Nonhomogenous, Linear, Second–Order, Differential Equations

Nonhomogeneous, Linear, Second-order, Differential Equations Outline Nonhomogenous, Linear, Second– Order, Differential Equations • Review last class • Second-order nonhomogenous equations with constant coefficients – Solution is sum of homogenous equation solution, yH , plus a particular solution, yP, for the nonhomogenous part – Method of undetermined coefficients – Variation of parameters 2 Review y’’ + αy’ + βy = 0 Review y’’ + αy’ + βy = 0 II • Three cases depending on 2 = β – α2/4 • Double root when β = α2/4: – y = (C1 + C2x) eαx/2 • Complex roots when β > α2/4, 2 > 0 – y = e–αx/2 [Acos x +Bsin x] • Distinct real roots when β < α2/4 – y = C1e1x + C2e2x     2  4         2 2 2 2  • Initial conditions y(0) = y0 and y’(0) = v0 • Double root when β = α2/4: – y = [(v0 + y0α/2)x + y0] eαx/2 • Complex roots when β > α2/4 – y = e–αx/2 [y0cos x +1(v0 + y0α/2)sin x] • Distinct real roots when β < α2/4 – y = C1e1x + C2e2x – C1 = (2y0 – v0)/(2 – ) – C2 = (v0 – y0)/(2 – ) 3 4 Example: Mechanical Systems Spring-mass Oscillations 2.5 • Spring/mass/damper with md2y/dt2 + cdy/dt + ky = 0, y(0) = y0, y’(0) = v0 • Solutions depended on parameters 2.0 v* = dimensionless initial velocity = v0/y0 1.5 1.0 – Pure oscillation for c = 0 – Underdamping when c2 < 4km – Critical damping when c2 = 4km – Overdamping when c2 > 4km – Solution for y/y0 depends on ct/m (or t for c = 0; 2 = k/m), km/c2, and mv0/cy0 – c2/m2 and k/m have dimensions (time)-2 y/y0 0.5 v* = 2 0.0 v* = 1 v* = 0 v* = -1 -0.5 v* = -2 -1.0 -1.5 -2.0 -2.5 0 5 1 2 3 4 5 6 7 t 8 9 10 11 12 13 6 Page 1 Nonhomogeneous, Linear, Second-order, Differential Equations 2 Overdamping 4km/c2 = 0.99 Overdamping with 4km/c = 0.75 1.6 v* = dimensionless initial velocity = mv0/cy0 1.4 1.2 1.2 0.6 displacemen 0.8 v* = dimensionles s initial velocity = mv /cy v* = 2.0 v* = 1.0 v* = 0.0 v* = -1.0 v* = -2.0 1.4 v* = 2.0 v* = 1.0 v* = 0.0 v* = -1.0 v* = -2.0 1 displacemen 1.6 0.4 1 0.8 0.6 0.4 0.2 0.2 0 0 -0.2 -0.2 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 dimensionless time ct/m 4 5 6 7 8 9 10 9 10 dimensionless time ct/m Underdamping 4km/c2 = 10 Critical Damping 1.6 1.2 v* = dimensionless initial velocity = mv0/cy0 1.4 1.2 1 v* = dimensionles s initial velocity = mv0/cy0 0.8 v* = 1.0 v* = 0.5 v* = 0.0 v* = -0.5 v* = -1.0 0.6 0.4 displacemen y/y0 1 0.8 0.4 0.2 0 0.2 -0.2 0 -0.4 -0.2 -0.6 0 1 2 3 4 5 6 7 8 9 10 dimensionless time ct/m v* = dimensionless initial velocity = mv0/cy0 0.2 2 3 4 5 6 7 8 • Proposed alternative with equation from trigonometry for cos(a – b) v* = 2.0 v* = 1.0 v* = 0.0 v* = -1.0 v* = -2.0 exp(-ct/2m) -exp(-ct/2m) 0.8 0.4 1 Alternative Formulations Underdamping 4km/c2 = 50 0.6 0 dimensionless time ct/m 1 displacemen 0.6 v* = 2.0 v* = 1.0 v* = 0.0 v* = -1.0 v* = -2.0 exp(-ct/2m) -exp(-ct/2m) y  C cost     C sin  sin t  C cos  cos t • How does this match original equation? 0 y  A sin t  B cos t -0.2 -0.4 -0.6 -0.8 0 1 2 3 4 5 6 dimensionless time ct/m 7 8 9 10 • Matches if A = C sin  and B = C cos  • This gives A2 + B2 = C2 sin2  + C2 cos2  = C2 and A/B = (C sin )/(C cos ) = tan  or  = tan-1(A/B) = tan-1(v0/y0 ) 12 Page 2 Nonhomogeneous, Linear, Second-order, Differential Equations Nonhomogeneous Equations • Solution to linear nonhomogeneous second-order equation, y = yH + yP Nonhomogeneous Equations II • What is first step in solving the general second-order nonhomogeneous ODE? d2y dy  p( x)  q( x) y  r ( x) dx 2 dx d2y dy  p( x)  q( x) y  r ( x) dx 2 dx • yH is general solution to corresponding homogenous equation, C1y1 + C2y2 • First step is finding yH, which is