ODE. Philippe Rukimbira. Department of Mathematics Florida International University PR (FIU) MAP / 92
|
|
- Emmeline Hilda Peters
- 6 years ago
- Views:
Transcription
1 ODE Philippe Rukimbira Department of Mathematics Florida International University PR (FIU) MAP / 92
2 4.4 The method of Variation of parameters 1. Second order differential equations (Normalized, standard form!). y + P(x)y + Q(x)y = f (x) Suppose y 1 and y 2 form a fundamental set of solutions on an interval I for y + P(x)y + Q(x)y = 0 We seek functions u 1 (x) and u 2 (x) such that: y p = u 1 y 1 + u 2 y 2 is a particular solution of the nonhomogeneous differential equation. PR (FIU) MAP / 92
3 In that case, y p = u 1 y 1 + y 1u 1 + u 2y 2 + y 2u 2 We can impose the additional condition on u 1 and u 2 : y 1 u 1 + y 2u 2 = 0 That is equivalent to y p = u 1 y 1 + u 2y 2 PR (FIU) MAP / 92
4 From there, y p = u 1 y 1 + u 1y 1 + u 2 y 2 + u 2y 2 Now, substitute for y p, y p and y p into the nonhomogeneous differential equation" which becomes: y p + Py p + Qy p = f (x) u 1 y 1 + u 1y 1 + u 2 y 2 + u 2y 2 + Pu 1y 1 + Pu 2y 2 + Qu 1y 1 + Qu 2 y 2 = f (x) PR (FIU) MAP / 92
5 Reorganizing leads to u 1 y 1 + u 1(y 1 + Py 1 + Qy 1) + u 2 y 2 + u 2(y 1 + Py 2 + Qy 2) = f (x). and finally, one obtains a second condition on u 1 and u 2 y 1 u 1 + y 2 u 2 = f (x) PR (FIU) MAP / 92
6 What we have now is a system of two equations involving (the derivatives of ) u 1 and u 2 y 1 u 1 + y 2u 2 = 0 y 1 u 1 + y 2 u 2 = f (x) PR (FIU) MAP / 92
7 Observe that the determinant of the linear system in no other than the Wronskian W (y 1, y 2 ) 0 by assumption. Hence, the system has a unique solution (u 1, u 2 ). ( ) 0 y det 2 u 1 = f (x) y 2 W (y 1, y 2 ) = y 2f (x) y 1 y 2 y 1 y 2 u 2 = f (x)y 1 y 1 y 2 y 1 y 2 From u 1 and u 2, we obtain u 1 and u 2 by integration. PR (FIU) MAP / 92
8 Example y 3y + 2y = e3x 1 + e x Notice that undetermined coefficient methods does not work in this example! Using the characteristic equation technique for instance, we find that the general solution to the associated homogeneous equation is y c = C 1 e x + C 2 e 2x. PR (FIU) MAP / 92
9 The method of variation of parameters tells us that we can find a particular solution y p = u 1 e x + u 2 e 2x by solving e x u 1 + e2x u 2 = 0 e x u 1 + 2e2x u 2 = e 3x 1 + e x PR (FIU) MAP / 92
10 e2x u 1 = 1 + e x u 2 = ex 1 + e x u 1 = (e x + 1) + ln (e x + 1) u 2 = ln (e x + 1) A particular solution is y p = ( (e x + 1) + ln (e x + 1))e x + ln (e x + 1)e 2x and, finally, the general solution is y = C 1 e x + C 2 e 2x + e 2x (ln (e x + 1)) + e x (ln (e x + 1) (e x + 1)). PR (FIU) MAP / 92
11 Variation of parameters for higher order equations y (n) + P 1 y (n 1) P n y = f (x). Let y 1,..., y n be n linearly independent solutions of y (n) + P 1 y (n 1) P n y = 0 Look for u 1,..., u n such that u 1 y u n y n is a particular solution of the nonhomogeneous equation. PR (FIU) MAP / 92
12 For that, one solves the following linear system: y 1 u y nu n = 0 y 1 u y nu n = 0. =. y (n 1) 1 u y (n 1) u n = f (x) u i = W i(y 1,..., y n ) W (y 1,..., y n ) where W i (y 1,..., y n ) is the Wronskian in which the column i has been replaced by the column (0,...0, f (x)). PR (FIU) MAP / 92
13 Example y + y = tan x y + y = sec x PR (FIU) MAP / 92
14 4.5 The Cauchy-Euler equation Standard form, n th order: Second order example: a n x n d n y dx n a 1x dy dx + a 0y = g(x). x 2 y 2xy + 2y = x 3 ln x Observation: The substitution t = ln x reduces Cauchy-Euler equation to an equation with constant coefficients! PR (FIU) MAP / 92
15 From dy dx = dy dt dt dx = 1 dy x dt and d 2 y dx 2 = 1 dy x 2 dt + 1 d 2 y dt x dt 2 dx = 1 x 2 ( dy dt + d 2 y dt 2 ) Substitution leads to d 2 y dt 2 3dy dt + 2y = te3t. This can be solved using the method of undetermined coefficients or variation of parameters! PR (FIU) MAP / 92
16 y(t) = C 1 e 2t + C 2 e t + ( 1 2 t 3 4 )e3t y(x) = C 1 x 2 + C 2 x + ( 1 2 ln x 3 4 )x 3 PR (FIU) MAP / 92
17 : Applications: Spring vibrations, electric circuits problems Second order differential equations with constant coefficients Consider a spring with natural length L. Suspend a mass with weight F g = mg. The spring is stretched by l. Choose an orientation vector i downward. Hooke s Law mg = Kl where K is the constant of the spring. PR (FIU) MAP / 92
18 Disturb the mass m with initial position x 0 and velocity v 0. If x denotes the displacement from equilibrium position, then mẍ i = Kx i mẍ = Kx ẍ + λ 2 x = 0 where λ 2 = K m. The mass executes a free, undamped motion. x(t) = C 1 cos λt + C 2 sin λt. PR (FIU) MAP / 92
19 The simple, harmonic motion. K λ = m is the angular velocity, x(t) = A cos (λt + Φ) T = 2π λ is the natural period of the motion f = λ 2π = 1 T is the natural frequency. A = C1 2 + C2 2 is the amplitude and Φ from tan Φ = C 2 C 1 is the phase angle. PR (FIU) MAP / 92
20 Solving t 0 = π 2 Φ λ is the phase shift. The motion can be graphed now! λt + Φ = π 2 PR (FIU) MAP / 92
