial 1

1 1 0( ) ....n n

n nf x A x A x A x A

with n as highest degree.

Try 1

1 1 0( ) ....r n n

p n ny x x C x C x C x C

Example 2.3.1: Solve 22 2 4y y y x x (i)

Solution: Step 1 : Find )(xy

h: 022 mm

( 1)( 2) 0m m 2,1m

xx

hBeAexy 2)(

Step 2: Find ( )py x : take r = 0; try

Step 3: Compare py and hy . Any similar term? YES/NO?

Step 4: If YES - do some modification on py by taking different values

of r. Then repeat Step 3. If NO – Proceed to find &p py y

Step 5: Substitute (ii,iii) into (i) and equate coefficient power of x.

2 2

2 2 1 2 1 0

2 2

2 1 0 2 1 2

2 2 2 2 4

(2 2 ) (2 2 ) 2 2 4

C C x C C x C x C x x

C C C x C C C x x x

2 1 0 2 1 22 2 0; (2 2 ) 4; 2 2C C C C C C

2

2 1 0( ) (ii)py x C x C x C

2 1

2

( ) 2(iii)

( ) 2

p

p

y x C x C

y x C

Notes: Choose the smallest r, (r = 0,1,2). Chosen to ensure that no term in ( )py x is in the

corresponding ( )hy x .


11

Therefore,

2

1C 1

1C 0

1

2C

2 1( )

2py x x x

Final Answer:

2

1)( 22 xxBeAexy xx

Example 2.3.2: Solve xyy '" 2

Solution: Step 1 :

x

hBeAy 2

Step 2 : Try

1 0

1

0

p

p

p

y C x C

y C

y

Step 3 : Choose new with 1;py r Try

1 0

( )p

y x x C x C

1 0

1 1;

4 4C C

21

( )4

py x x

Final Answer:

)(4

1)( 22 xxBeAxy x

Look! This term is already

appears in hy . We have to

modify this py . Choose

new !!py

Compare to hy again

and now, this py is

correct!!

What is the ( )py x

for this f (x)


12

Case 2: For )(xf is the form mxCe .

Seek r mx

py x De .

Example 2.3.3:

Solve 32 5 .xy y y e (i)

Solution: Step 1: Find )(xy

h.

022 mm

0)1)(2( mm 2,1m ,

xx

hBeAexy 2)( (ii)

Step 2: Find ( )py x . Try r = 0;

3

3

3

3 (iii)

9

x

p

x

p

x

p

y De

y De

y De

Step 3: Substitute (iii) into (i) xxxx eDeDeDe 3333 5239 Equate coefficients: 3 3(9 3 2 ) 5x xD D D e e

1

10 52

D D

31

2

x

py e

Final Answer:

phyyxy )( xxx eBeAe 32

2

1

Is there any term in py

appears inhy ?? NO!!

Then, continue..

Notes: Choose the smallest r, (r = 0,1,2).

Chosen to ensure that no term in ( )py x

is in the corresponding ( )hy x .

What is the ( )py x

for this f (x)


13

Example 2.3.4:

Solve 22 3 .xy y e

Solution: Identify that f(x) in form 2xCe

. Step 1: Find yh : 022 mm

0)2( mm

2,0 m 2( ) x

hy x A Be

Step 2: Find yp Try r = 0;

x

pDey 2

Step 3: Change

py ; Try r = 1;

py = xDxe 2

2 22x x

py De Dxe

2 24 4x

py De Dxe

D = 2

3

Final Answer:

xx xeBeAxy 22

2

3)(

Look! This term is

already appears in hy . We

have to modify this py .

Choose new !!py

Compare to hy again

and now, this py is

correct!!

