MA 266 Review Topics - Exam # 2 (updated)

Transcription

1 MA 66 Reiew Topics - Exam # updated Spring First Order Differential Equations Separable, st Order Linear, Homogeneous, Exact Second Order Linear Homogeneous with Equations Constant Coefficients The differential equation ay + by + cy = has Characteristic Equation ar + br + c = Call the roots r and r The general solution of ay + by + cy = is as follows: a If r, r are real and distinct y = C e r t + C e r t b If r = λ + iµ hence r = λ iµ y = C e λt cos µt + C e λt sin µt c If r = r repeated roots y = C e r t + C te r t 3 Theory of nd Linear Order Equations y t Wronskian of y, y is W y, y t = y t y t y t a The functions y t and y t are linearly independent oer a < t < b if W y, y for at least one point in the interal b THEOREM Existence & Uniqueness If pt, qt and gt are continuous in an open y + pt y + qt y = gt interal α < t < β containing t, then the IVP yt = y y t = y has a unique solution y = ϕt defined in the open interal α < t < β c Superposition Principle If y t and y t are solutions of the nd order linear homogeneous equation P ty + Qty + Rty = oer the interal a < t < b, then y = C y t + C y t is also a solution for any constants C and C d THEOREM Homogeneous If y t and y t are solutions of the linear homogeneous equation P ty + Qty + Rty = in some interal I and W y, y for some t in I, then the general solution is y c t = C y t+c y t This is usually called the complementary solution and we say that y t, y t form a Fundamental Set of Solutions FSS to the differential equation e THEOREM Nonhomogeneous The general solution of the nonhomogeneous equation P ty + Qty + Rty = Gt is yt = y c t + y p t, where y c t = C y t + C y t is the general solution of the corresponding homogeneous equation P ty + Qty + Rty = and y p t is a particular solution of the nonhomogeneous equation P ty + Qty + Rty = Gt f Useful Remark : If y p t is a particular solution of P ty + Qty + Rty = G t and if y p t is a particular solution of P ty + Qty + Rty = G t, then y p t = y p t + y p t is a particular solution of P ty + Qty + Rty = [G t + G t]

2 4 Reduction of Order If y t is one solution of P ty + Qty + Rty =, then a second solution may be obtained using the substitution y = t y t This reduces the original nd order equation to a st equation using the substitution w = d Sole that first order equation for dt w, then since w = d dt, sole this st order equation to determine the function 5 Finding A Particular Solution y p t to Nonhomogeneous Equations You can always use the method of Variation of Parameters to find a particular solution y p t of the linear nonhomogeneous equation y + pt y + qt y = gt Variation of Parameters may require integration techniques If the coefficients of the differential equation are constants rather than functions and if gt has a ery special form see table below, it is usually easier to use Undetermined Coefficients : a Undetermined Coefficients - IF ay + by + cy = gt AND gt is as below: gt Form of y p t P m t = a m t m + a m t m + + a t s A m t m + A m t m + + A } e αt P m t t s e αt A m t m + A m t m + + A } e αt P m t cos βt or e αt P m t sin βt t s e αt [F m t cos βt + G m t sin βt]} where s = the smallest nonnegatie integer s =, or such that no term of y p t is a solution of the corresponding homogeneous equation In other words, no term of y p t is a term of y c t F m t, G m t are both polynomials of degree m b Variation of Parameters - If y t and y t are two independent solutions of the homogeneous equation y + pt y + qt y =, then a particular solution y p t of the nonhomogeneous equation y + pt y + qt y = gt has the form where y p t = u t y t + u t y t y y u gt y = y y, u y gt = y y y y y y Remember: Coefficient of y in must be in order to use the aboe formulas

3 6 Spring-Mass Systems m u + γ u + k u = F t u = u, u = u m = mass of object, γ = damping constant, k = spring constant, F t = external force Weight w = m g, Hooke s Law : F s = k d, L L L equilibrium: d m F = F s g kd=mg d ut m I Undamped Free Vibrations : m u + k u = Simple Harmonic Motion Note that A cos ω t + B sin ω t = R cosω t δ, where R = A + B = amplitude, π ω = frequency, = period and δ = phase shift determined by tan δ = B ω A II Damped Free Vibrations : m u + γ u + k u = III i γ 4km > ii γ 4km = oerdamped distinct real roots to CE critically damped repeated roots to CE iii γ 4km < underdamped complex roots to CE motion is oscillatory Forced Vibrations : F t = F cos ωt or F t = F sin ωt, for example i m u + γ u + k u = F t Damped In this case if you write the general solution as ut = u T t + u t, then u T t = Transient Solution ie the part of ut such that u T t as t and u t = Steady-State Solution the solution behaes like this function in the long run k ii m u + k u = F cos ωt Undamped If ω = ω = m Resonance occurs and the solution is unbounded; while if ω ω then motion is a series of beats solution is bounded 7 n th Order Linear Homogeneous Equations With Constant Coefficients a y n + a y n + + a n y + a n y = This differential equation has n independent solutions Characteristic Equation : a r n + a r n + + a n r + a n = will hae n characteristic roots that may be real and distinct, repeated, complex, or complex and repeated

