Differential equations

Transcription

1 Differential equations First order differential equation Imlicit F(, y, y ) = or Elicit y = f ( y, ) In general = hysical laws involving a rate of change (with time) of an event = function

2 Solution of DE = Function y= h Derivable y h = and satisfy DE on interval a< < b Conditions for DE to have a solution are general Many simle DE do not have a solution E. ( y ) = no solution E. ( y ) y + y = General solution: y= c c family of straight lines; Singular solution: y = arabola 4 General solution: many solutions differing by constant family of curves Singular solution: additional solution not obtained from general solution E. Radioactive decay Observed decay with time of radioactive substance to amount resent dy ky dt = dy k is a constant (half life) deending on radioactive substance; since dt < k < General solution: y kt = ce y = cke kt = ky t Verification: Particular solution: at t =, y = Y y = ce = c c= Y yt = Ye kt

3 Geometrical interretations y of y- y = f ( y, ) is the sloe of y( ) if y( ) solution assing through (, ) lane then sloe at that oint is f (, y ) f (, ) solution curves Isoclines: (, ) y = f y = k= constant E. y = y, the isoclines are hyerbolas y = direction field (or sloe field) of General solution eist: y = ce ; verification: = = y ce ( ) y E. van der Pol equation of electronics y.( ) = y Direction fields give all solutions with limited accuracy More accurate method = numeric method Allows to solve DE for which we have no solution E Euler-Cauchy method and Runge- Kutta method 3

4 Methods to find general solutions Case : Searable DE g y y = f g y dy = f d dy Solution: g( y) d= f d+ C d Integrals eist rovided g and f continuous dy d = dy g y dy = f d + C d but since E. 9yy + 4 = 9ydy+ 4d = Solution: or 9ydy+ 4d+ C = y + = C 4 9 family of ellises centered on (,) E. dy y = + y = d+ C + y Solution: arctan y = + C y = tan( + C) Note y = tan + C is not a solution E. 3 y = ky with k > dy k+ C Solution: = k d+ C ln y = k+ C y = e = Ce y C C Where C =+ e for y > and C = e y < k 4

5 E. 4 Initial value roblem initial condition (IC) y DE: y = IC: y = Solution: dy = d ln y = ln + C = ln + C C y = y C IC = C = y = E. 5 Initial value roblem DE: y y Solution: y = = IC: dy = ln = + = = y c e =+ C for y > c Where e = C for y < C = for y = IC: y y e + C d y C y e Ce = = Bell-shaed curves 5

6 Case : Reduction to searable form y y = g Solution use transformation y u = y = u y = u + u u + u= g u u = g u u du d = g u u E. y y yy = y y = = y y u + u + u = u u = u+ = u u u udu d = ln+ u = ln + C + u C y C + u = + = + y = C C C + y = 4 Family of circles, tangent to origin, with centers on -ais 6

7 Construction of mathematical models Radio carbon dating 4 Based on isotoe C 6 Rendered radioactive by cosmic rays bombardment 4 In the air the ratio of the isotoes C C is constant Princile of method: When living organism dies the absortion (breathing and eating) of The ratio of C 4 ends 6 4 C C measured comared to value in air to date of death Half life of 4 6C is 573 yrs 4 ± yrs (error of 7%) Mathematical model: kt y = ky yt = ye where y is the initial amount of C 4 6 From definition of half life time at which the level is ½ the original one: k 573 k ye = y e = k = ln. yr 573 Measurement: if remains contain 5% of original level.t ln.5 ye =.5 y t = yrs 46± 8yrs. Comarisons with other dating methods suggest t 3 yrs value given by model is too low, why? 7

8 Miing roblem A thank contains gal of water in which 4 lb of salt is dissolved. Each minute 5gal of water, with lb/gal of salt is added, and the same amount is retrieved yt lb Problem: amount of salt as a function of time: Simlification: the miing rocess is instantaneous and erfect y t = dy = dt Mathematical model: rate of change inflow outflow Inflow: lb/min Outflow: since one gal of water contains y( t) lb gal yt 5yt lb 5 gal coming out will contain = =.5yt 4 min y =.5yt Rate of change: IC: y = 4 lb Solution: y =.5( y 4) dy =.5dt ( y 4) = + = +.5t ln y 4.5t C y Ce 4 IC: y = 4 C = 36.5t yt 4 36e = Solution increases with time from 4 to 4. Equilibrium reached when amount going in = amount going out y = ; 8