21 Example #1, page 197. Example: A 16 lb weight is placed upon the lower end of a coil spring suspended vertically from a fixed support. The weight comes to rest in its equilibrium position, thereby stretching the spring 6 in. Determine the resulting displacement as a function of time in each of the following cases: a) If the weight is then pulled down 4 in. below its equilibrium position and released at t=0 with initial velocity of 2 ft/sec directed downward. b) If the weight is then pulled down 4 in. below its equilibrium position and released at t=0 with an initial velocity of 2 ft/sec directed upward. c) If the weight is then pushed up 4 in. above its equilibrium position and released at t=0 with an initial velocity of 2 ft/sec. directed downward. PR (FIU) MAP / 92
22 16 = K 6 12 = K 2, so K = 32. ẍ + K 16 x = 0, M = M 32 = 1 2 slugs Finish the example! ẍ + 64x = 0 PR (FIU) MAP / 92
23 Damped motion Mẍ = Kx βẋ; F R = βẋ ẍ + λ 2 x + 2bẋ = 0 where 2b = β M. ẍ + 2bẋ + λ 2 x = 0 Free, damped motion The motion is not necessarily periodic anymore. PR (FIU) MAP / 92
24 m 2 + 2bm + λ 2 = 0 b 2 λ 2 > 0: Over-damped motion. m = b ± b 2 λ 2. b x(t) = e bt [C 1 e 2 λ 2t + C 2 e b 2 λ 2t ]. Observe the damping factor e bt! PR (FIU) MAP / 92
25 b 2 λ 2 = 0: Critically damped motion. x(t) = e bt (C 1 + C 2 t). PR (FIU) MAP / 92
26 b 2 λ 2 = 0: Critically damped motion. x(t) = e bt (C 1 + C 2 t). b 2 λ 2 < 0: Under-damped motion. x(t) = e bt [C 1 cos λ 2 b 2 t + C 2 sin λ 2 b 2 t]. Observe a periodic factor and a damping factor! PR (FIU) MAP / 92
27 Example #2, page 208. PR (FIU) MAP / 92
28 Forced motion Mẍ + βẋ + Kx = F cos ωt ẍ + 2bẋ + λ 2 x = f cos ωt where f = F M, 2b = β M, λ2 = K M. PR (FIU) MAP / 92
29 Assume we are in the under-damped situation, so that x c (t) = Ae bt cos ( λ 2 b 2 t + φ). We look for a particular solution using the undetermined coefficients method: x p (t) = C cos ωt + D sin ωt PR (FIU) MAP / 92
30 We find that: ω 2 C + 2bωD + λ 2 C = f ω 2 D 2bωC + λ 2 D =0 PR (FIU) MAP / 92
31 We find that: ω 2 C + 2bωD + λ 2 C = f ω 2 D 2bωC + λ 2 D =0 or equivalently (λ 2 ω 2 )C + 2λωD = f 2bωC + (λ 2 ω)d =0 PR (FIU) MAP / 92
32 By Cramer s method for instance, f (λ 2 ω 2 ) C = (λ 2 ω 2 ) 2 + 4b 2 ω 2 2bωf D = (λ 2 ω 2 ) 2 + 4b 2 ω 2 PR (FIU) MAP / 92
33 x p = f (λ 2 ω 2 ) (λ 2 ω 2 ) 2 + 4b 2 ω cos ωt + 2bωf 2 (λ 2 ω 2 ) 2 sin ωt + 4b 2 ω2 We will express x p as x p = A cos (ωt + φ) PR (FIU) MAP / 92
34 f (λ 2 ω 2 ) A cos φ = (λ 2 ω 2 ) 2 + 4b 2 ω 2 2bωf A sin φ = (λ 2 ω 2 ) 2 + 4b 2 ω 2 So and finally A = f [(λ 2 ω 2 ) 2 + 4b 2 ω 2 ] 1/2 x p (t) = f [(λ 2 ω 2 ) 2 + 4b 2 ω 2 cos (ωt + φ) ] 1/2 PR (FIU) MAP / 92
35 da dω = 2(λ2 ω 2 )ω 4b 2 ω [(λ 2 ω 2 ) 2 + 4b 2 ω 2 ] 3/2 f The critical ω s are ω = 0 and ω 2 = λ 2 2b 2. x p (t) achieves the maximum amplitude when ω = ω R = K λ 2 2b 2 = M β2 2M 2 PR (FIU) MAP / 92
36 ω R is called the resonance frequency. Observe that the resonance frequency ω R λ 2 b 2 = smaller than the frequency of the free motion! The graph of y = A(ω) is called the resonance curve! K M β2 4M 2 is PR (FIU) MAP / 92
37 x(t) = x c + x p The function x c (t) eventually becomes negligible as time goes on, this is the transient state. x p (t) which remains forever, is called the steady state solution. PR (FIU) MAP / 92
38 In the undamped situation, b = 0, x p (t) = f λ 2 cos (ωt + φ) ω2 We can see that as ω approaches the natural frequency λ, the amplitude of the steady state blows up! This phenomenon is known as Pure resonance. It always has destructive effects on the systems. PR (FIU) MAP / 92
39 Assignments: page 189, #10-12, Page 217 (Ross), # 4-7, Page 224, #1-3 PR (FIU) MAP / 92
40 Example: Forced motion A mass weighting 4 lb stretches a spring 1.5 in. The mass is displaced 2 in. in the positive direction and released with zero initial velocity. assuming that there is no damping, and that the mass is acted upon by an external force of 2 cos 3t lb, formulate the initial value problem describing the motion of the mass and solve it. PR (FIU) MAP / 92
41 4 = 1 8K so K = 32 x(0) = 1 6 ẋ(0) = 0 4 = m.32 so, m = x = 2 cos 3t 8ẍ PR (FIU) MAP / 92
42 ẍ + 256x = 16 cos 3t x(0) = 0 ẋ(0) = 0 Solving the above initial value problem leads to x(t) = cos 16t + cos 3t PR (FIU) MAP / 92
43 Analogous systems Mass on a spring: mẍ + βẋ + Kx = f (t) Inductor-Resistor-Capacitor (L-R-C) series circuit: L q + R q + 1 C q = E(t) (E(t) is the electric potential) PR (FIU) MAP / 92
44 Put E(t) = E 0 sin ωt Write the circuit equation as LI + RI + 1 C I = ωe 0 cos ωt Solving: I(t) = I tr + I sp where lim I tr = 0 t + I sp = Steady periodic solution PR (FIU) MAP / 92
45 I sp = E 0 cos (ωt α) R 2 + (ωl 1 ωc )2 α = tan 1 Z = ωrc 1 LCω 2, 0 α π R 2 + (ωl 1 ωc )2 (in Ohms) is called the impedance of the circuit. The amplitude of the signal is I 0 = E 0 Z PR (FIU) MAP / 92
46 Electrical resonance I 0 = E 0 Z = E 0 R 2 +(ωl 1 ωc )2 attains its maximum when ω = 1 LC This is the resonance frequency!. Tuning a radio receiver consists essentially in modifying the value of C so as to match the frequency of the incoming radio signal! PR (FIU) MAP / 92
47 6. Series solutions Examples Solve dy dx 2xy = 0 We look for a solution of the form y = c n x n n=0 Then and dy dx = nc n x n 1 n=0 n=0 dy dx 2xy = nc n x n 1 2 c n x n+1 = 0 n=0 PR (FIU) MAP / 92
48 (m + 1)c m+1 x m 2 c m 1 x m = 0 m=0 m=1 c 1 + [(m + 1)c m+1 2c m 1 ]x m = 0 m=1 PR (FIU) MAP / 92
49 c 1 = 0 and (m + 1)c m+1 2c m 1 = 0 which implies that for m 2. c 1 = 0 and c m+1 = 2 c m 1 m + 1 We will use the recurrence formula for c m to generate all coefficients in the power series. PR (FIU) MAP / 92