4 a For each real root r that is not repeated get a solution of : b For each real root r that is repeated m times get m independent solutions of : e rt, te rt, t e rt,, t m e rt c For each complex root r = λ + iµ repeated m times get m solutions of : e λt cos µt, te λt cos µt,, t m e λt cos µt and e λt sin µt, te λt sin µt,, t m e λt sin µt don t need to consider its conjugate root λ iµ 8 Undetermined Coefficients for n th Order Linear Equations This can only be used to find y p t of a y n + a y n + + a n y + a n y = gt and gt one of the 3 ery SPECIAL FORMS in table in 5 aboe The particular solution has the same form as before : y p t = t s [ ], where s = the smallest nonnegatie integer such that no term of y p t is a term of y c t, except this time s =,,,, n 9 Laplace Transforms a Be able to compute Laplace transforms using definition : Lft} = F s = e st ft dt and using a table of Laplace transforms see table on page 37 and using linearity : Lft + gt} = Lft} + Lgt}, Lc ft} = c Lft} b Computing Inerse Laplace Transforms: Must be able to use a table of Laplace transforms usually together with Partial Fractions or Completing the Square, to find inerse Laplace transforms: ft = L F s} c Soling Initial Value Problems: Recall that e rt Ly } = s Ly} y Ly } = s Ly} s y y Ly } = s 3 Ly} s y s y y d Discontinuous Functions : i Unit Step Function Heaiside Function : If c, u c t = y y = u t c, t < c, t c c t Lu c t} = e c s s

5 ii Unit Pulse Function : u a t u b t = y, a t < b, otherwise y = u t u t a b a b t iii Translated Functions : y = gt =, t < c ft c, t c = u c t ft c y y=ft y y=gt f f t c t Lu c t ft c} = e c s F s, where F s = Lft} Thus, L e c s F s} = u c t ft c, where ft = L F s} A useful formula NOT in the book : Lu c t ht} = e c s Lht + c} i Unit Impulse Functions : If y = δt c c, then L δt c} = e cs e Conolutions: t } Lf gt} = L ft τ gτ dτ = Lft} Lgt}

6 Systems of Linear Differential Equations : x t = A xt a Rewrite a single n th order equation p ty n + p ty n + + p n ty = gt as a system of st order equations Use the substitution : Let x = y = y x n = y n to get st Order System : x = x = x 3 x n = x n x n = p p n x p n p x n + gt} b Existence & Uniqueness Theorem for Systems If Pt and gt are continuous on an x interal α < t < β containing t, then the IVP t = Pt xt has a unique solution xt xt = x defined on the interal α < t < β c The set of ectors x, x,, x m} is linearly independent if the equation k x + k x + + k m x m = is satisfied only for k = k = = k m = This means you cannot write any one of these ectors as a linear combination of the others d Sole systems of st x a b x order equations = ie, x c d = A x using : i Elimination Method : Basic idea - eliminate one of the unknowns either x or from the original system to get an equialent single nd order differential equation ii Eigenalues & Eigenectors Method : See below for solutions ia this method and corresponding phase portraits a b Eigenalue : If A =, then the eigenalues of A are the roots of c d e If x t = Eigenector : = x t t A λ I = and if x t = a λ b c d λ = is a nonzero solution to A λ I = x t t W [x, x x ] = t t, then the Wronskian is t t If x t and x t are solutions of x = A x and W [x, x ]t, then the set x t, x t} forms a Fundamental Set of Solutions of the system and a Fundamental Matrix is x t x Φt = t t t

7 Eigenalue & Eigenector Method and Phase Portraits : x = A x The following describes how to find the general solution to and plot solutions trajectories A plot of the trajectories of a gien homogeneous system a b x = x c d is called a phase portrait To sketch the phase portrait, we need to find the corresponding eigenalues and eigenectors of the matrix A = and then consider 3 cases : a b c d a λ < λ, real and distinct : If, are e-ectors corresponding to λ and λ, respectiely x t = e λ t and x t = e λ t are solutions and hence general solution of is xt = C x t + C x t and hence if λ < λ : xt = C e λt + C }} e λt }} dominates as t dominates as t < λ < λ x λ < λ < x x λ < < λ x x

