Solving Linear 2nd Order ODEs with Analytical and Numerical Methods, Slides of Calculus for Engineers

Download Solving Linear 2nd Order ODEs with Analytical and Numerical Methods and more Slides Calculus for Engineers in PDF only on Docsity!

Engr/Math/Physics 25

Chp9: Ordinary

Differential Eqns

Learning Goals

• List Characteristics of Linear, MultiOrder,

NonHomgeneous Ordinary Differential

Equations (ODEs)

• Solve ANALYTICALLLY, Linear, 2nd^ Order,

NonHomogeneous, Constant Coefficient

ODEs

• Use MATLAB to determine Numerical

Solutions to Ordinary Differential Equations

(ODEs)

Solving 1 st^ Order ODEs - 1

• Given the Simple ODE

with

• No Zero Order (i.e., “y”) term
• An INITIAL Condition

= 2 t ;I.C. y ( ) 0 = 7. 37

dt

dy

 Can Solve by

SEPARATING the

VARIABLES

t dt dy tdt dt

dy (^2) ⋅ ⇒ = 2 

 



 Now use the IC in

the Limits of Integ.

 AND Integrating

Both Sides

∫ dy^ =^ ∫^2 td^ t

( )

( ) ∫ ∫

=

=

=

=

=

y t t

y

d d

β

β

α

α

α β β 0 0

2

• Note the use of DUMMY VARIABLES of INTEGRATION α and β

Solving 1 st^ Order ODEs - 2

• Integrating (^)  Separating The

Variables sometimes

works for 1 st^ Order

Eqns

 The Function on the

RHS of the 1 st^ Order

ODE is the

FORCING Function

• Function only of t
• Can be a CONSTANT

( )

( )

∫ ∫

=

=

=

=

=

y t t

y

d d

β

β

α

α

α β β 0 0

1 2

[ ] (^) ( )

( ) [ ]

y t t y 0

2 α (^0) = 7. 37 = β

( )

Or ( ) 7. 37

1. 37 0 2

2 2

= +

− = −

y t t

y t t

1 st^ Order Response Eqns

• Given y (^) p and yc then the TOTAL Solution to the ODE

y ( ) t = yp ( ) t + yc ( ) t

 Consider the Case Where the Forcing Function is a Constant

• f(t) = A

 Now Solve the ODE in Two Parts for y (^) p & yc

 For the Particular Soln, Notice that a CONSTANT Fits the Eqn:

( ) ( )

( )

• ( ) = 0

• =

y t dt

dy t

y t A dt

dy t

c

c

p

p

τ

τ

( )

[ ( )] (^) [ ] 0

and

1

1

dt

d K dt

d y t

y t K

p

p

1 st^ Order Response Eqns cont

• Sub Into the General (Particular) Eqn y (^) p and dy (^) p/dt

K A

K A

1

1

or

 Next, Divide the Homogeneous Eqn by τ·yc to yield

 Next Separate the Variables & Integrate

( ) ( )

0

1

• = y t τ

dy t dt

c

c

( )

dy ( ) t dt ∫ (^) y (^) c t c ∫

τ

 Recognize LHS as a Natural Log; so

( ( ))

where const

ln

D

y (^) c t t τ D

 Next Take “e” to The Power of the LHS & RHS

2 nd^ Order Linear Equation

• Need Solutions to the 2 nd Order ODE

 As Before The Solution Should Take This form

 If the Forcing Fcn is a Constant , A , Then Discern a Particular Soln

 Verify y (^) p

2 ( ) ( ) ( ) (^ )

2 t ky t f t dt

dy t c dt

d y m + + =

y ( t ) = yp ( t ) + yc ( t )

 Where

• y (^) p ≡ Particular Solution
• y (^) c ≡ Complementary Solution

( ) k

A f t = A ⇒ yp =

A k

ky k A

dt

d y dt

dy k

y A

p

p p p

⇒ = =

= ⇒ = 2 = 0

2

 For Any const Forcing Fcn, f(t) = A

( ) y ( t ) k

A

y t = + c



The Complementary Solution

• The Complementary Solution Satisfies the HOMOGENOUS Eqn

 Look for Solution of this type

 Need y (^) c So That the “0th”, 1 st^ & 2nd Derivatives Have the SAME FORM so they will CANCEL (i.e., Divide-Out) in the Homogeneous Eqn

2 ( ) ( ) ( )^0

2

+ t + ky t =

dt

dy

t c

dt

d y

m

st y ( t ) = Ge  Sub Assumed Solution (y = Ge st) into the Homogenous Eqn

mGest^ s^2 + cGests + kGest = 0  Canceling Ge st

0

2 ms + cs + k =

 The Above is Called the Characteristic Equation

Complementary Solution cont.

• Given the “Roots” of the Homogeneous Eqn

 In the Unstable case the response will grow exponentially toward ∞

• This is not terribly interesting  If the Solution is Stable, need to Consider three Sets of values for s based on the sign of γ
• 1. γ > 0 → s 1 , s 2 REAL and UNequal roots
• 1. γ = 0 → s 1 = s 2 = s; ONE REAL root
• 1. γ < 0 → Two roots as COMPLEX CONJUGATES

 Can Generate STABLE and UNstable Responses

• Stable

m

c

s

1 , 2

− ± γ

m

c

• UNstable (^0)

m

c

Complementary Soln Cases 1&

• For the Linear, 2nd^ Order, Constant Coeff,

Homogenous Eqn

 By the Methods of MTH4 & ENGR43 Find

Solutions to the ODE by discriminant case:

2 ( ) ( ) ( )^0

2

+ t + ky t =

dt

dy

t c

dt

d y

m

1. Real & Unequal Roots (Stable for Neg Roots)

s t s t

yc t A e A e 1 2 ( ) = 1 + 2

2. Single Real Root (Stable for Neg Root)

( )

st

yc ( t ) = A 1 + A 2 t e

2 nd^ Order Solution

• For the Linear, 2nd

Order, Constant Coeff,

Homogenous Eqn

 Can Find Solution

based Upon the

nature of the Roots

of the Characteristic

Eqn

2 ( ) ( ) ( )^0

2

+ t + ky t =

dt

dy

t c

dt

d y

m

ms^2 + cs + k = 0

y ( t = 0 ) = VALUE

 To Find the Values of

the Constants Need

TWO Initial

Conditions (ICs)

• The ZERO Order IC
• The 1st^ Order IC

y ( ) VALUE dt

dy

t

= =

=

0 0

Properly Apply Initial conditions

• The IC’s Apply ONLY to the TOTAL Solution
• Many times It’s EASY to forget to add the

PARTICULAR solution BEFORE applying the IC’s

• Do NOT neglect yp(t) prior to IC’s

y ( t ) y ( t ) y ( t ) p c

= +

2 nd^ Order ODE Example - 2

• From the

Zero Order IC

 To Use the 1 st^ Order

IC need to take

Derivative

 Now Have 2 Eqns

for A 1 & A 2

 Then at t = 0

y ( ) 0 = A 1 e^0 + A 2 e^0 = 0. 5

[ ( ) ]

t t

t t

Ae A e

dt

dy

y t Ae A e

dt

d

6 2

3 1

6 2

3 1

− −

2

17 3 1 0 6 2 0 0

− = − − = =

Ae A e dt

dy

t

3 6 8. 5

1. 5

1 2

1 2 − − = −

• =

A A

A A

 Solve w/ MATLAB

BackDivision

2nd Order ODE Example - 3

• MATLAB session

 Or

 The Response Curve

**>> C = [1,1; -3,-6]; >> b = [0.5; -8.5]; >> A = C\b A = -1.

>> A_6 = 6A A_6 = -11. 14.*

( ) (^) ; 0 6

14 6

(^11 3 ) y t = − e −^ t + e − t t >

-0.4 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

-0.

-0.

-0.

0

t

y(t)

Be sure to check for correct IC’s  Starting-Value & Slope