Week 8 [compatibility mode]
- 1. KNF1023 Engineering Mathematics II Second Order ODEs Prepared By Annie ak Joseph Prepared By Annie ak Joseph Session 200782009
- 2. Learning Objectives Explain about variation of parameters method in the inhomogeneous ODEs Explain about Spring Mass System and Electric Circuits
- 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. Method of Variation of Parameters The solution of the second order inhomogeneous ODEs 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. Method of Variation of Parameters Thus, the variation parameters will involved two 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. Example 1 Consider the inhomogeneous ODE y − y = 3 exp(x ) '' Notice that y '' − y = 0 has two linearly independent solutions given by y1 = exp(− x ) and y 2 = exp( x ) Hence y1 y2 − y1' y2 = exp ( − x ) exp ( x ) − ( − exp ( − x ) ) exp ( x ) = 2 '
- 7. Continue… If we look for a particular solution of the inhomogeneous ODE in the form y p = u ( x )e − x + v ( x )e x then u and v are given by 3exp( x) exp( x) 3 u ( x) = − ∫ dx = − exp(2 x) 2 4 and 3 exp( x ) exp( − x ) 3 v( x) = ∫ dx = x 2 2
- 8. Continue… Notice that we can ignore the constants of integration, since we are only interested in getting a particular solution. So, a particular solution of the inhomogeneous ODE 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. Continue… The required general solution of the inhomogeneous ODE is 3 3 y = − exp( x) + x exp( x) + C exp(− x) + D exp( x) 4 2 Where 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 3 Where , A = C and B = D − ,are arbitrary constants 4
- 10. APPLICATIONS OF SECOND-ORDER EQUATIONS - Spring –Mass System Simple harmonic motion may be defined as Motion in a straight line for which the acceleration Is proportional to the displacement and in the opposite direction. Example of this type of motion are a weight on a spring. Damped System If we considered a mass-spring system and modeled it by the homogeneous linear ODE, my"+cy '+ ky = 0
- 11. APPLICATIONS OF SECOND-ORDER EQUATIONS - Spring –Mass System Here y(t) as a function of time t is the displacement of the body of mass m from rest. Forces within this system are inertia my”, the damping force cy’ (if c>0) and the spring force ky acting as a restoring forces. The corresponding characteristics equation is 2 c k λ + λ+ =0 m m The roots are c 1 λ1, 2 =− ± c 2 − 4mk 2m 2m
- 12. APPLICATIONS OF SECOND-ORDER EQUATIONS - Spring –Mass System Using the abbreviated notations c 1 α= and β= c 2 − 4mk 2m 2m We can write λ1 = −α + β λ2 = −α − β
- 13. APPLICATIONS OF SECOND-ORDER EQUATIONS - Spring –Mass System The form of the solution will depend on the damping, 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. APPLICATIONS OF SECOND-ORDER EQUATIONS - Spring –Mass System
- 15. APPLICATIONS OF SECOND-ORDER EQUATIONS - Spring –Mass System Case 1. Overdamping. When the damping constant c is so large that c2 > 4mk, then λ1,λ2 are distinct real roots, the general solution is − (α − β ) t − (α + β ) t y (t ) = c1e + c2e Case II.Underdamping. If the damping constant c is so small that c2< 4mk then β is pure imaginary. The roots of the characteristics equation are complex conjugate,
- 16. APPLICATIONS OF SECOND-ORDER EQUATIONS - Spring –Mass System λ1 = −α + jω λ2 = −α − jω The general solution is y (t ) = e −α t ( A cos ωt + B sin ωt )
- 17. APPLICATIONS OF SECOND-ORDER EQUATIONS - Spring –Mass System Case III. Critical damping. If c2 = 4mk, then β=0, λ1 = λ2 = − α , and the general solution is y (t ) = (c1t + c2 )e −αt Undamped System If the damping of the system is so small and can be disregarded, then my"+ky = 0 The general solution is y (t ) = A cos ω0t + B sin ω0t
- 18. Example 2 In testing the characteristics of a particular type of spring, it is found that a weight of 4.90 N stretches the spring 0.49 m when the weight and spring are placed in a fluid which resists the motion with a force equal to twice the velocity. If the weight is brought to rest and then given a velocity of 12 m/s, find the equation of motion.
- 19. Continue… Using Newton’s second law, we have my"+cy '+ ky = 0 my"+2.00 y '+ ky = 0 my" = −2.00 y '− ky The weight is 4.90 N and the acceleration due to gravity is 9.80 m/s2. Thus, the mass, m is m=4.90/9.80 = 0.500 kg The constant k is found from the fact that the spring stretches 0.49 m for a force of 4.9 N, Thus, using Hooke’s Law,
- 20. Continue… 4.90 = k (0.490) k = 10.0 N/m This means that the differential equation to be solved is 0.500 y"+2.00 y '+10 y = 0 1.00 y"+4.00 y '+20.0 y = 0 Solving this equation, we have
- 21. Continue… 2 1 . 00 m + 4 . 00 m + 20 . 0 = 0 − 4 . 00 ± 16 . 0 − 4 ( 20 . 0 )( 1 . 00 ) m = 2 . 00 m = − 2 . 00 ± 4 . 00 j − 2 . 00 t y = e ( c 1 cos 4 . 00 t + c 2 sin 4 . 00 t )
- 22. Continue… Since the weight started from the equilibrium position with a velocity of 12.0 m/s, we know that y= 0 and y’ = 12.0 for t = 0. Thus, 0 0 = e (c1 + 0c2 ) c1 = 0 Thus, since c1=0, we have −2.00t y = c2e sin 4.00t y' = 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. Continue… This means that the equation of motion is −2.00 t y = 3.00e sin 4.00t
- 24. Application on Electric Circuit Given 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. 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. 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. Continue… i '' + 5i ' + 4i = ( −2α e − t + α te− t ) + 5 (α e − t − α te − t ) + 4 (α te− t ) −t = 3α e Compare to 10e − t 3α = 10 10 α= 3 10 − t Thus i p = te 3
- 28. Continue… Thus, the general solution is −4 t −t 10 −t i = i p + I = Ae + Be + te 3 Given i ( 0 ) = 0 and i ' ( 0 ) = 0 −4( 0 ) −0 10 −0 0 = Ae + Be + 0e , A + B = 0 − − − (1) 3 ' −4 t −t 10 − t 10 − t i = −4 Ae − Be + e − te 3 3 −4( 0 ) 10 −0 10 10 0 = −4 Ae − Be + e − ( 0 ) e , 4 A + B = − − − (2) −0 −0 3 3 3
- 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. Prepared By Annie ak Joseph Prepared By Annie ak Joseph Session 2007/2008