50 c 0 c 1 = 0 arbitrary c 2 = 2 c 0 2 = c o c 3 = 2 c 1 3 = 0 c 4 = 2 c 2 4 = c 0 2 c 5 = 0 PR (FIU) MAP / 92
51 In general, c 2n+1 = 0 for n = 0, 1, 2... and c 2n = 2 c 2(n 1) 2n = 2. =. = c 0 n! c 2(n 2) (n 2)(n 1) = c 2(n 2) n(n 1) PR (FIU) MAP / 92
52 y = c 0 (1 + m=1 ) ( x 2m = c 0 m! m=0 y = c 0 e x 2. (x 2 ) m ) m! PR (FIU) MAP / 92
53 y = c 0 (1 + m=1 ) ( x 2m = c 0 m! m=0 y = c 0 e x 2. (x 2 ) m ) Check this solution using the separable equation technique! m! PR (FIU) MAP / 92
54 Example 2 (x 1)y xy + y = 0, y(0) = 2, y (0) = 6 y = n 0 c n x n y = n 0 nc n x n 1 y = n 0 n(n 1)c n x n 2 0 = (x 1)y xy + y = xy y xy + y = [(n + 1)nc n+1 (n + 2)(n + 1)c n+2 + (1 n)c n ]x n n 0 PR (FIU) MAP / 92
55 So c n+2 = (1 n)c n + (n + 1)nc n+1 (n + 2)(n + 1) c 0 = y(0) = 2 c 1 = y (0) = 6 Using the above recurrence formula, we see that c 2 = 1, c 3 = 1 3, c 4 = 1 4.3, c 5 = Claim: for n 2, c n = 2 n!. PR (FIU) MAP / 92
56 proof True for c 2 and c 3. Assume it is true for n 1 and n 2, then c n = (1 n)c n 2 + (n 1)(n 2)c n 1 n(n 1) = 2 2 (1 n) (n 2)! + (n 1)(n 2) (n 1)! n(n 1) = 2 n! PR (FIU) MAP / 92
57 y = 2 + 6x 2 n 2 x n n! = 8x 2 2x 2 n 2 = 8x 2( n 0 y = 8x 2e x x n n! ) x n n! Check by substitution that this is indeed the solution to the initial value problem. PR (FIU) MAP / 92
58 Example 2 (x 1)y (2 x)y + y = 0, y(0) = 2, y (0) = 1 The recurrence relation is: c 0 = y(0) = 2, c 1 = y (0) = 1 c n+2 = c n + (n 2)c n+1 n + 2 y = 2 x + 2x 2 x 3 + x 4 2 x x x PR (FIU) MAP / 92
59 Series solutions around ordinary points y + P(x)y + Q(x)y = 0 The standard form for a second order linear differential equation. Definition A point x 0 is said to be an ordinary point for the above differential equation if P(x) and Q(x) are analytic at x 0 ; that is, both have power series in (x x 0 ) with positive radius of convergence. PR (FIU) MAP / 92
60 A point that is not an ordinary point is said to be a singular point. Note: For power series solutions at an ordinary point, the radius of convergence is at least equal to the distance to the nearest singular point. PR (FIU) MAP / 92
61 Example takes the standard form (x 2 1)y + 4xy + 2 x 2 1 y = 0 y + 4x x 2 1 y 2 + (x 2 1) 2 y = 0 1 and 1 are singular points for this equation. Any other point is a regular point, according to the above definition. PR (FIU) MAP / 92
62 Theorem If x = x 0 is an ordinary point of the differential equation y + P(x)y + Q(x)y = 0, we can always find two linearly independent power series solutions of the form y = c n (x x 0 ) n n=0 PR (FIU) MAP / 92
63 Example Find two linearly independent solutions for y xy + 2y = 0 PR (FIU) MAP / 92
64 Example Find two linearly independent solutions for y xy + 2y = 0 Look for y = c n x n n=0 Then: y = y = nc n x n 1 n=1 n(n 1)c n x n 2 n=2 PR (FIU) MAP / 92
65 Therefore, y xy + 2y = 0 implies that 0 = n=2 n(n 1)c nx n 2 n=1 nc nx n + n=0 2c nx n = n=0 (n + 2)(n + 1)c n+2x n + n=0 2c nx n n=1 nc nx n = 2c 2 + 2c 0 + n=1 [(n + 2)(n + 1)c n+2 + (2 n)c n ]x n PR (FIU) MAP / 92
66 We deduce that c n+2 = c 2 = c 0 (n 2)c n (n + 2)(n + 1) PR (FIU) MAP / 92
67 We deduce that c n+2 = c 2 = c 0 (n 2)c n (n + 2)(n + 1) c 0 : Arbitrary c 2 = c 0 c 4 = 0 c 2n = 0, n > 2 PR (FIU) MAP / 92
68 c 1 is also arbitrary. c 3 = c c 2n+1 = (2n 3)c 2n 1 (2n + 1)(2n), 2n + 1 > 3 y 1 = c 0 c 0 x 2 = c 0 (1 x 2 ) y 2 = c 1 x c 1 6 x = c 1 (x 1 6 x ) PR (FIU) MAP / 92
69 Another example y xy y = 0, x 0 = 1 Find two linearly independent power series solutions. Look for for y = c n (x 1) n n=0 y (x 1)y y y = 0 PR (FIU) MAP / 92
70 6.2. Solutions about singular points: Frobenious Method Definition For a differential equation Py + Qy + Ry = 0 A singular point x 0 is said to be a regular singular point if (x x 0 ) Q P and (x x 0 ) 2 R P are analytic at x 0. Otherwise, the regular point x 0 is said to be irregular singular. PR (FIU) MAP / 92
71 Example: The Euler Equation x 2 y + αxy + βy = 0 x = 0 is a regular singular point. As an alternate method of solution, we look for y = x r Then x r (x r ) αx(x r ) + βx r = 0 = x r (r(r 1) + αr + β) Any solution of F(r) = r(r 1) + αr + β = 0 leads to a solution y = x r of the Euler equation. PR (FIU) MAP / 92
72 Real, Distinct Roots If r 1 and r 1 are the 2, real distinct roots, then y = Ax r 1 + Bx r 2 Example: 2x 2 y + 3xy y = 0, x > 0 PR (FIU) MAP / 92
73 Double Roots F(r) = (r r 1 ) 2 In this case y 1 = x r 1 is a solution and the method of reduction of order shows that y 2 = x r 1 ln x is also a solution. (Check this directly!) Example: x 2 y + 5xy + 4y = 0, x > 0 PR (FIU) MAP / 92
74 Complex-Conjugate Roots r 1 = λ + iµ, r 2 = λ iµ z 1 = x λ+iµ = e (λ+iµ) ln x = x λ (cos (µ ln x) + i sin (µ ln x)) z 2 = x λ (cos (µ ln x) i sin (µ ln x)) PR (FIU) MAP / 92
75 y 1 = 1 2 (z 1 + z 2 ) = x λ cos (µ ln x) and y 2 = 1 2i (z 1 z 2 ) = x λ sin (µ ln x) are two linearly independent real solutions. Example: x 2 y + xy + y = 0 PR (FIU) MAP / 92
76 More Generally Example: 2x 2 y xy + (1 + x)y = 0 0 is a regular singular point. We look for y = x r n=0 a n x n = a n x n+r Substituting into the differential equation and grouping like terms from lowest power of x to higher powers: n=0 F(r)x r+k + n 1[G(r, n)]x n+r+k = 0 PR (FIU) MAP / 92