the general solution to corresponding homogenous equation dy d 2 yH  p( x) H  q( x) yH  0 dx dx 2 d 2 yH dy  p( x) H  q( x) y H  0 dx 2 dx • yP is particular solution 13 Nonhomogeneous Equations III • Substitute solution y = yH + yP into the nonhomogeneous equation Nonhomogeneous Equations IV • Solve nonhomogeneous equations by first finding homogeneous solution d 2  yH  yP  d  yH  yP   p( x)  q ( x ) y H  y P   r ( x ) dx 2 dx 2 2 d yH d yP dy dy   p ( x ) H  p ( x ) P  q ( x ) y H  q ( x) y P  r ( x ) dx 2 dx 2 dx dx d 2 yH dy H d 2 yP dy  p( x)  q ( x) y H   p ( x ) P  q ( x) y P  r ( x ) dx 2 dx dx 2 dx Homogeneous Equation = 0 14 d 2 yP dy  p( x) P  q( x) y P  r ( x) dx 2 dx dy d 2 yH  p( x) H  q( x) yH  0 dx dx 2 • Do not solve for C1 and C2 in homogenous solution yH = C1y1 + C2y2 • Solve d 2 yP dy  p( x) P  q( x) y P  r ( x) for yP dx 2 dx • Apply initial/boundary conditions to y = C1y1 + C2y2 + yP to get C1 and C2 15 Solving 16 Example: solve d 2 yP dy  3 P  2 yP  x 2 dx dx 2 d 2 yP dy   P   y P  r ( x) dx 2 dx • Method of undetermined coefficients applies for constant coefficient equation – Assume a solution for yP based on the form of r(x) with constants • Process for assuming yP to be described later – E.g., if r(x) = x2 assume a solution of the form yP = a0 + a1 x + a2 x2 – Substitute proposed solution into the differential equation for yP – Set coefficient sum of like terms in yP to zero – Solve for the unknown ak coefficients in yP 17 • Assume yP = a0 + a1 x + a2 x2 • Substitute yP, dyP/dx = a1 + 2a2x and d2yP/dx2 = 2a2 into differential equation   2a2  32a2 x  a1   2 a2 x 2  a1 x  a0  x 2 • Set coefficient sums of like terms (like powers of x in this example) to zero • 2a2 + 3a1 + 2a0 = 0 for x0 terms 3 equations for a0, • 6a2 + 2a1 = 0 for x1 terms a1, and a2 solved • 2a2 = 1 for x2 terms on next slide 18 Page 3 Nonhomogeneous, Linear, Second-order, Differential Equations Example: solve d 2 yP dy  3 P  2 yP  x 2 dx 2 dx • • • • • • 2a2 + 3a1 + 2a0 = 0 => a0 = -a2 - 3a1/2 6a2 + 2a1 = 0 => a1 = -3a2 2a2 = 1 => a2 = ½ so a1 = -3a2 = -3/2 a0 = -a2 - 3a1/2 = -1/2 – 3(-3/2)/2 = 7/4 So yP= a0 + a1 x + a2 x2 = 7/4 – 3x/2 + x2/2 Check this: yP’’ + 3yP’ + 2yP = 1 + 3(x – 3/2) + 2(7/4 – 3x/2 + x2/2) = 1 – 9/2 + 7/2 + x(3 – 3) + x2 = x2 proving solution for yP 19 Example: solve • • • • • • • • • Example: solve • Just found yP = 7/4 – 3x/2 + x2/2 • Characteristic equation for homogenous ODE has two distinct real roots      2  4   3  32  4(2)   2,1 2 2 • y = yH + yP = C1e-2x + C2e-x + 7/4 – 3x/2 + x2/2 • Find C1 and C2 from initial conditions such as y(0) = y0 and y’(0) = v0 20 Example: solve d2y dy  3  2 y  x2 dx 2 dx y = C1e-2x + C2e-x + 7/4 – 3x/2 + x2/2 y’ = -2C1e-2x - C2e-x – 3/2 + x y0 = C1e0 + C2e0 + 7/4 – 3(0)/2 + 02/2 y0 = C1 + C2 + 7/4 => C2 = y0 – 7/4 – C1 v0 = –2C1e0 – C2e0 – 3/2 + 0 v0 = –2C1 – C2 – 3/2 y0 + v0 = C1 + C2 + 7/4 + (– 2C1 – C2 – 3/2) C1 = ¼ – y0 – v0 C2 = y0 – 7/4 – C1 = 2y0 + v0 – 2 d2y dy  3  2 y  x2 dx 2 dx d2y dy  3  2 y  x2 dx 2 dx • Summary of process to solve nonhomogeneous ODE with given initial conditions • • • • First found yH = C1e-2x + C2e-x Next found yP = 7/4 – 3x/2 + x2/2 Wrote y = yH + yP (including C1 and C2) Then used initial conditions on y and dy/dx to find C1 = ¼ – y0 – v0 and C2 = 2y0 + v0 – 2 • Substitute C1 and C2 equations into solution for y to get desired result • y = (¼ – y0 – v0)e-2x + (2y0 + v0 – 2)e-x + 7/4 – 3x/2 + x2/2 21 Finding Particular Solution 22 Table of Trial yP Solutions • Used for constant coefficient equation y’’ + ay’ + by = r(x) • Solution is y = yP + yH, where yH is solution of yH’’ + ayH’ + byH = 0 • Postulate a solution for yP following gidelines on next two charts • Plug postulated yP solution into ODE and solve for unknown coefficients For these r(x) r(x) = Aeax r(x) = Axn r(x) = Asin x r(x) = Acos x r(x) = Aeaxsin x – Overall coefficients of like terms on both sides of ODE must vanish r(x) = Aeaxcos x 23 Start with this yP yP = Beax yP = a0 + a1x + … + anxn yP = B sin x + C cos x yP = eax (B sin x + C cos x) 24 Page 4 Nonhomogeneous, Linear, Second-order, Differential Equations Exercise: solve Special Rules • If the right-hand-side, r(x) consists of more than one term from the previous table, use a yP that contains all the corresponding yP terms d2y dy  3  2 y  2 cos(5 x) dx 2 dx • Previous example with new right hand side • Keep same initial conditions y(0) = y0 and dy/dx(0) = v0 • Will have same homogenous solution found previously: yH = C1e-2x + C2e-x • Find new particular solution for 2cos(5x) • Use initial conditions on y and dy/dx to find C1 and C2 • Substitute C1 and C2 equations into solution for y to get desired result – For r(x) = Acos bx + Cedx, use yP = E sin bx + F cos bx + Gedx • If r(x) is proportional to a solution for the homogenous equation, use yP equal to x times the yP shown in the table – For a double root, multiply table yP by x2 25 26 Variation of Parameters Variation of Parameters II • Background – we previously stated the solution to the general first-order linear equation dy/dx + f(x)y = g(x) p   f ( x)dx • Homogenous equation has separable solution found below  y  e  p C   e p g ( x)dx  • Now we derive this solution as an example of variation of parameters • Start with solution to homogenous equation: dyH/dx + f(x)yH = 0 dy H   f ( x) y H dx ln y H    f ( x)dx   dy H   f ( x)dx yH yH  e  p p   f ( x)dx • Try a solution to the nonhomogenous equation of the form y(x) = u(x)yH(x) • This gives y’ = u’yH + uyH’ dy/dx + f(x)y = g(x) • Substitute this into original equation 27 Variation of Parameters III 28 Variation of Parameters IV • Rearrange equation to get homogenous differential equation with yH that is zero dy du dy  f ( x) y  g ( x)  yH  u H  f ( x )uy H   g ( x ) dx dx dx du du  dy H  u  f ( x ) yH   yH  g ( x)  yH  g ( x) dx dx  dx  • Since we know yH(x), we now have a separable ODE to solve for u as shown on the next chart ln • Rearrange final (separable) differential equation to solve for u by integration yH du  g ( x)  dx du g ( x )  dx yH Substitute y  uyH  ue  p y  uy H  y H   u g ( x) dx  C yH p   f ( x )dx  f ( x ) dx g ( x)  f ( x ) dx g ( x )dx  C dx  C  e  e yH • Can we apply this to second-order? 29 30 Page 5 Nonhomogeneous, Linear, Second-order, Differential Equations Variation of Parameters V Variation of Parameters VI • Nonhomogenous d2y dy  p ( x )  q( x ) y  r ( x ) and homogenous dx 2 dx equations d 2 y1 dy1 • y1(x) and y2(x) are dx 2  p( x ) dx  q( x ) y1  0 LI homogenous d 2 y dy 2  p ( x ) 2  q( x ) y2  0 solutions 2 dx dx • Try solution y = u(x) y1(x) + v(x) y2(x) – Two functions but one equation – Create another (arbitrary) equation • First derivative of y(x) is = u’(x) y1(x) + u(x) y1’(x) + v’(x) y2(x) + v(x) y2’(x) • Pick arbitrary second equation to simplify result: u’(x) y1(x) + v’(x) y2(x) = 0 • This gives y’(x) = u(x) y1’(x) + v(x) y2’(x) • y’’ = uy1’’ + u’y1’ +vy2’’ + v’y2’ • Substitute results for y, y’, and y’’ into the original differential equation 31 Variation of Parameters VII 2 Variation of Parameters VIII 2 d y dy d y du dy1  p  qy  r  u 21   dx 2 dx dx dx dx 2 d y2 dv dy2 dy   dy   p u 1  v 2   q (uy1  vy2 )  r v dx 2 dx dx dx   dx • Two equations in two unknowns dy1 du dy2 dv  r dx dx dx dx du dv y1  y2 0 • