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# Week 8 [compatibility mode]

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### Transcript of "Week 8 [compatibility mode]"

1. 1. KNF1023 Engineering Mathematics II Second Order ODEs Prepared By Annie ak JosephPrepared ByAnnie ak Joseph Session 200782009
2. 2. Learning Objectives Explain about variation of parameters method in the inhomogeneous ODEs Explain about Spring Mass System and Electric Circuits
3. 3. Method of Variation of Parameters There is a more systematic approach for finding a particular solution of the 2nd order linear inhomogeneous ODE. y "+ f ( x) y + g ( x) y = r ( x) The approach known as the method of variation of parameters makes use of two linearly independent solutions of the corresponding homogeneous ODE.
4. 4. Method of Variation of ParametersThe solution of the second order inhomogeneousODEs is y= y p ( x) + Y ( x) { { a particular solution general solution of the of in hom ogeneous ODE corresponding hom ogeneous ODE y p = u ( x)Y1 ( x) + v( x)Y2 ( x)
5. 5. Method of Variation of ParametersThus, the variation parameters will involvedtwo equations which is r (x )y 2 (x ) u(x ) = −∫ dx ( y1 y 2 − y1 y 2 )and r ( x ) y1 ( x ) v( x ) = ∫ dx ( y1 y 2 − y1 y 2)
6. 6. Example 1Consider the inhomogeneous ODEy − y = 3 exp(x ) Notice that y − y = 0 has two linearlyindependent solutions given by y1 = exp(− x ) and y 2 = exp( x )Hencey1 y2 − y1 y2 = exp ( − x ) exp ( x ) − ( − exp ( − x ) ) exp ( x ) = 2
7. 7. Continue…If we look for a particular solution of theinhomogeneous ODE in the form y p = u ( x )e − x + v ( x )e xthen u and v are given by 3exp( x) exp( x) 3 u ( x) = − ∫ dx = − exp(2 x) 2 4and 3 exp( x ) exp( − x ) 3 v( x) = ∫ dx = x 2 2
8. 8. Continue…Notice that we can ignore the constants ofintegration, since we are only interested in gettinga particular solution.So, a particular solution of the inhomogeneousODE y "− y = 3exp( x) is y p = u ( x) exp(− x) + v( x) exp( x) 3 3 = − exp(2 x) exp(− x) + x exp( x) 4 2 3 3 = − exp( x) + x exp( x) 4 2
9. 9. Continue…The required general solution of theinhomogeneous ODE is 3 3 y = − exp( x) + x exp( x) + C exp(− x) + D exp( x) 4 2Where C and D are arbitrary constants.This can again be rewritten as 3  3 y = x exp( x) + C exp(− x) +  D −  exp( x) 2  4 3 = x exp( x) + A exp(− x) + B exp( x) (as before) 2 3Where , A = C and B = D − ,are arbitrary constants 4
10. 10. APPLICATIONS OF SECOND-ORDER EQUATIONS -Spring –Mass SystemSimple harmonic motion may be defined asMotion in a straight line for which the accelerationIs proportional to the displacement and in theopposite direction. Example of this type of motionare a weight on a spring.Damped SystemIf we considered a mass-spring system andmodeled it by the homogeneous linear ODE, my"+cy + ky = 0
11. 11. APPLICATIONS OF SECOND-ORDER EQUATIONS -Spring –Mass SystemHere y(t) as a function of time t is thedisplacement of the body of mass m from rest.Forces within this system are inertia my”, thedamping force cy’ (if c>0) and the spring force kyacting as a restoring forces.The corresponding characteristics equation is 2 c k λ + λ+ =0 m mThe roots are c 1 λ1, 2 =− ± c 2 − 4mk 2m 2m
12. 12. APPLICATIONS OF SECOND-ORDER EQUATIONS -Spring –Mass SystemUsing the abbreviated notations c 1 α= and β= c 2 − 4mk 2m 2mWe can write λ1 = −α + β λ2 = −α − β
13. 13. APPLICATIONS OF SECOND-ORDER EQUATIONS -Spring –Mass SystemThe form of the solution will depend on thedamping, and,Case I. c2> 4mk. Distinct real roots λ1, λ2(Overdamping)Case II. c2< 4mk. Complex conjugate(Underdamping)Case III. c2 = 4mk. A real double root(Critical damping)
14. 14. APPLICATIONS OF SECOND-ORDER EQUATIONS -Spring –Mass System
15. 15. APPLICATIONS OF SECOND-ORDER EQUATIONS -Spring –Mass SystemCase 1. Overdamping.When the damping constant c is so large that c2 >4mk, then λ1,λ2 are distinct real roots, the generalsolution is − (α − β ) t − (α + β ) t y (t ) = c1e + c2eCase II.Underdamping.If the damping constant cis so small that c2< 4mk then β is pure imaginary.The roots of the characteristics equation arecomplex conjugate,
16. 16. APPLICATIONS OF SECOND-ORDER EQUATIONS -Spring –Mass System λ1 = −α + jω λ2 = −α − jωThe general solution is y (t ) = e −α t ( A cos ωt + B sin ωt )
17. 17. APPLICATIONS OF SECOND-ORDER EQUATIONS -Spring –Mass SystemCase III. Critical damping.If c2 = 4mk, then β=0, λ1 = λ2 = − α , and thegeneral solution is y (t ) = (c1t + c2 )e −αtUndamped SystemIf the damping of the system is so small and can bedisregarded, then my"+ky = 0The general solution is y (t ) = A cos ω0t + B sin ω0t
18. 18. Example 2In testing the characteristics of a particulartype of spring, it is found that a weight of 4.90 Nstretches the spring 0.49 m when the weight andspring are placed in a fluid which resists the motionwith a force equal to twice the velocity. If theweight is brought to rest and then given a velocityof 12 m/s, find the equation of motion.