8 b λ = α + i β : If w = a + i b is a complex e-ector corresponding to λ then x t = Re e λt w } = e α t a cos βt b sin βt and x t = Im e λt w } = e α t a sin βt + b cos βt are real-alued solutions and hence general solution of is xt = C x t + C x t If say α < : OR x x Test a point to decide which c λ = λ : If there is only one linearly independent eigenector corresponding to λ, then solutions to x = A x are x t = e λt and x t = t e λt + e λt a, where A λ I = A λ I a = is an eigenector of A, while a is called a generalized eigenector of A The general solution of the system is xt = C x t + C x t and hence: xt = C e λt [ + C t e λt } } + eλt ] a dominates as t ± If say λ < : OR x x Test a point to decide which

9 Particular Solutions to Nonhomogeneous Linear Systems : x = A x + gt g t a Undetermined Coefficients for Systems The column ector gt = must hae g t each component function g t and g t as one of the three special forms like those for Undetermined Coefficients for regular nd order equations and A must be a constant matrix The main difference is if say gt = u e λt and λ is also an eigenalue of A, then try a particular solution of the form x p = a te λt + b e λt b Variation of Parameters for Systems : x = At x + gt: x p t = Φt Φ t gt dt, where Φt is a Fundamental Matrix of the homogeneous system x = At x+ gt can hae any form and A need not be a constant matrix Practice Problems y y y = [] For what alue of α will the solution to the IVP y = α y = satisfy y as t? [] a Show that y = x and y = x are solutions of the differential equation y + xy y = b Ealuate the Wronskian W y, y at x = c Find the solution of the initial alue problem y + xy y =, y =, y = 4 [3] Find the largest open interal for which the initial alue problem 3 y + y + x y = x 3, y = 3, y =, has a solution In Problems 4, 5, and 6 find the general solution of the homogeneous differential equations in a and use the method of Undetermined Coefficients to find a particular solution y p in b and find the form of a particular solution c [4] a y 5y + 6y = b y 5y + 6y = t c y 5y + 6y = e t + cos3t [5] a y 6y + 9y = b y 6y + 9y = te 3t c y 6y + 9y = e t + cos3t [6] a y y + y = b y y + y = e x + cos3x c y y + y = e x cos3x [7] Find the general solution to a y + y 6y = 7e 4t b y + y 6y = 7e 4t sin t [8] Sole this IVP: y y = 4t, y =, y = [9] Find the general solution to y + y = tan t, < x < π [] The differential equation y xy + y = has solutions y x = x and y = Use the method of Variation of Parameters to find a solution of y xy + y = [] The differential equation y +xy y = has one solution y x = x Use the method of Reduction of Order to find a second linearly independent solution of y + xy y = [] For what nonnegatie alues of γ will the the solution of the initial alue problem u + γu + 4u =, u = 4, u = oscillate? [3] a For what positie alues of k does the solution of the initial alue problem u + ku = 3 cost, u =, u =, become unbounded Resonance?

10 b For what positie alues of k does the solution of the initial alue problem u + u + ku = 3 cost, u =, u =, become unbounded Resonance? [4] Find the steady state solution of the IVP y + 4y + 4y = sin t, y =, y = [5] A 4-kg mass stretches a spring 39 m If the mass is released from m below the equilibrium position with a downward elocity of m/sec, what is the maximum displacement? In Problems 6 and 7 find the general solution of the homogeneous differential equations in a and use the method of Undetermined Coefficients to find the form of a particular solution of the nonhomogeneous equation in b [6] a y y = b y y = t + e t [7] a y y y + y = b y y y + y = e t + cos t [8] Find the solution of the initial alue problem y y + y =, y =, y =, y = [9] Find the general solution of the differential equation y + y = t [] Find the general solution of y + 4y = cos t [] Find a fundamental set of solutions of y 5 4y = [] Find the Laplace transform of these functions: a ft = 3 e t b gt = t 5 c ht = cosh πt d kt = t 3 e 5t [3] Find the inerse Laplace transform of 9 a F s = b F s = s s [4] Sole these initial alue problems: a s s c F s =, t < 5 c y y =, 5 t < ; y = y = t, t < d y + 4y =, < t < ; y = y = e y + y = gt, y = and where gt : y 3 y y 6y = y = y = y = g t 8 d F s = 3s + s + 4 s + s + 5 b y y + y = cos t y = y = 4 t f y + 4y = δt 3, y = y = t } [5] L e τ cos πt τ dτ =? [6] } Gs If gt = L Gs}, then L s 3 =? [7] Use the Elimination Method to sole the system x = x + x = 4x + [8] Rewrite the nd order differential equation y + y + 3ty = cos t with y =, y = 4 as a system of st order differential equations [9] Find eigenalues and corresponding eigenectors of a A = b A = 4