What is the ( )py x

for this f (x)


14

Case 3: f(x) = xA sin

1 or xA cos

1

Try sin cosr

py x C x D x

Example 2.3.5:

Solve 2 sin . (i)y y y x

Solution: Identify that f(x) in form sin x . Step 1: Find yh :

2( ) x x

hy x Ae Be

Step 2: Find yp , try r = 0;

sin cos

cos sin (ii)

sin cos

p

p

p

y C x D x

y C x D x

y C x D x

Step 3: Substitute (ii) into (i)

sin cos cos sin 2( sin cos ) sinC x D x C x D x C x D x x

Equate coefficients:

3 1, cos

10 10C D x

3 1

sin cos10 10

py x x

Final Answer:

2 3 1( ) sin cos

10 10

x xy x Ae Be x x

Notes: Choose the smallest r, (r = 0,1,2).

Chosen to ensure that no term in ( )py x

is in the corresponding ( )hy x .

What is the ( )py x

for this f (x)


15

Case 4: Sums or products of the Case 1,2 and 3 1) Sums of the Case 1,2 and 3 : we have

1 2 3( ) ( ) ( ) ( ) ..... ( )nf x f x f x f x f x

then

1 2 3( ) ( ) ( ) ( ) ..... ( )p p p p pny x y x y x y x y x .

The solution become:

1 2 3( ) ( ) ( ) ( ) ( ) ..... ( )h p p p pny x y x y x y x y x y x

2) : Products of the Case 1,2 and 3; ( )py x is given by the table

( )f x ( )py x

( ) mx

nP x e 1

1 1 0....r n n mx

n nx C x C x C x C e

cos( )

sinn

xP x

x

1

1 1 0.... cosn n

n n

r C x C x C x C xx

1

1 1 0.... sinn n

n n

r C x C x C x C xx

cos

sin

mxx

Cex

sin cosr mxx e C x D x

cos( )

sin

mx

n

xP x e

x

1

1 1 0.... cosr n n mx

n nx C x C x C x C e x

1

1 1 0.... sinr n n mx

n nx C x C x C x C e x

Notes: Choose the smallest r, (r = 0,1,2). Chosen to ensure that no term in ( )py x is in the

corresponding ( )hy x .


16

Example 2.3.6:

Solve

22 8

2

d y dy xy xedtdt

(i)

Solution:

Step 1: Find ( )hy x

0822 mm 2 4 0 2, 4m m m

2 4( )h

x xy x Ae Be (ii)

Step 2: Find ( )py x

Try: 1 0( ) ( )p

xy x e C x C (iii)

Step 3: Solve using undetermined coefficient method.

1 1 0( )p

x x xy x e C x e C C e

1 1 1 0( )p

x x x xy x e C x e C e C C e

Substitute into (i):

1 1 1 0

1 1 0 1 02( ) 8 ( )

x x x xe C x e C e C C e

x x x x xe C x e C C e e C x C xe

1 0

1, 0.

9C C

Final Answer :

12 4( )9

x x xy x Ae Be xe

What is the ( )py x

for this f (x)


17

Example 2.3.7:

Solve 2 xy y y e x (i)

Solution: Step1: Characteristic Equation:

xx

hBeAexy 2)( (ii)

Step 2: Write 21 ppp yyy

Step 3: Find 1py

Try 1

x

py De

Try new yp

1

1

12

x

p

x x

p

x x

p

y xDe

y De xDe

y De xDe

Subst. this into (i) and solve it, gives

x

p xey 3

11

Step 4: Find 2p

y , try

2 1 0 1 0

1 1,

2 4py C x C C C

Step 5: Final Answer

1 2

2 1 1 1

3 2 4

h p p

x x x

y x y x y x y x

Ae Be xe x

Look!! This is same with hy .