9 Heating roblem (Newton s law of cooling) The time rate of change of temerature of a body is described by the following equation: dt T = Tbody TMedium dt Model assuming ambient temerature constant at T medium = 3 dt dt = k( T 3 body ) = kdt dt T 3 ( body ) ln T 3 = kt + C T = 3+ Ce body body kt F IC: T() = T = 66 F 3 F+ C = 66 F C = 34 F T t = 3 F+ 34 F e kt Observation: after t = h T = 63 F body 63 3 T = + e = e = k ( h ) = ln.9765=.4687 h k k h 3 F 34 F 63 F.9765 Rate of change of temerature decreases with time as the body temerature aroach temerature of ambient Equilibrium = temerature with the ambient t = h For T e.4687 h = 3 F+ 34 F = 53.4 F 9

10 Velocity escae from Earth MODEL = Newton s law of gravitationma = GMm r At the surface of Earth with radius R : At any radius a( r) gr = r a = g = GM R Alying definition of acceleration and chain rule dr v R Searation of variable DE vdv= gr = g + C r r v IC: on surface of the Earth r = R and v= v C = gr gr General solution: v = + v gr r dv dv dr dv R a r = = = v= g dt dr dt dr r When v = the rojectile stos and reverse trajectory (fall back to Earth) v However, if we take v = gr, then gr = and ositive reaching only at the infinite v gr = > and v remains r Escae velocity: v = gr

11 Eact DE General form: M( yd ) Solution: u(, y), + N, y dy = = C, such that u u du = d+ dy= Md+ Ndy = y Condition (necessary and sufficient): M and N have continuous first artial derivatives N u u M and = = = y y y Solution: if M( yd, ) + N(, y) dy = then u = Md+ k ( y) role of constant to determine k( y ) derive u y Note: we could have started with u = Ndy+ l 3 3 E. + 3y d+ 3y + y dy = M N Verification: = = 6y y where k( y ) lays the to get dk and integrate. dy 3 First integration: u = ( 3 + 3y ) d+ k ( y) 4 + y + k( y) Determination of k: General solution: 4 4 u dk 3 dk y = + = + = = + y y y y k C y dy dy y + y = C 4 4

12 E. (sin cosh yd ) (cossinh ydy ) Verification: y = 3 =, with IC M N = sinsinh y = y Integration: u = ( sin cosh y) d+ k ( y) = coscosh y + k ( y) u dk Determination of k: = cossinh y + = N = cossinh y y dy dk = k = C dy te u, y = coscosh y+ C and since du = u = C coscosh y = C IC: coscosh3= C.7 coscosh y =.7

13 Reduction to eact form using integrating factors If P( y) d, + Q, y dy = not EDE Multilication by integrating factor (, ) Note: may eist more than one integrating factor F y makes DE eact FPd+ FQdy = Conditio n: FP = FQ FP y + FPy = FQ + FQ y df F = F Fy = F = d Easy to solve if and df df P Q FPy = Q+ FQ = R d F d Q = y Th. if right side deends only on then F = e R d Note: similarly we could have worked with F( y ) for convenience sake E. + = with IC: y sin y d ycosy dy = π Verification: M y N = 4ycosy = ycos y not eact DE We have P = sin y P = 4y cos y and y Q= ycos y Q = ycos y Since R P Q 3ycosy 3 = Q y R = 4ycosy ycos y ycosy = = ycos y Therefore F = e R d e e[ 3ln ] 3 F = d = = 3 3