77 Solve the indicial equation F(r) = 0 and obtain recurrence relations from G(r, n) = 0 If r 1 r 2 is not an integer, we always obtain two linearly independent solutions. PR (FIU) MAP / 92
78 Example 2 x 2 y + (x 2 + 4x)y + (2x + 2)y = 0 y = c n x n+r n=0 y = (n + r)c n x n+r 1, y = (n + r)(n + r 1)c n x n+r 2 PR (FIU) MAP / 92
79 x 2 y = (n + r)c n x n+r+1 = (m 1 + r)c m 1 x m+r n=0 m=1 2xy = 2c n x n+r+1 = 2c m 1 x m+r n=0 m=1 PR (FIU) MAP / 92
80 (n + r)(n + r 1)cn x n+r + (n + r)c n x n+r+1 + 4(n + r)c n x n+r + 2c n x n+r+1 + 2c n x n+r = 0 c 0 (r(r 1) + 4r + 2)x r + m=1 [[(m + r)(m + r + 3) + 2]c m + (m + r + 1)c m 1 ] x m+r = 0 F(r) = r(r + 3) + 2 = 0 = r 2 + 3r + 2 (m + r + 1)c m 1 c m = (m + r)(m + r + 3) + 2 PR (FIU) MAP / 92
81 Work with the smaller root 2 first: r 1 = 2, r 2 = 1 c n = c n 1 n Inductively, we see that c n = ( 1)n c 0 n! This leads to ( 1) n y 1 = c 0 x n 2 = c 0 x 2 ( 1) n x n n! n! n=0 n=0 = c 0 x 2 e x PR (FIU) MAP / 92
82 Form r = 1, we obtain Inductively: c n = c n 1 n + 1 c n = ( 1) n c 0 (n + 1)! Y 2 = c 0 ( 1) n (n + 1)! x n 1 = c 0 x 2 n=0 Check directly that x 2 and x 2 e x are solutions! The general solution is y = Ax 2 e x + Bx 2 ( 1) n+1 (n + 1)! x n+1 = c 0 x 2 (e x 1) PR (FIU) MAP / 92
83 6.3 Bessel s Equation and Bessel s functions Bessel s Equation of order p. x 2 y + xy + (x 2 p 2 )y = 0 Any solution is called a Bessel function of order p. Theses functions occur in connection with problems of Physics and Engineering. PR (FIU) MAP / 92
84 The case p = 0 0 is a regular, singular point. xy + y + xy = 0 Look for y = c n x n+r n=0 r 2 c 0 x r 1 + (1 + r)c 1 x r + [(n + r) 2 c n + c n 2 ]x n+r 1 = 0 n=2 The indicial equation is r 2 = 0 with double root r 1 = 0 = r 2. PR (FIU) MAP / 92
85 Then and (1 + r) 2 c 1 = 0 (n + r) 2 c n + c n 2 = 0, n 2 r = 0 c 1 = 0 r = 0 n 2 c n + c n 2 = 0 c n = c n 2 n 2 PR (FIU) MAP / 92
86 c 2n+1 = 0 c 2n = ( 1)n c 0 (n!) 2 2 2n, n 1 y = c 0 n=0 J 0 (x) = ( 1) n (n!) 2 (x 2 )2n = y 1 (x). n=0 ( 1) n (n!) 2 (x 2 )2n Bessel Function of the first kind of order zero. J 0 (x) = 1 x x 4 64 x A second solution must be of the form (See Theorem 6.3) y = x cnx n + J 0 (x) ln x PR (FIU) n=0 MAP / 92
87 8.3. Successive approximations Picard s Method y = f (x, y), y(x 0 ) = y 0. y = φ(x), the d.e. is just specifying the slope of the tangent line to the graph of the solution! Zeroth approximation: φ 0 = y 0. First approximation: φ 1 (satisfying a different d.e.) φ 1 (x) = f (x, φ 0(x)), φ 1 (x 0 ) = y 0. x φ 1 (x) = y 0 + f (t, φ 0 )dt x 0 PR (FIU) MAP / 92
88 Second approximation: φ 2. (Satisfying a different d.e.) φ 2 (x) = f (x, φ 1), φ 2 (x 0 ) = y 0.. n th approximation: x φ 2 (x) = y 0 + f (t, φ 1 (t))dt x 0 φ n (x) = x x 0 f (t, φ n 1 (t))dt PR (FIU) MAP / 92
89 One has a sequence of functions φ 0, φ 1,..., φ n,... The exact solution is given by: φ = lim n φ n Picard used this method to prove existence of solutions! Observe that φ (x) = lim n φ n(x) = lim n f (x, φ n 1 (x)) = f (x, φ). PR (FIU) MAP / 92
90 Example y = xy, y(0) = 1 Let y = φ(x) where u = x 2 2. φ 0 = 1 x φ 1 = 1 + tdt = 1 + x x φ 2 = 1 + t(1 + t2 0 2 )dt = 1 + x ( x 2 2 φ 3 = 1 + u + u2 2 + u3 3! 2 )2 PR (FIU) MAP / 92
91 . Where u = x 2 2. φ n (x) = n i=1 u i i! φ(x) = lim φ n (x) = e x 2 n 2. PR (FIU) MAP / 92
92 8.4. Numerical Methods: The Euler Method Approximating the solution of y = f (x, y), y(x 0 ) = y 0. Let x 1 = x 0, x 2 = x 0 + h,... x N = x N 1 + h If φ(x) is the exact solution, let φ(x 1 ),..., φ(x N ) be the evaluation of φ at points x k. PR (FIU) MAP / 92
93 A numerical method will use the IVP to estimate φ(x k ), k = 1, 2,..., N. Let y 1, y 2,..., y N be approximations to φ(x 1 ),, φ(x 2 ),..., φ(x N ). A one-step method uses y k 1 to find y k using the differential equation. The method has also an alternate name of starting method". PR (FIU) MAP / 92
94 The multi-step methods (using several previous approximations to find y k ) are also known as "continuing methods". PR (FIU) MAP / 92
95 The Euler Method φ the exact solution of y = f (x, y), y(x 0 ) = y 0. y n+1 = y n + hf (x n, y n ). Geometrically, the segment of then graph of y = φ(x) between (x n, φ(x n )) and (x n+1, φ(x n+1 )) is replaced by the line segment joining (x n, y n ) and (x n+1, y n + hf (x n, y n )). y 0 = φ(x 0 ) PR (FIU) MAP / 92
96 PR (FIU) MAP / 92
97 Example y = 2x + y; y(0) = 1 Use h = 0.2. x 0 = 0 x 1 = 0.2 x 3 = 0.4 x 3 = 0.6 x 4 = 0.8 y 0 = 1 y 1 = y 2 = y 3 = y 4 = PR (FIU) MAP / 92
dx n a 1(x) dy
HIGHER ORDER DIFFERENTIAL EQUATIONS Theory of linear equations Initial-value and boundary-value problem nth-order initial value problem is Solve: a n (x) dn y dx n + a n 1(x) dn 1 y dx n 1 +... + a 1(x)
More information4.2 Homogeneous Linear Equations
4.2 Homogeneous Linear Equations Homogeneous Linear Equations with Constant Coefficients Consider the first-order linear differential equation with constant coefficients a 0 and b. If f(t) = 0 then this
More informationSeries Solution of Linear Ordinary Differential Equations
Series Solution of Linear Ordinary Differential Equations Department of Mathematics IIT Guwahati Aim: To study methods for determining series expansions for solutions to linear ODE with variable coefficients.