11 x [3] Find the solution of the IVP = Find a fundamental matrix Φt x x [3] Sole =, x = 4 x, x [3] Find the general solution of the system x t = A xt, where A = = 3 [33] Tank # initially holds 5 gals of brine with a concentration of lb/gal, while Tank # initially holds 5 gals of brine with a concentration of 3 lb/gal Pure H O flows into Tank # at 5 gal/min The well-stirred solution from Tank # then flows into Tank # at 5 gal/min The solution in Tank # flows out at 5 gal/min Set up and sole an IVP that gies x t and t, the amount of salt in Tanks # and #, respectiely, at time t [34] Tank # initially holds 5 gals of brine with concentration of lb/gal and Tank # initially holds 5 gals of brine with concentration 3 lb/gal The solution in Tank # flows at 5 gal/min into Tank #, while the solution in Tank # flows back into Tank # at 5 gal/min Set up an IVP that gies x t and t, the amount of salt in Tanks # and #, respectiely, at time t x x 3 [35] Find the general solution of = + e t x [36] Find a particular solution of = 3 x 6e t [37] Find the general solution of = [38] Match the phase portraits shown below that best corresponds to each of the gien systems of differential equations: i x = x ; Solution : xt = C e t + C e t ii x = x ; Solution : xt = C e t + C e 3t } iii x = x ; Solution : xt = C e t + C e t t + cos t sin t i x = x ; Solution : xt = C e sin t t + C e cos t t A B

12 C D E F G H

13 Answers [] α = [] b W x, x = 4 ; c y = 3x x [3] < x < [4] a y = C e t + C e 3t b y = At + Bt + C c y = Ate t + B cos3t + C sin3t [5] a y = C e 3t + C te 3t b y = t At + Be 3t c y = Ae t + B cos3t + C sin3t [6] a y = C e x cos3x + C e x sin3x b y = Ae x + B cos3x + C sin3x c y = xa cos3x + B sin3xe x [7]a y = C e 3t + C e t + e4t b y = C e 3t + C e t + e4t + cos t + 4 sin t [8] y = 4 + 4e t t 4t [9] y = C cos t + C sin t cos t lnsec t + tan t [] y = ln x or y = ln x + C x + C [] y = x or y = Ax + Bx, A [] γ < 4 [3] a k = 8 resonance b NO alue of k, all solutions are bounded [4] y = 3 sin t 4 cos t 5 [5] ut = cos 5t + sin 5t = 5 cos 5t δ, δ = tan Thus amplitude = 5 [6] a y = C + C e t + C 3 e t b y = tat + B + Cte t [7] a y = C e t + C te t + C 3 e t b y = At e t + B cos t + C sin t [8] y = 3 e t + te t [9] y = C + C cos t + C 3 sin t + 3 t3 t [] y = C + C e 4t + cos t sin t [], t, t, e t, e t } s 6 [] a b s c d 6 s s s 6 s π s 5 4 [3] a 3e t e t b e t + te t c 4 3 t3 e t d 3e t cos t e t sin t [4] a y = 5 e3t + 4e t b y = cos t sin t + 5 4et cos t e t sin t c y = + et + e t + u 5 t + et 5 + e t 5, or y = + cosh t + u 5 t + cosht 5 d y = sin t + t u 8 4 t t sin t + u 8 4 t cos t 4 4 e y = 3 e t 3 u t e t + 3 u 4 t e t 4 f y = u 3tt sin t 3 [5] s s + s + π [6] t t τ e 3t τ gτ dτ or t τ e 3τ gt τ dτ [7] x t = C e 3t + C e t, t = C e 3t C e t x [8] Let x = y, = y, then = x = 3tx + cos t, where x =, = 4 [9] a λ = 3, = ; λ =, = [9] b λ =, = ; λ =, = e [3] xt = e 3t + e t 3t e, Φt = t e 3t e t sin t cos t [3] xt = e t e cos t t [3] xt = C sin t e t + C e t + te t x x x [33] = 5, = x 5 75 x Solution : = 5e t + 5e t 5 }

14 x x x [34] = 5 5, = x 5 75 Solution : = 5 3 3t e 3 [35] xt = C e t + C e t + e t [36] x p t = [37] xt = C e t + C e t + e t + [38] i C ii A iii B i D 3

15 ft = L F s} F s = Lft} e at s a 3 t n n! s n+ 4 t p p > 5 sin at 6 cos at 7 sinh at 8 cosh at 9 e at sin bt e at cos bt s Γp + s p+ a s + a s s + a a s a s s a b s a + b s a s a + b t n e at n! s a n+ t n e at n! s a n+ u c t e cs 3 u c tft c e cs F s 4 e ct ft F s c 5 fct 6 t ft τ gτ dτ s s c F, c > c F s Gs 7 δt c e cs 8 f n t s n F s s n f sf n f n 9 t n ft F n s