14 3 4 Multilication by integrating factor: Verification: M sin y d+ ycosy dy = N 3 = 4ycos y = y, eact Solution: u = 3 sin( y ) d+ k ( y) = 4 sin y + k( y) = + = = = u 4 cos 4 cos y y y k y y y k y k y C 4 u(, y) = sin y = C π = = = = IC: 4 4 sin 8 C u, y siny 8 4

15 Linear DE Linear DE: y + y= r Homogeneous linear DE: r( ) = Solution: searation of variables If non homogeneous: ( y r) d+ dy = Pd+ Qdy = df P Q = F e d F d Q = = y Integrating factor: d d d e ( y + y) = e y e = r d d e y = e rd+ C Putting h = h h d a general solution is: y = e erd+ C E. y y = e = and = + e y e e d C r = e h = d = { } { } y = e e e d+ C = e e + C = Ce + e More simly using integration factor = y ye = e e ye = e ye = e + C y= e + Ce ( ) e 5

16 E. y + y= e ( 3sin+ cos) = h= = + + = + = + 3 y e e e 3sin cos d C e e sin C Ce e sin Integration: + = + By art: udv = uv vdu The second integral yields: e 3sin cos d e 3sind e cosd 3 3 = = 3 and = cos = sin u e du e dv v + = + = e 3sind e cosd e 3sind e sin e 3sind e sin = + = + 3 y e e sin C Ce e sin E. 3 y tan sin y = + = with IC: tan tan ln sec h h = h = d = e = sec e = cos r = sin= sincos h er = sec sin cos = sin = + = IC: y cos sin d C Ccos cos = = C C 3 = y 3cos cos Reduction to linear form Bernoulli equation: + = y y g y Where a is any real number for a = or a = yields linear DE a Transformation: u = y a u ( a) y a = y = ( a) y a ( gy a y ) u = ( a)( g y a ) ( ) ( ) u + a u = a g 6

17 E. Verhulst equation (logistic oulation model) y Ay = By where A and B are ositive constant Bernoulli form with a = u = y u = y y = y By + Ay = B Ay u + ua= B, which is linear Using revious method: = A h= A r = B B A A A A A A u = e Be d+ C = e e + C = Ce + General solution: y = = u B + Ce A A B A Interretation: utting For B = the oulation grows eonentially The breaking term Initially small oulations y oulations y = t we find the logistic law of oulation growth At y = e (law of Malthus) C By revent the oulation to grow without bound A < < increases monotone to A B B A > decrease to same value B while large 7

18 Autonomous DE In the logistic equation, time, the indeendent variable, does not aear elicitly y = f ( t, y) = y f ( y) = = autonomous DE Autonomous DE has constant solutions = equilibrium solutions (or equilibrium oints), te determined by zeros (critical oints) of f ( y ), because f ( y) = y = y = C An equilibrium solution can be stable solutions close to it for some t remain close to it for all further t (e y = 4 in logistic equation) unstable solutions initially close to it do not remain close to it as t increases (e. y = in logistic equation) E. hase line lot y = y y Stable equilibrium solution at y = Unstable equilibrium at y = We can condense the direction field (A) to a hase line lot (C) giving y and y and direction of arrows (B) showing equilibrium 8

19 Orthogonal trajectories Imortant class of roblem in hysics = find family of curves that intersect a given family of curves at right angle = orthogonal trajectories E. Isotherms, heat flows, equiotential curves, curves of steeest descent, etc. Ste : find DE for which the given curves are solution curves y = f ( y, ) Ste : write down DE of orthogonal trajectories Ste 3: solve DE of orthogonal trajectories y = f ( y, ) E. one arameter family curves Fyc,, = y c = or y = c family of arabola Ste : y = c not good because we must eliminate arameter y y = c y = y = y 3 Ste : DE of orthogonal trajectories y = y Ste 3: orthogonal trajectories ellises + y = c = = + one arameter family of ydy d y c 9