More informationODE Homework Series Solutions Near an Ordinary Point, Part I 1. Seek power series solution of the equation. n(n 1)a n x n 2 = n=0
ODE Homework 6 5.2. Series Solutions Near an Ordinary Point, Part I 1. Seek power series solution of the equation y + k 2 x 2 y = 0, k a constant about the the point x 0 = 0. Find the recurrence relation;
More information17.8 Nonhomogeneous Linear Equations We now consider the problem of solving the nonhomogeneous second-order differential
ADAMS: Calculus: a Complete Course, 4th Edition. Chapter 17 page 1016 colour black August 15, 2002 1016 CHAPTER 17 Ordinary Differential Equations 17.8 Nonhomogeneous Linear Equations We now consider the
More informationSeries Solutions Near a Regular Singular Point
Series Solutions Near a Regular Singular Point MATH 365 Ordinary Differential Equations J. Robert Buchanan Department of Mathematics Fall 2018 Background We will find a power series solution to the equation:
More informationSection 3.4. Second Order Nonhomogeneous. The corresponding homogeneous equation. is called the reduced equation of (N).
Section 3.4. Second Order Nonhomogeneous Equations y + p(x)y + q(x)y = f(x) (N) The corresponding homogeneous equation y + p(x)y + q(x)y = 0 (H) is called the reduced equation of (N). 1 General Results
More informationSection 3.4. Second Order Nonhomogeneous. The corresponding homogeneous equation
Section 3.4. Second Order Nonhomogeneous Equations y + p(x)y + q(x)y = f(x) (N) The corresponding homogeneous equation y + p(x)y + q(x)y = 0 (H) is called the reduced equation of (N). 1 General Results
More informationApplications of Second-Order Differential Equations
Applications of Second-Order Differential Equations ymy/013 Building Intuition Even though there are an infinite number of differential equations, they all share common characteristics that allow intuition
More information2. Determine whether the following pair of functions are linearly dependent, or linearly independent:
Topics to be covered on the exam include: Recognizing, and verifying solutions to homogeneous second-order linear differential equations, and their corresponding Initial Value Problems Recognizing and
More information3.4.1 Distinct Real Roots
Math 334 3.4. CONSTANT COEFFICIENT EQUATIONS 34 Assume that P(x) = p (const.) and Q(x) = q (const.), so that we have y + py + qy = 0, or L[y] = 0 (where L := d2 dx 2 + p d + q). (3.7) dx Guess a solution
More informationMATH 23 Exam 2 Review Solutions
MATH 23 Exam 2 Review Solutions Problem 1. Use the method of reduction of order to find a second solution of the given differential equation x 2 y (x 0.1875)y = 0, x > 0, y 1 (x) = x 1/4 e 2 x Solution
More informationMath 266 Midterm Exam 2
Math 266 Midterm Exam 2 March 2st 26 Name: Ground Rules. Calculator is NOT allowed. 2. Show your work for every problem unless otherwise stated (partial credits are available). 3. You may use one 4-by-6
More informationSecond-Order Linear ODEs
C0.tex /4/011 16: 3 Page 13 Chap. Second-Order Linear ODEs Chapter presents different types of second-order ODEs and the specific techniques on how to solve them. The methods are systematic, but it requires
More informationElementary Differential Equations, Section 2 Prof. Loftin: Practice Test Problems for Test Find the radius of convergence of the power series
Elementary Differential Equations, Section 2 Prof. Loftin: Practice Test Problems for Test 2 SOLUTIONS 1. Find the radius of convergence of the power series Show your work. x + x2 2 + x3 3 + x4 4 + + xn
More informationMATH 251 Week 6 Not collected, however you are encouraged to approach all problems to prepare for exam
MATH 51 Week 6 Not collected, however you are encouraged to approach all problems to prepare for exam A collection of previous exams could be found at the coordinator s web: http://www.math.psu.edu/tseng/class/m51samples.html
More informationThe Method of Frobenius
The Method of Frobenius Department of Mathematics IIT Guwahati If either p(x) or q(x) in y + p(x)y + q(x)y = 0 is not analytic near x 0, power series solutions valid near x 0 may or may not exist. If either
More informationMA22S3 Summary Sheet: Ordinary Differential Equations
MA22S3 Summary Sheet: Ordinary Differential Equations December 14, 2017 Kreyszig s textbook is a suitable guide for this part of the module. Contents 1 Terminology 1 2 First order separable 2 2.1 Separable
More informationENGI Second Order Linear ODEs Page Second Order Linear Ordinary Differential Equations
ENGI 344 - Second Order Linear ODEs age -01. Second Order Linear Ordinary Differential Equations The general second order linear ordinary differential equation is of the form d y dy x Q x y Rx dx dx Of
More informationMath Assignment 5
Math 2280 - Assignment 5 Dylan Zwick Fall 2013 Section 3.4-1, 5, 18, 21 Section 3.5-1, 11, 23, 28, 35, 47, 56 Section 3.6-1, 2, 9, 17, 24 1 Section 3.4 - Mechanical Vibrations 3.4.1 - Determine the period
More informationA( x) B( x) C( x) y( x) 0, A( x) 0
3.1 Lexicon Revisited The nonhomogeneous nd Order ODE has the form: d y dy A( x) B( x) C( x) y( x) F( x), A( x) dx dx The homogeneous nd Order ODE has the form: d y dy A( x) B( x) C( x) y( x), A( x) dx
More information2. Second-order Linear Ordinary Differential Equations
Advanced Engineering Mathematics 2. Second-order Linear ODEs 1 2. Second-order Linear Ordinary Differential Equations 2.1 Homogeneous linear ODEs 2.2 Homogeneous linear ODEs with constant coefficients
More informationIntroductory Differential Equations
Introductory Differential Equations Lecture Notes June 3, 208 Contents Introduction Terminology and Examples 2 Classification of Differential Equations 4 2 First Order ODEs 5 2 Separable ODEs 5 22 First
More informationMATH 312 Section 6.2: Series Solutions about Singular Points
MATH 312 Section 6.2: Series Solutions about Singular Points Prof. Jonathan Duncan Walla Walla University Spring Quarter, 2008 Outline 1 Classifying Singular Points 2 The Method of Frobenius 3 Conclusions
More informationENGI 9420 Lecture Notes 1 - ODEs Page 1.01
ENGI 940 Lecture Notes - ODEs Page.0. Ordinary Differential Equations An equation involving a function of one independent variable and the derivative(s) of that function is an ordinary differential equation
More informationMB4018 Differential equations