What is the ( )py x

for this f (x)


18

Exercise 2.3.1: 1. Decide whether or not the method of undetermined coefficients

can be applied to find a particular solution of the given equation.

a) 1'' 2 xy y y x e

[Ans: NO]

b) 35 '' 3 2 cos5y y y x x

[Ans: Yes]

c) '' 4 3cos( )xy y e

[Ans: N0]

d) 2'' 3 cosy y y x

[Ans: Yes]

2. Find the form for py to

2 3 ( )y y y f x

where f(x) equals

a) 7cos3x

[Ans: sin3 cos3 ]p

y C x D x

b) 2 sinxxe x

[Ans: 1 0 1 0

cos sin ]x x

py C x C e x B x B e x

c) 2 xx e

[Ans: 2

2 1 0]x

py x C x C x C e

d) sin cos2x x

[Ans: sin cos sin 2 cos2 ]p

y A x B x C x D x

e) 3xxe

[Ans: 1 0

]x

py x C x C e


19

3. Solve 2 6.y y

[Ans: 2 3 ]xy A Be x

4. Solve 3 .xy y e

[Ans: 31

cos sin ]10

xy x A x B x e

5. Solve 2'' ' 2 sin .y y y x

[Hint : use trigonometry identity]

6. Solve '' 4 3cos 2 .y y x [Hint:

sin 2 cos 2 ,

sin 2 cos 2 ]

p

h

y x C x D x

y A x B x


20

2.3.2 Method of Variation of Parameters

Variation of parameters can be used to determine a particular solution. This method applies even when the coefficients of the d.e. are a function of x. Obviously it can solve non-homogeneous second ode with constant coefficients.

ay'' + by' + cy = f(x)

Solve all types of f(x) such as 2tan , sec ,1 , ln , sin( )

xx x x x e and etc.

Method of variation of parameters: Step 1 : Identify a and f(x) Step 2 : Consider homogeneous equation. Choose y1 and y2. Step 3 : Find Wronskian, W for y1 and y2.

1 2

1 2 2 1

1 2

y y

W y y y y

y y

Step 4 : Find u and v, where

2 ( )( )

y f xu x dx C

aW ;

1 ( )( )

y F xv x dx D

aW

Step 5 : The general solution for particular solution of

differential equation is given by

1 2y uy vy


21

Example 2.3.8:

Find the solution for secy y x (i)

Solution : Step 1 : Identify a = 1, f(x) = sec x.

Step 2: Find root 0 yy :

012 m 22 1 im im .

cos sinh

y x A x B x (ii)

From eqn (ii), we choose )(1 xy and )(2 xy .

Let 1( ) cosy x x , xxy sin)(2 (iii)

Step 3: Find Wronskian, W

1 2

1 2

cos sin

sin cos

y yx x

Wx x

y y

2 2cos cos sin ( sin ) cos sin 1x x x x x x (iv)

Step 4: Find xu and )(xv .

2 ( ) sin sec( ) tan

1

y F x x xu x dx u dx xdx

aW

sin

ln cos .cos

xu x dx x C

x

1 ( )( ) cos .sec 1 .

y F xv x dx v x xdx dx x D

aW

Step 5 : The general solution

1 2( ) cos sin ln cos cos siny x uy vy A x B x x x x x Notes: Constant a is the coefficients for y” in actual equations. We can let the actual equations in easier form so that a = 1. For example, if we are given this function

3 secy y x , we can write the equation in this form1 sec

3 3

xy y .

Both are arbitrary choices, where we can let xxy sin)(1

and 2 ( ) cosy x x

and we can let arbitrary constants A and B equal to 1.


22

Example 2.3.9:

Find the solution for xexyyy 2144 by using variation

of parameters. Solution: Step 1 : a = 1 and f(x) = (x+1)e2x

Step 2 : Identify xy1 and xy2 .

Characteristic eqn: 0442 mm 2,2m

x

h eBxAxy 2 .

Set xexy 2

1 and xxexy 2

2

Step 3 : Find the Wronskian, W(x):

)(xW

1221 yyyy

2 2 2 2 2 4. 2 .2x x x x x xe e xe xe e e

Step 4 : Find xu and xv :

2 2

2 2

4

11

x x

x

y f x xe x eu x dx dx x x dx x x dx

W e

3 2

3 2

x xu x C

2 2

1

4

1( 1)

x x

x

y f x e x ev x dx dx x dx

W e

2

2

xv x x D

Step 5 : General solution:

3 22 2 2

1 26 2

x x xx xy uy vy e Ce Dxe

Check your answer by using the method of undetermined coefficients.