20 Eistence and uniqueness of solutions For DE, in most cases, general solutions eists For an IVrb there eist 3 ossibilities: ) No solution ) Only one solution 3) Infinity of solution Problem of eistence: conditions for at least one solution Problem of uniqueness: conditions for at most one solution Th. eistence theorem If f ( y, ) continuous at all oints (, ) y in R: < a and y y < b and bounded in R: f (, y) K then the IVrb has at least one solution defined for all in interval < α where α smaller than two numbers a and b K Th. Uniqueness theorem If f (, ) f M y y and f y continuous for all (, ) y in R and bounded f K and then the IVrb has at most one solution y defined at least for all in interval < α. a) b a α = a; b) b < a α = b, for larger or smaller s nothing can be deduced K K K

21 Linear DE of nd order Many imortant alications in hysics Theory tyical of that of linear DE in general etension to higher order is straightforward General form: y + y + q y= r Homogeneous: y + y + q y= Solution: function h( ) in interval I: a< < b and with continuous derivatives h and h in I E. y y = with y= e or y= e as solutions General solution = linear combination: y = ce + ce Suerosition or linearity rincile Th. Fundamental theorem for the homogeneous linear DE nd For homogeneous linear DE of nd any linear combinations of solutions (or sum and constant multiles) on an oen interval are solutions

22 IVrb - General solution as basis Solution: y= cy + cy IC conditions: y( ) = k and y = k E. y y = with IC: y = 4 and Alying IC to solution: y= ce + ce y = c + c = 4 y = ce ce y = c c = c = and c = 3 y = General solution: y= e + 3e A general solution of homogeneous linear DE of nd order is a solution of the tye y= cy + cy, where y ky The two solutions form a basis y ( ) and y ky ky + = imlies k = k = are linearly indeendent E. 3 y y = basis: y = e and y= e solution: y = ce + ce y + y = basis: y = cos and y = sin solution: y= ccos+ csin

23 Reduction of order Method to find y when y is known y + y + qy = Set y = uy y= uy+ uy y = uy+ uy+ uy Substitute in DE: y + y + qy = ( ) ( ) uy u ( y y ) u( y y qy ) uy + uy + uy + uy + uy + quy = = y + y u + u = y y Or U + + U = y du y Searation of variables: = d lnu ln y d U e + = U y = y Because U = u and y = uy y = y Ud y Since u Ud y = =, the ratio is not constant and the two solutions form a basis. E. 4 y y + = or y + = y y y By insection y Then = since y = and y = = d= ln ln And U = e = y = Ud = ln General solution: y = c + c ln d A articular solution describes the unique behavior of a given hysical system. If, q and r are continuous function on I then there always eist a solution on I which is unique (no singular solution eists) 3

24 Homogeneous linear DE of nd order with constant coefficients General form: y + ay + by = As a solution we try: y= e y = λe y = λ e λ λ λ Relacing in the equation: ( λ λ ) Solution ( λ aλ b) + a + b e λ = + + = (characteristic equation) Roots: ( a a 4b) ( a a 4b) λ = + λ = So the basis are: y = e( λ ) and y = e( λ ) Deending on the sign of the discriminant ) > distinct real roots ) = real double root 3) < comle conjugate roots a 4b 3 solutions ossible: λ λ Case distinct real roots y = ce + ce E. DE: y y y + = IC: y = 4 and λ = ( + 9) = λ = 9 = y = ce + ce y = ce ce y = 5 IC: y = c + c = 4 y = c c = 5 c = and c = 3 y= e + 3e 4

25 Case real double root a λ = λ = λ = A first solution y e a = Second solution: y = uy y = uy + uy y = uy + uy + uy ( ) ( ) uy + uy + uy + auy + uy + buy =, which reduces to uy + u y + ay = a But since y = y uy = u = u = c + c Possible second solution is y y = and general solution is = ( + ) y c ce a E. y + 8y + 6y = Characteristic equation: General solution: = ( + ) λ y c ce + 8λ+ 6= double root: λ = 4 4 E. 3 IVrob: DE: y 4y 4y Characteristic equation: + = CI: y = 3 and λ 4λ + 4 = λ λ = General solution: = ( + ) = + ( + ) y c ce y ce c ce = + = = 5 IC: y = c = 3 y c c c y= 3 5 e Particular solution: y = 5