MB4018 Differential equations Part II http://www.staff.ul.ie/natalia/mb4018.html Prof. Natalia Kopteva Spring 2015 MB4018 (Spring 2015) Differential equations Part II 0 / 69 Section 1 Second-Order Linear
More informationAMATH 351 Mar 15, 2013 FINAL REVIEW. Instructor: Jiri Najemnik
AMATH 351 Mar 15, 013 FINAL REVIEW Instructor: Jiri Najemni ABOUT GRADES Scores I have so far will be posted on the website today sorted by the student number HW4 & Exam will be added early next wee Let
More informationApplications of Second-Order Linear Differential Equations
CHAPTER 14 Applications of Second-Order Linear Differential Equations SPRING PROBLEMS The simple spring system shown in Fig. 14-! consists of a mass m attached lo the lower end of a spring that is itself
More informationChapter 2 Second-Order Linear ODEs
Chapter 2 Second-Order Linear ODEs Advanced Engineering Mathematics Wei-Ta Chu National Chung Cheng University wtchu@cs.ccu.edu.tw 1 2.1 Homogeneous Linear ODEs of Second Order 2 Homogeneous Linear ODEs
More informationSeries solutions of second order linear differential equations
Series solutions of second order linear differential equations We start with Definition 1. A function f of a complex variable z is called analytic at z = z 0 if there exists a convergent Taylor series
More information17.2 Nonhomogeneous Linear Equations. 27 September 2007
17.2 Nonhomogeneous Linear Equations 27 September 2007 Nonhomogeneous Linear Equations The differential equation to be studied is of the form ay (x) + by (x) + cy(x) = G(x) (1) where a 0, b, c are given
More informationJim Lambers MAT 285 Spring Semester Practice Exam 2 Solution. y(t) = 5 2 e t 1 2 e 3t.
. Solve the initial value problem which factors into Jim Lambers MAT 85 Spring Semester 06-7 Practice Exam Solution y + 4y + 3y = 0, y(0) =, y (0) =. λ + 4λ + 3 = 0, (λ + )(λ + 3) = 0. Therefore, the roots
More informationReview for Exam 2. Review for Exam 2.
Review for Exam 2. 5 or 6 problems. No multiple choice questions. No notes, no books, no calculators. Problems similar to homeworks. Exam covers: Regular-singular points (5.5). Euler differential equation
More informationLecture 1: Review of methods to solve Ordinary Differential Equations
Introductory lecture notes on Partial Differential Equations - c Anthony Peirce Not to be copied, used, or revised without explicit written permission from the copyright owner 1 Lecture 1: Review of methods
More informationBessel s Equation. MATH 365 Ordinary Differential Equations. J. Robert Buchanan. Fall Department of Mathematics
Bessel s Equation MATH 365 Ordinary Differential Equations J. Robert Buchanan Department of Mathematics Fall 2018 Background Bessel s equation of order ν has the form where ν is a constant. x 2 y + xy
More informationAPPLIED MATHEMATICS. Part 1: Ordinary Differential Equations. Wu-ting Tsai
APPLIED MATHEMATICS Part 1: Ordinary Differential Equations Contents 1 First Order Differential Equations 3 1.1 Basic Concepts and Ideas................... 4 1.2 Separable Differential Equations................
More informationSeries Solutions of Linear ODEs
Chapter 2 Series Solutions of Linear ODEs This Chapter is concerned with solutions of linear Ordinary Differential Equations (ODE). We will start by reviewing some basic concepts and solution methods for
More informationMath 240: Spring-mass Systems
Math 240: Spring-mass Systems Ryan Blair University of Pennsylvania Tuesday March 1, 2011 Ryan Blair (U Penn) Math 240: Spring-mass Systems Tuesday March 1, 2011 1 / 15 Outline 1 Review 2 Today s Goals
More informationM A : Ordinary Differential Equations
M A 2 0 5 1: Ordinary Differential Equations Essential Class Notes & Graphics D 19 * 2018-2019 Sections D07 D11 & D14 1 1. INTRODUCTION CLASS 1 ODE: Course s Overarching Functions An introduction to the
More informationSection 3.4. Second Order Nonhomogeneous. The corresponding homogeneous equation. is called the reduced equation of (N).
Section 3.4. Second Order Nonhomogeneous Equations y + p(x)y + q(x)y = f(x) (N) The corresponding homogeneous equation y + p(x)y + q(x)y = 0 (H) is called the reduced equation of (N). 1 General Results
More informationPractice Problems For Test 3
Practice Problems For Test 3 Power Series Preliminary Material. Find the interval of convergence of the following. Be sure to determine the convergence at the endpoints. (a) ( ) k (x ) k (x 3) k= k (b)
More informationMath 4B Notes. Written by Victoria Kala SH 6432u Office Hours: T 12:45 1:45pm Last updated 7/24/2016
Math 4B Notes Written by Victoria Kala vtkala@math.ucsb.edu SH 6432u Office Hours: T 2:45 :45pm Last updated 7/24/206 Classification of Differential Equations The order of a differential equation is the
More informationMath Assignment 11
Math 2280 - Assignment 11 Dylan Zwick Fall 2013 Section 8.1-2, 8, 13, 21, 25 Section 8.2-1, 7, 14, 17, 32 Section 8.3-1, 8, 15, 18, 24 1 Section 8.1 - Introduction and Review of Power Series 8.1.2 - Find
More informationDifferential Equations
Differential Equations A differential equation (DE) is an equation which involves an unknown function f (x) as well as some of its derivatives. To solve a differential equation means to find the unknown
More information2 Series Solutions near a Regular Singular Point
McGill University Math 325A: Differential Equations LECTURE 17: SERIES SOLUTION OF LINEAR DIFFERENTIAL EQUATIONS II 1 Introduction Text: Chap. 8 In this lecture we investigate series solutions for the
More informationPractice Problems For Test 3
Practice Problems For Test 3 Power Series Preliminary Material. Find the interval of convergence of the following. Be sure to determine the convergence at the endpoints. (a) ( ) k (x ) k (x 3) k= k (b)
More informationSecond order linear equations
Second order linear equations Samy Tindel Purdue University Differential equations - MA 266 Taken from Elementary differential equations by Boyce and DiPrima Samy T. Second order equations Differential
More informationLinear Second Order ODEs
Chapter 3 Linear Second Order ODEs In this chapter we study ODEs of the form (3.1) y + p(t)y + q(t)y = f(t), where p, q, and f are given functions. Since there are two derivatives, we might expect that
More informationUnderstand the existence and uniqueness theorems and what they tell you about solutions to initial value problems.