23

Exercise 2.3.2 Solve the following differential equations 1. 4 tan 2y y x

2. 2 3 6y y y

3. '' 4 2siny y x

4. '' 2 ' 2 cosxy y y e x

5. 32 '' 2 ' 4 2 tx x x e

6. 2'' 4 ' 4 tx x x te

7. '' 3 ' 2 sin( )xy y y e


24

2.4 Second ODE : Applications 2.4.1 MECHANICAL SYSTEM

The governing equation is

( )my by ky f t

where

m = inertia b = damping k=stiffness

( )f t = force vibration.

Undamped; 0b Damped; 0b

Free vibration;0)( tf

->Homogeneous

0my ky

0my by ky

Force vibration;0)( tf

->Non-

Homogeneous

( )my ky f t

( )my by ky f t

Do you know what

type of equations

are these?

y b


25

Case 1: UNDAMPED FREE VIBRATIONS (When b = 0 ; f(t) = 0) The equation is given by:

02

2

kydt

ydm (i)

Divide by m; 2

20

d y ky

mdt

Rewrite,

22

20

d yy

dt where

m

k . (ii)

Step 1: Using characteristic equation:

2 2 0r r i

Step 2: General solution to (ii) is

tktkty sincos)( 21

We also can express y(t) in the more convenient form

)sin()( 2

2

2

1 tkkty

is an angular

frequency and

period = 2 /

2 2

1 2k k is

an amplitude

The motion of a

mass in case 1 is

called simple

harmonic motion

y


26

Notes:

Given tktk sincos 21 , we can write this as

t

kk

kt

kk

kkk sincos

2

2

2

1

2

2

2

2

1

12

2

2

1

i.e

2 2 1 2

1 2 1 22 2 2 2

1 2 1 2

cos sin cos sink k

k t k t k k t tk k k k

where

1 2

2 2 2 2

1 2 1 2

sin , cosk k

k k k k

1

2 2

1 2

2

2 2

1 2

sintan

cos

k

k k

k

k k

1

2

arctank

k

tkk

ttkk

tktk

sin

sincoscossin

sincos

2

2

2

1

2

2

2

1

21

Notes: Alternative representation:

2

2

2

1

2

2

2

2

1

1 sin,coskk

k

kk

k

tkk

ttkk

tktk

cos

sinsincoscos

sincos

2

2

2

1

2

2

2

1

21


27

Case 2: DAMPED FREE VIBRATIONS (When b ≠ 0 ; f(t) = 0) The equation is given by:

02

2

kydt

dyb

dt

ydm (i)

Step 1: Characteristic equation for (i) is

02 kbrmr Its roots are

224 1

42 2 2

b b mk br b mk

m m m

Step 2: There are 3 cases related to these roots:

1) 2 4b mk Underdamped or Oscillatory Motion

The roots are i , where

m

b

2:

242

1: bmk

m

The general solution to (i) is

)sincos()( 21 tktkety t

can express in

)sin()( 2

2

2

1 tekkty t

2) 2 4b mk Overdamped Motion

The roots are

1

2

br

m

,

2

2

14

2 2

br b mk

m m


1 2

1 2( )

r t r t

y t k e k e

tmbekk )2/(2

2

2

1

is

damping factor

Complex roots

Two distinct roots


28

3) 2 4b mk Critical damped Motion

The roots are

1 2

2

br r r

m


1 2( ) ( ) rty t k k t e

Two repeated roots


29

Case 3: UNDAMPED FORCED VIBRATIONS (When b = 0 ; f(t) ≠ 0) Example 2.4.1 Solve the initial boundary problem for undamped mechanical system.

tFky

dt

ydm cos

2

2

(i)

000 yy where mk

Solution: Find homogeneous solution of (i). Rewrite (i) as

tm

Fy

m

k

dt

ydcos

2

2

(ii)

Set mk

mk ,2

Step 1: Using characteristic Equation:

imm 022

tBtAtyh sincos (iii)

Step 2: Find particular solution of (ii)

Try tDtCtyp sincos (iv) a

2sin cospy C t D t (iv) b

2 2cos sinpy C t D t (iv) c

Note: Since , system not at resonance.