26 Case 3: a E. Basis: 4b < comle roots y + y = λ + = λ = λ =± =± i i e and e i Using Euler formula: i cos = e + e i sin = e e i i i i e = cos+ isin i e = cos isin cos and sin are also solutions Comle eonential function ( cos sin ) z s+ it s it s e = e = ee = e t+ i t Obtained from Maclaurin series of 3 4 e it = + it + it + it + it +! 3! 4! t t t t t t e it = i t + +! 4! 6! 3! 5! 7! e Since 4 6 t t t cost = + + and! 4! 6! t t t sint = t + + 3! 5! 7! it e = cost+ isin t 6

27 Case 3 (continue): comle roots To make radicand ositive: λ = a+ a 4b = a+ i 4b a = a + i b a 4 λ = a+ iω and λ = a iω where ω = b a 4 a a + iω ( ω ω ) λ e = e = e cos + isin a a iω ( ω ω ) λ e = e = e cos isin a Adding and dividing by : y = e cosω a Subtracting and dividing byi : y = e sinω General solution: y= e ( A cos ω+ B sin ω) a E. DE: y +.y + 4.y = IC: y = y = Characteristic equation: λ +.λ+ 4.= Comle roots:.± i ω = General solution:. y= e Acos+ Bsin IC: y = A= (.sin cos ) = +. y Be y = B = B = Particular solution:. y= e sin Interretation: damed vibration amlitude of oscillation decreases eonentially 7

28 Summary of cases to 3 Case Roots Basis General solution Distinct real λ, λ e, e y = ce + ce λ λ λ λ Real double a a λ = a e, e = ( + ) y c c e a Comle conjugate 3 λ = a+ iω λ = a iω e e a a cosω sinω ( cosω sinω ) a y= e A + B Note: in mechanics or electrical circuits, these 3 cases corresond to 3 different forms of motions or flows current 8

29 Differential oerator The differential can be seen as a transformation writing DE as Dy = y differential oerator D that oerates on y E. D = or D( sin) = cos D Dy = D y = y or Higher derivative: Second order DE oerator: L y = D + ad+ b y = y + ay + by Dy L= P D = D + ad+ b The oerator is linear: L[ αy+ βw] = αl( y) + βl( w) Homogeneous differential equation: L( y ) = Since λ D e = e D e = λ e λ λ λ λ = ( λ + λ+ ) = ( λ) P D e a b e P e Since λ λ λ e λ is solution P ( λ ) = Case : P ( λ ) different real roots Case : P ( λ ) double real root needs second indeendent solution 9

30 Start with = ( λ + λ+ ) = ( λ) P D e a b e P e Differentiating: Because P dp dλ For double root: λ λ λ = ( λ) + ( λ) = λ λ λ λ P D e P e P e P D e = ( λ) + ( λ) = ( λ+ ) + ( λ + λ+ ) P e P e a e a b e λ λ λ λ ( λ λ ) ( λ ) = + e + a + e + be = D + ad+ b e a e λ is second solution λ λ λ λ λ = P( λ) P ( λ) P( D)( e λ ) Since P ( λ ) is a olynomial P( D) = = = is olynomial oerator E. Factorize P( D) = D + D 6 and solve P( D)[ y ] = D D D D + 6= + 3 D y= y y D+ 3 D y= D+ 3 y y = y + y 6y Since D+ 3 y= y + 3y D D+ 3 y= D y + 3y = y + y 6y Similarly: 3 Solutions: ( D+ 3y) = y = e and D y = y = e + 3

31 Modeling: free oscillation (mass-sring system) Physical assumtions: mass of body >> mass of sring Orientation: ositive dislacements downward Hooke s law: F = ks where k > sring modulus and s stretch the smaller k the larger s Assume static equilibrium: F balance the weight W Alying Newton law: F + W = ks+ mg = = mg Putting y = at osition of equilibrium y measures dislacement of body from equilibrium osition Eeriment: ulling weight downward restoring force F Daming: dissiation of energy of mechanical system = ky 3