Review Outline To review for the final, look over the following outline and look at problems from the book and on the old exam s and exam reviews to find problems about each of the following topics.. Basics
More informationThe Harmonic Oscillator
The Harmonic Oscillator Math 4: Ordinary Differential Equations Chris Meyer May 3, 008 Introduction The harmonic oscillator is a common model used in physics because of the wide range of problems it can
More informationSECOND-ORDER DIFFERENTIAL EQUATIONS
Chapter 16 SECOND-ORDER DIFFERENTIAL EQUATIONS OVERVIEW In this chapter we extend our study of differential euations to those of second der. Second-der differential euations arise in many applications
More informationDefinition of differential equations and their classification. Methods of solution of first-order differential equations
Introduction to differential equations: overview Definition of differential equations and their classification Solutions of differential equations Initial value problems Existence and uniqueness Mathematical
More informationMath 211. Substitute Lecture. November 20, 2000
1 Math 211 Substitute Lecture November 20, 2000 2 Solutions to y + py + qy =0. Look for exponential solutions y(t) =e λt. Characteristic equation: λ 2 + pλ + q =0. Characteristic polynomial: λ 2 + pλ +
More informationSection 3.4. Second Order Nonhomogeneous. The corresponding homogeneous equation. is called the reduced equation of (N).
Section 3.4. Second Order Nonhomogeneous Equations y + p(x)y + q(x)y = f(x) (N) The corresponding homogeneous equation y + p(x)y + q(x)y = 0 (H) is called the reduced equation of (N). 1 General Results
More informationMATH 2250 Final Exam Solutions
MATH 225 Final Exam Solutions Tuesday, April 29, 28, 6: 8:PM Write your name and ID number at the top of this page. Show all your work. You may refer to one double-sided sheet of notes during the exam
More information2. Higher-order Linear ODE s
2. Higher-order Linear ODE s 2A. Second-order Linear ODE s: General Properties 2A-1. On the right below is an abbreviated form of the ODE on the left: (*) y + p()y + q()y = r() Ly = r() ; where L is the
More information7.3 Singular points and the method of Frobenius
284 CHAPTER 7. POWER SERIES METHODS 7.3 Singular points and the method of Frobenius Note: or.5 lectures, 8.4 and 8.5 in [EP], 5.4 5.7 in [BD] While behaviour of ODEs at singular points is more complicated,
More informationPower series solutions for 2nd order linear ODE s (not necessarily with constant coefficients) a n z n. n=0
Lecture 22 Power series solutions for 2nd order linear ODE s (not necessarily with constant coefficients) Recall a few facts about power series: a n z n This series in z is centered at z 0. Here z can
More informationMA 266 Review Topics - Exam # 2 (updated)
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
More informationAPPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS
APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS Second-order linear differential equations have a variety of applications in science and engineering. In this section we explore two of them: the vibration
More informationLECTURE 14: REGULAR SINGULAR POINTS, EULER EQUATIONS
LECTURE 14: REGULAR SINGULAR POINTS, EULER EQUATIONS 1. Regular Singular Points During the past few lectures, we have been focusing on second order linear ODEs of the form y + p(x)y + q(x)y = g(x). Particularly,
More informationCh 3.7: Mechanical & Electrical Vibrations
Ch 3.7: Mechanical & Electrical Vibrations Two important areas of application for second order linear equations with constant coefficients are in modeling mechanical and electrical oscillations. We will
More informationEx. 1. Find the general solution for each of the following differential equations:
MATH 261.007 Instr. K. Ciesielski Spring 2010 NAME (print): SAMPLE TEST # 2 Solve the following exercises. Show your work. (No credit will be given for an answer with no supporting work shown.) Ex. 1.
More informationSection Mass Spring Systems
Asst. Prof. Hottovy SM212-Section 3.1. Section 5.1-2 Mass Spring Systems Name: Purpose: To investigate the mass spring systems in Chapter 5. Procedure: Work on the following activity with 2-3 other students
More informationForced Mechanical Vibrations
Forced Mechanical Vibrations Today we use methods for solving nonhomogeneous second order linear differential equations to study the behavior of mechanical systems.. Forcing: Transient and Steady State
More information=================~ NONHOMOGENEOUS LINEAR EQUATIONS. rn y" - y' - 6y = 0. lid y" + 2y' + 2y = 0, y(o) = 2, y'(0) = I
~ EXERCISES rn y" - y' - 6y = 0 3. 4y" + y = 0 5. 9y" - 12y' + 4y = 0 2. y" + 4 y' + 4 y = 0 4. y" - 8y' + 12y = 0 6. 25y" + 9y = 0 dy 8. dt2-6 d1 + 4y = 0 00 y" - 4y' + By = 0 10. y" + 3y' = 0 [ITJ2-+2--y=0
More informationOrdinary differential equations
Class 7 Today s topics The nonhomogeneous equation Resonance u + pu + qu = g(t). The nonhomogeneous second order linear equation This is the nonhomogeneous second order linear equation u + pu + qu = g(t).
More information12d. Regular Singular Points
October 22, 2012 12d-1 12d. Regular Singular Points We have studied solutions to the linear second order differential equations of the form P (x)y + Q(x)y + R(x)y = 0 (1) in the cases with P, Q, R real
More informationThe most up-to-date version of this collection of homework exercises can always be found at bob/math365/mmm.pdf.