19. 19. Continue…Using Newton’s second law, we have my"+cy + ky = 0 my"+2.00 y + ky = 0 my" = −2.00 y − kyThe weight is 4.90 N and the acceleration due togravity is 9.80 m/s2. Thus, the mass, m ism=4.90/9.80 = 0.500 kgThe constant k is found from the fact that thespring stretches 0.49 m for a force of 4.9 N, Thus,using Hooke’s Law,
20. 20. Continue…4.90 = k (0.490)k = 10.0 N/mThis means that the differential equation to besolved is 0.500 y"+2.00 y +10 y = 0 1.00 y"+4.00 y +20.0 y = 0Solving this equation, we have
21. 21. Continue… 21 . 00 m + 4 . 00 m + 20 . 0 = 0 − 4 . 00 ± 16 . 0 − 4 ( 20 . 0 )( 1 . 00 )m = 2 . 00m = − 2 . 00 ± 4 . 00 j − 2 . 00 ty = e ( c 1 cos 4 . 00 t + c 2 sin 4 . 00 t )
22. 22. Continue…Since the weight started from the equilibriumposition with a velocity of 12.0 m/s, we know thaty= 0 and y’ = 12.0 for t = 0. Thus, 0 0 = e (c1 + 0c2 ) c1 = 0Thus, since c1=0, we have −2.00ty = c2e sin 4.00ty = c2e− 2.00t (cos 4.00t )(4.00) + c2 sin 4.00t (e− 2.00t )(−2.00)12.0 = c2e0 (1)(4.00) + c2 (0)(e0 )(−2.00)c2 = 3.00
23. 23. Continue…This means that the equation of motion is −2.00 t y = 3.00e sin 4.00t
24. 24. Application on Electric CircuitGiven the RLC circuit as below,(a) Show that the circuit can be modeled as d 2i di i L 2 + R + = E (t ) dt dt C ()(b) Given that E t = −0.01e − t Solve the differential equation for the initial conditions, when and t = 0,i = 0 di = 0 dt
25. 25. Solution(a) KVL : VL + VR + VC = E ( t ) di 1 L + iR + ∫ idt = E ( t ) differential with respect dt C to t. 2 d i di i L 2 + R + = E (t ) dt dt C(b) Given E ( t ) = −0.01e , hence −t E ( t ) = 0.01e − t d 2i di i 0.001 2 + 0.005 + = 0.01e − t dt dt 250 d 2i di + 5 + 4i = 10e − t dt 2 dt
26. 26. Continue… i + 5i + 4i = 0 2 λ + 5λ + 4 = 0 λ = −4@− 1 −4 t −t I = Ae + Be For Particular Solution i p = α te − t i p = α e − t − α te −t i = −α e − (α e − α te p −t −t −t ) = −2α e −t + α te −t
27. 27. Continue…i + 5i + 4i = ( −2α e − t + α te− t ) + 5 (α e − t − α te − t ) + 4 (α te− t ) −t = 3α eCompare to 10e − t 3α = 10 10 α= 3 10 − tThus i p = te 3
28. 28. Continue…Thus, the general solution is −4 t −t 10 −t i = i p + I = Ae + Be + te 3Given i ( 0 ) = 0 and i ( 0 ) = 0 −4( 0 ) −0 10 −00 = Ae + Be + 0e , A + B = 0 − − − (1) 3 −4 t −t 10 − t 10 − ti = −4 Ae − Be + e − te 3 3 −4( 0 ) 10 −0 10 100 = −4 Ae − Be + e − ( 0 ) e , 4 A + B = − − − (2) −0 −0 3 3 3
29. 29. Continue… 10 10 A= ,B = − 9 9 10 −4t 10 −t 10 − t i ( t ) = e − e + te 9 9 3 Try to use variation parameter to solve this application question and check whether the answer is the same or not…
30. 30. Prepared By Annie ak JosephPrepared ByAnnie ak Joseph Session 2007/2008
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