32 k Ignoring daming: my + ky= y + y = m yt = Acosω t+ Bsinω t where ω = Solution: Harmonic oscillation Putting B = sinδ and A = cosδ Solution transformed into yt = Ccos( ω t δ) k m B C = A + B and tanδ = A Period: π ω ω Frequency: π (Hertz) 3 ossible cases deending on initial velocity: y = or y =± V 3

33 Including daming dy daming force y = F = cy dt Newton s law: c k my + cy + ky= y + y + y = m m Characteristic equation: Roots: c λ λ, = ± c m m c k + λ + = m m 4mk c Putting α = and m β = c m λ = α+ β and λ = α β 4mk Formal solution deends on amount of daming: Case : over-daming Case : critical daming c > 4km c = 4km Case 3: under-daming c < 4km 33

34 Solution in case (Over-daming): ( α β ) t ( α β ) yt = ce + ce + t Both terms aroach zero as t no oscillation 34

35 Solution in case (Critical daming): = ( + ) Term ( c ct) yt c ct e αt + allows one zero system ass at most by y = osition Solution in case 3 (Under-daming): αt αt = ( cosω + sinω ) = cos ( ω δ) y t e A t B t Ce t Where k c m m 4m ω = 4mk c = Amlitude varies between curves y=± Ce αt. ω The smaller is c, the higher the frequency. At the limit c ω ω π 35

36 Euler-Cauchy equation General form: + + = where a and b are constant y ay by Trying the solution: = = = ( ) m m m mm + am + b = y y m y m m m m m m m m m m + am + b = Droing m + a m+ b= m factor m m Case : distinct real roots y = and y = m m General solution: y = c + c E..5. = y y y m m = 3.5. m =.5 and m = 4 c General solution: y= + c 4 36

37 Case : double root ( a) m ( a) = m + ( a ) m+ b 4 where b = ( a) First solution: y = a for second solution, substituting y = uy in y+ ay+ by = ( ) ( ) = uy uy uy a uy uy buy uy + u y + ay + u y + ay + by = y = a = a y y + ay = y ( a) Since last term is zero and ( ) ( ) u ( u + u ) y = = ln u = ln u = u= ln u Second solution: y = y ln General solution: y= ( c + c ) ( ) ln a E = y y y ( a) = m =, and solution = ( + ) y c cln 37

38 Case 3 comle conjugate roots: m = µ + iν and m ln i iν ν iνln = e = e = µ iν Using Euler formula: ( cos( νln ) sin( ν ln )) m µ iν µ iν ln µ = = e = + i ( cos( ν ln ) sin( νln )) m µ iν µ iν ln µ = = e = i Usual transformation yields real solutions: µ µ = ( ν ) and y = ( ν ) y cos ln sin ln E = y y y + + = = ± = ± m 6m 3 m, i 3 General solution: y = Acos( ln) + Bsin( ln) 38

39 Eistence and Uniqueness theory: Wronskian Homogeneous linear DE of nd : y + y + q y= General solution: y= cy + cy IVrob: y( ) = K and y = K Th. Eistence and uniqueness for IVrob If ( ) and q continuous on I and in I then eist solution y( ) on I Linear indeendent solutions Wronskian ky + ky = k = k = Linear indeendence: Wronskian determinant: (, ) y y W y y = = yy yy Th. linear deendence and indeendence Linear deendent W = for = Linear indeendent W Th. 3 general solution y y If ( ) and q continuous on oen interval I eists general solution Th. 4 Form of general solution Under conditions Th. 3, general solution Y cy cy basis and coefficients are suitable constants. No singular solution eists. = + where y and y are 39