Millersville University Department of Mathematics MATH 365 Ordinary Differential Equations January 23, 212 The most up-to-date version of this collection of homework exercises can always be found at http://banach.millersville.edu/
More informationChapter 2 Second Order Differential Equations
Chapter 2 Second Order Differential Equations Either mathematics is too big for the human mind or the human mind is more than a machine. - Kurt Gödel (1906-1978) 2.1 The Simple Harmonic Oscillator The
More informationTable of contents. d 2 y dx 2, As the equation is linear, these quantities can only be involved in the following manner:
M ath 0 1 E S 1 W inter 0 1 0 Last Updated: January, 01 0 Solving Second Order Linear ODEs Disclaimer: This lecture note tries to provide an alternative approach to the material in Sections 4. 4. 7 and
More informationUnforced Mechanical Vibrations
Unforced Mechanical Vibrations Today we begin to consider applications of second order ordinary differential equations. 1. Spring-Mass Systems 2. Unforced Systems: Damped Motion 1 Spring-Mass Systems We
More informationOld Math 330 Exams. David M. McClendon. Department of Mathematics Ferris State University
Old Math 330 Exams David M. McClendon Department of Mathematics Ferris State University Last updated to include exams from Fall 07 Contents Contents General information about these exams 3 Exams from Fall
More informationSection 3.7: Mechanical and Electrical Vibrations
Section 3.7: Mechanical and Electrical Vibrations Second order linear equations with constant coefficients serve as mathematical models for mechanical and electrical oscillations. For example, the motion
More informationLecture 13: Series Solutions near Singular Points
Lecture 13: Series Solutions near Singular Points March 28, 2007 Here we consider solutions to second-order ODE s using series when the coefficients are not necessarily analytic. A first-order analogy
More informationENGI 9420 Lecture Notes 4 - Stability Analysis Page Stability Analysis for Non-linear Ordinary Differential Equations
ENGI 940 Lecture Notes 4 - Stability Analysis Page 4.01 4. Stability Analysis for Non-linear Ordinary Differential Equations A pair of simultaneous first order homogeneous linear ordinary differential
More informationChapter 3. Reading assignment: In this chapter we will cover Sections dx 1 + a 0(x)y(x) = g(x). (1)
Chapter 3 3 Introduction Reading assignment: In this chapter we will cover Sections 3.1 3.6. 3.1 Theory of Linear Equations Recall that an nth order Linear ODE is an equation that can be written in the
More informationMath 3313: Differential Equations Second-order ordinary differential equations
Math 3313: Differential Equations Second-order ordinary differential equations Thomas W. Carr Department of Mathematics Southern Methodist University Dallas, TX Outline Mass-spring & Newton s 2nd law Properties
More informationChapter 3: Second Order Equations
Exam 2 Review This review sheet contains this cover page (a checklist of topics from Chapters 3). Following by all the review material posted pertaining to chapter 3 (all combined into one file). Chapter
More informationSecond-Order Linear ODEs
Chap. 2 Second-Order Linear ODEs Sec. 2.1 Homogeneous Linear ODEs of Second Order On pp. 45-46 we extend concepts defined in Chap. 1, notably solution and homogeneous and nonhomogeneous, to second-order
More informationSTABILITY. Phase portraits and local stability
MAS271 Methods for differential equations Dr. R. Jain STABILITY Phase portraits and local stability We are interested in system of ordinary differential equations of the form ẋ = f(x, y), ẏ = g(x, y),
More informationHonors Differential Equations
MIT OpenCourseWare http://ocw.mit.edu 18.034 Honors Differential Equations Spring 009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. LECTURE 7. MECHANICAL
More informationMath53: Ordinary Differential Equations Autumn 2004
Math53: Ordinary Differential Equations Autumn 2004 Unit 2 Summary Second- and Higher-Order Ordinary Differential Equations Extremely Important: Euler s formula Very Important: finding solutions to linear
More informationThe Method of Frobenius
The Method of Frobenius R. C. Trinity University Partial Differential Equations April 7, 2015 Motivating example Failure of the power series method Consider the ODE 2xy +y +y = 0. In standard form this
More informationM A : Ordinary Differential Equations
M A 2 0 5 1: Ordinary Differential Equations Essential Class Notes & Graphics C 17 * Sections C11-C18, C20 2016-2017 1 Required Background 1. INTRODUCTION CLASS 1 The definition of the derivative, Derivative
More information4.1 Analysis of functions I: Increase, decrease and concavity
4.1 Analysis of functions I: Increase, decrease and concavity Definition Let f be defined on an interval and let x 1 and x 2 denote points in that interval. a) f is said to be increasing on the interval
More informationMA Ordinary Differential Equations
MA 108 - Ordinary Differential Equations Santanu Dey Department of Mathematics, Indian Institute of Technology Bombay, Powai, Mumbai 76 dey@math.iitb.ac.in March 21, 2014 Outline of the lecture Second
More informationB Ordinary Differential Equations Review
B Ordinary Differential Equations Review The profound study of nature is the most fertile source of mathematical discoveries. - Joseph Fourier (1768-1830) B.1 First Order Differential Equations Before
More informationIntroduction to First Order Equations Sections
A B I L E N E C H R I S T I A N U N I V E R S I T Y Department of Mathematics Introduction to First Order Equations Sections 2.1-2.3 Dr. John Ehrke Department of Mathematics Fall 2012 Course Goals The
More informationMATH 391 Test 1 Fall, (1) (12 points each)compute the general solution of each of the following differential equations: = 4x 2y.
MATH 391 Test 1 Fall, 2018 (1) (12 points each)compute the general solution of each of the following differential equations: (a) (b) x dy dx + xy = x2 + y. (x + y) dy dx = 4x 2y. (c) yy + (y ) 2 = 0 (y
More informationThursday, August 4, 2011
Chapter 16 Thursday, August 4, 2011 16.1 Springs in Motion: Hooke s Law and the Second-Order ODE We have seen alrealdy that differential equations are powerful tools for understanding mechanics and electro-magnetism.
More informationExam Basics. midterm. 1 There will be 9 questions. 2 The first 3 are on pre-midterm material. 3 The next 1 is a mix of old and new material.
Exam Basics 1 There will be 9 questions. 2 The first 3 are on pre-midterm material. 3 The next 1 is a mix of old and new material. 4 The last 5 questions will be on new material since the midterm. 5 60
More informationSecond In-Class Exam Solutions Math 246, Professor David Levermore Thursday, 31 March 2011
Second In-Class Exam Solutions Math 246, Professor David Levermore Thursday, 31 March 211 (1) [6] Give the interval of definition for the solution of the initial-value problem d 4 y dt 4 + 7 1 t 2 dy dt
More informationApplied Differential Equation. October 22, 2012
Applied Differential Equation October 22, 22 Contents 3 Second Order Linear Equations 2 3. Second Order linear homogeneous equations with constant coefficients.......... 4 3.2 Solutions of Linear Homogeneous
More informationPolytechnic Institute of NYU MA 2132 Final Practice Answers Fall 2012
Polytechnic Institute of NYU MA Final Practice Answers Fall Studying from past or sample exams is NOT recommended. If you do, it should be only AFTER you know how to do all of the homework and worksheet
More informationMath K (24564) - Lectures 02
Math 39100 K (24564) - Lectures 02 Ethan Akin Office: NAC 6/287 Phone: 650-5136 Email: ethanakin@earthlink.net Spring, 2018 Contents Second Order Linear Equations, B & D Chapter 4 Second Order Linear Homogeneous
More information2.4 Models of Oscillation
2.4 Models of Oscillation In this section we give three examples of oscillating physical systems that can be modeled by the harmonic oscillator equation. Such models are ubiquitous in physics, but are
More information1 Solution to Homework 4
Solution to Homework Section. 5. The characteristic equation is r r + = (r )(r ) = 0 r = or r =. y(t) = c e t + c e t y = c e t + c e t. y(0) =, y (0) = c + c =, c + c = c =, c =. To find the maximum value
More informationREFERENCE: CROFT & DAVISON CHAPTER 20 BLOCKS 1-3
IV ORDINARY DIFFERENTIAL EQUATIONS REFERENCE: CROFT & DAVISON CHAPTER 0 BLOCKS 1-3 INTRODUCTION AND TERMINOLOGY INTRODUCTION A differential equation (d.e.) e) is an equation involving an unknown function
More informationsuch as the ratio test. However, even without knowing the formula for a n and y 4
5.2 Series Solutions near an Ordinary Point, Part I 247 such as the ratio test. However, even without knowing the formula for a n we shall see in Section 5.3 that it is possible to establish that the series
More information