30

To find const and D and C, substitute (iv) a,b,c into (ii). Equate

coefficients of tcos and tsin 2 2

0( cos sin ) ( cos sin ) cosm C t D t k C t D t F t

2 2

0( )cos ( )sin cosCm Ck t Dk Dm t F t

2 20 ,

FD C

m

2 2

cos sin cosF

y x A t B t tm

(v)

Find A & B using initial condition.

00 y

220

m

FA

2 2

FA

m

00 y , Find B?

B=0.

Final Answer:

2 2 2 2

cos cosF F

y t t tm m


31

Case 4: DAMPED FORCED VIBRATIONS (When b≠0, f(t)≠0)

The equation is given by:

2

2cos

d y dym b ky F t

dt dt

The general solution is

2

( /2 )

2 2 2

4sin sin( )

2 ( )

b m t Fmk by t Ae t t

m k m b

______________________________________________________________________________ Exercise 2.4.1: 1. A 4 pound weight is attached to a spring whose spring constant is

16 lb/ft. what is the period of simple harmonic motion?

[ans : period = 2 k

m

2

period8 ]

2. A 24-pound weight, is attached to the end of a spring, stretches it

4 inches. Find the equation of motion if the weight is released from rest from a point 3 inches above the equilibrium position.

[ans : 1

( ) cos4 64

y t t ]

3. A 10 - pound weight attached to a spring stretches it 2 feet. The

weight is attached to a dashpot damping device that offers a resistance numerically equal to ( 0) times the instantaneous

velocity. Determine the values of the damping constant so that

the subsequent motion is a) overdamped; b)critically damped; c) underdamped.

[ans : 5 5 5

) ; ) ; )02 2 2

a b c ]


32

4. A 16 - pound weight stretches a spring 8/3 feet. Initially the weight starts from rest 2 feet below the equilibrium position and the subsequent motion takes place in a medium that offers a damping force numerically equal to ½ the instantaneous velocity. Find the equation of motion if the weight is driven by an external

force equal to ( ) 10cos3f t t .

[ans : 24 47 64 47 10

( ) cos sin cos3 sin33 2 2 33 47

t

y t e t t t t

]

5. The motion of a mass-spring system with damping governed by:

2

216 0;

d yby y

dt

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

Find the equation of motion and sketch its graph for b = 0, 6, 8 and 10.

6. Determine the equation of motion for an undamped system at

resonance governed by: 2

25cos ;

d yy t

dt

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

Sketch the solution. 7. The response of an overdamped system to constant force is

governed by: 2

28 6 18;

d y dyy

dt dt

(0) 0, (0) 0.y y

Compute and sketch the displacement y(t). What is the limiting value of y(t) at t ?


33

2.4.2 RLC – Circuit

The equation for RLC circuit is given by:

;dI q

L RI E tdt c

with the initial conditions : IIqq 0,0

where .dq

Idt

Example 2.4.3:

The series RLC circuit has a voltage source given by 100VE t ,

a resistor 20R , an inductor 10HL , and a capacitor

1

6260c

. If the initial current and the initial charge on the

capacitor are both zero, determine the current in the circuit for t>0. Solution:

The equation :

(i)dI q

L RI E tdt c

The initial conditions: 0 0 , 0 0q I

The information:

16260,100,20,10

ctERL

Series RLC Circuit notations:

V - the voltage of the power source (measured in volts V)

I - the current in the circuit (measured in amperes A)

R - the resistance of the resistor (measured in ohms = V/A);

L - the inductance of the inductor (measured in henrys = H = V·s/A)

C - the capacitance of the capacitor (measured in farads = F = C/V =

A·s/V)

q - the charge across the capacitor (measured in coulombs C)

http://en.wikipedia.org/wiki/Volt

http://en.wikipedia.org/wiki/Ampere

http://en.wikipedia.org/wiki/Electrical_resistance

http://en.wikipedia.org/wiki/Ohm_%28unit%29

http://en.wikipedia.org/wiki/Inductance

http://en.wikipedia.org/wiki/Henry_%28unit%29

http://en.wikipedia.org/wiki/Second

http://en.wikipedia.org/wiki/Capacitance

http://en.wikipedia.org/wiki/Farad

http://en.wikipedia.org/wiki/Coulomb

http://en.wikipedia.org/wiki/Coulomb


34

Method 1: Change the equation into homogeneous equation

Step 1: Differentiate eqn (i) with t, we have

tEdt

d

dt

dq

cdt

dIR

dt

IdL

12

2

tEdt

dI

cdt

dIR

dt

IdL

12

2

(ii)

In our case 16260,100,20,10

ctERL

Eqn. (ii) becomes

0626020102

2

Idt

dI

dt

Id

Or 2

22 626 0

d I dII

dt dt

(iii)

Step 2: Find I(t),

2 2 626 0m m

im 251

cos 25 sin 25tI t e A t B t (iv)

Step 3: Need to find A & B . Substitute the Initial condition given into (iv):

0 0 0I A

tBetI t 25sin (v)

This is a homogenous equation. Can you remember how to solve this?? Use characteristic equation!!!

This is the general solution of

I(t). Use the initial conditions

given to find A and B.


35

What is B? How to find B?

From I.C. 0 0,q we have to find ? at 0.dI

tdt

From (i), at t = 0,

1000

0 c

qRI

dt

dIL

ot

10

100100

Ldt

dI

dt

dIL

otot

From (iv),

teteB

dt

dI tt 25cos2525sin

5

2

25

1025100 BB

Final answer:

2

sin 255

tI t e t

Method 2 : Solve the non-homogeneous equation Step 1:

2

2We know , substitute to equation (i):

dq dI d qI

dt dt dt

2

2 (vi)

d q dq qL R E t

dt dt c

1

In our case 10 , 20 , 100 , 6260 ,L R E t c

equation (vi) become

New initial

condition t = 0,

I = 0.


36

2

210 20 6260 100

d q dqq

dt dt

Or

2

210 626 10

d q dqq

dt dt

Step 2:

Then find and h pq q

626

10,25sin25cos

p

t

h qtBtAeq

10

cos 25 sin 25626

tq t e A t B t (vii)

Step 3: i)Given initial charge is 0

0at0 tq

626

10

626

100 AA

ii)Initial current is 0

0at0 tdt

dqI

Differentiate (vii)

t

t

etBtA

tBtAedt

dq

25sin25cos

25sin2525cos25

tBAtBAedt

dq t 25sin2525sin25

(iii)

This is a nonhomogenous equation. Can you remember how to solve this?? You have

to find and h pq q !!!

This is the general

solution of q(t). Use

the initial conditions

given to find A and B.


37

From I.C: 0a0 tt

dt

dq

1505

1

3130

2

25625

10

25

250

AB

AB

The final answer :

tetI t 25sin

5

2

Exercise 2.4.2

1. An RLC series circuit has a voltage source given by E(t)=20 V , a resistor of 100 Ω, an inductor of 4 H and a capacitor of 0.01 F. If the initial current is zero and the initial charge on the capacitor is 4 C, determine the current in the circuit for t>0.

2. An RLC series circuit has a voltage source given by of E(t)=10cos 20t V , a resistor of 120 Ω, an inductor of 4 H and a capacitor of (2200)-1 F. Find the steady state current (solution) for this circuit.

3. An RLC series circuit has a voltage source given by of E(t)=30sin 50t V , no resistor, an inductor of 2 H and a capacitor of 0.02 F. What is the current in this circuit for t>0 if at t =0, I(0)=q(0)=0?

chapter 2: second order differential equations …chapter 2: second order differential equations...

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