40 Nonhomogeneous DE of nd General form: y + y + q y= r Th. Relation between nonhomogeneous DE and homogeneous DE a) The difference of solutions of an nonhomogeneous DE on some oen interval I is a solution of the equivalent homogenous DE on I b) The sum of a solution of a nonhomogeneous DE on I and of a solution of the equivalent homogenous DE on I is a solution of the nonhomogeneous DE General solution of nonhomogeneous DE on oen interval I is of the form: y = y + y h Where y = cy + cy = general solution of homogeneous DE and h any solution of nonhomogeneous DE without arbitrary constants Particular solution: c and c are values defined in y ( ) h y = A general solution includes all solutions If coefficients of nonhomogeneous DE of nd and r( ) are continuous in I Solution eists because both yh and y Once y established, IVrb solution is unique eist on I Th. Suose that coefficient and r continuous on I then every solution of nonhomogeneous DE is obtained by assigning suitable values to arbitrary constants in general solution of nonhomogeneous DE Practical conclusion: to solve a nonhomogeneous DE of nd order or IVrb: ) Solve the homogenous DE ) Find articular solution y ( ) 4

41 E. DE: y y y.4e + + = IC: y =. and y =.9 Ste solve homogenous DE λ λ λ, + + = = ± i ( cos sin ) = + yh e A B Ste determine articular solution of nonhomogeneous DE We try y = Ce + + Ce =.4e C =. y= y + y = e Acos+ Bsin +.e General solution: h Ste 3 articular solution y = A+.=. A= y = e cos Bsin sin+ Bcos +.e y = + B+.=.9 B= Particular solution: y= e cos+.e 4

42 Solution by undetermined coefficients Considering the form y + ay + by = r Choose for substituting in solution. Rules: y a form similar to r involving unknown coefficients to be determined by A) Basic If r( ) is one of the function in table below choose of table and determine coefficients by substituting y in second column y + derivatives in DE Terms in r( ) Choice for y ke γ n k ( n =,,,... ) ke ke kcosω ksinω α α cosω sinω Ce γ n K + K K+ K n n n K cosω+ Msinω ( cosω + sinω ) α e K M B) Modification rule if a term in y haens to be a solution of homogeneous DE then multily by and (double root) C) Sum rule if r( ) is a sum of a functions in table then use sum of y 4

43 E. y + 4y= 8 y = K + K+ K y = K Substituting: ( ) K + 4 K + K+ K = 8 Equating coefficients: 4K = 8, 4K =, K + 4K = K =, K = and K = y = General solution: y= y + y = A + B + h cos sin E. y 3y + y= e Characteristic equation: λ λ λ λ 3 + = = and = y = ce + ce h Choice for y cannot be Ce because this is solution of homogenous DE Trying (rule B) y = Ce y = C ( e + e ) and y = C( e + e ) Substituting: C + e 3C + e + Ce = e C= General solution: y = ce + ce e 43

44 E. 3 IVrb DE: y + y + y= ( D+ ) y = e y = and Characteristic equation: ( λ ) h y = c + ce Rule B (double root) + = λ = (double root) y = C e y = C( ) e = ( 4 + ) Substituting: and terms dro out y C e C 4+ e + C e + C e = Ce = e C = y c ce e General solution: = ( + ) + IC: y = c = and y c c c = y e = = = y = 44

45 E. 4 DE:.5 y + y + 5y=.5e + 4cos4 55sin4 IC: y =. y = 6. Characteristic equation: λ + λ + 5= λ = ± i ( cos sin ) = + yh e A B, From the table:.5 y = Ce + Kcos4+ Msin4 = +.5 y.5 Ce 4Ksin4 4M cos4 =.5 y.5 Ce 6Kcos4 6Msin4.5 Substituting: ( + + ) + ( + + ) + ( + ).5 5 Ce 6K 8M 5K cos4 6M 8K 5M sin4 Equating coefficients: 6.5C =.5 C =. K+ 8M = 4 K = and M = 5 8K M = 55.5 y= e Acos+ Bsin +.e + 5sin4 General solution: y = A+.=. A= IC: = y e Bsin Bcos.e cos4 y = B+.+. = 6. B= Particular solution:.5 y= e sin+.e + 5sin4 45

46 Solution by variation of arameters Standard form: y + y + q y= r With, q, and r arbitrary variable functions continuous in interval I yr yr y y d y d W W Particular solution: = + Where y and y solutions of homogeneous equation and W = yy yy E. y + y = sec Solution to homogeneous equation: y + y= y = cos and y = sin W = coscos sin sin = y = cos sinsecd+ sin cossecd= cos ln cos + sin General solution: = ( + ) + ( + ) y c ln cos cos c sin 46

47 Modeling: forced oscillations. Resonance Free motion of mass on sring homogenous DE: my + cy + ky = Driving force (or inut) r( t ) nonhomogeneous DE: my + cy + ky= r( t) Particular interest: r t = F cosωt Solution = outut or resonse: y = yh + y Method of underdetermined coefficients = cosω + sin sin cos y t = ωa ωt+ ωb ωt = y t ω acosωt ω bsinωt y t a t b ωt Substituting: ( ω ) ω ω ( ω ) k m a+ cb= F ca + k m b = k mω a ωcb cosωt ωca k mω b sinωt = F cosωt Linear system of algebraic equations in two unknowns Solution by elimination (Cha 6) a = F k mω ( ω ) k m + ω c and b = F ωc ( ω ) k m + ω c k Putting ω = with ω > m a = F m m( ω ω ) ( ) + ω ω ω c and b = F m ωc ( ) ω ω + ω c 47

48 Case I: undamed forced oscillation No daming c = F F = = m ( ω ω ) ω k ω y t cosωt cosωt F y= Ccos( ω t δ ) + cosωt ω k ω Interretation: outut = suerosition of harmonic oscillations, one with natural ω frequency π and the other with the frequency of the inut ω π F Maimum amlitude: a = k ρ where ρ = ω ω Ratio of amlitude of ρ a y and inut: = k F is the resonance factor 48

49 Resonance henomenon: For ω ω ρ a F In case of resonance DE: y + ω y= cosωt m From modification rule: F y t = tsinωt becomes larger and larger mω Without daming, vibrations could destroy the system 49

50 Beats: but very close, for IC: y y If ω ω Particular solution: Or ( ω ω ) = = F yt = t t m ( cosω cosω ) ω ω F ω + ω ω ω = m yt sin t sin t Since ω ω small, eriod of last sine becomes very large this roduces the beat henomenon (amlitude varies harmonically) 5

51 Case II: damed forced oscillation Daming c > h αt ( cosω sinω ) y t = e A t+ b t with c ω = m The solution aroach zero after a sufficiently long time Solution of nonhomogeneous DE = transient solution: y = yh + y Aroach steady state solution harmonic oscillation y after sufficiently long time outut becomes This is the normal behavior of real hysical systems no daming is ideal case Practical resonance: in damed case amlitude stay finite as ω ω, becomes maimum for some frequency ω ( c) some inut may destroy the system Amlitude of y cos( ω η) y t = C t where C = a + b = F m ( ) ω ω + ω c b and tanη = = a m dc dω ωc ( ω ω ) Maimum m ( ) ( ω ω ) m = c = + = ω ω c ω For large daming ω increases c > m ω = mk no solution C decreases in monotone way as 5

52 If c m ω one solution c For ω = ωma = ω increases as c decreases aroaching ω as c m Amlitude C ( ω ) = Since C ( ω) ma mf c 4m ω c always finite for c > increases as c decreases and aroaches as c Phase lag η : π ω < ω η < π ω = ω η = π ω > ω η > Note: Whereas DE of nd order have various alications, higher order DE occur much more rarely in hysics (E. binding of elastic beams) The theory is much similar to that of DE of nd order 5

53 ) Transformation v = ay+ b+ k 4y+ 5 y + y+ 3= E. 7 The terms 4y and y v= y y= v y = ( v ) = Substitution: ( v) ( v ) ( v) [ v+ 5] ( v ) + v + 3= ( v+ 5) ( v+ 5) v = v 6 4v+ 4v+ ( v+ 5) v = 4v+ dv d dv d 4v = 4v = + + dv= d+ C v ln 4v+ = + C 4v + 4 Substitution ( ) v= y y ln 4 8y+ = + C 4 4 8y ln 4 8y+ = 8+ C 4+ 8y+ ln 4 8y+ = C 53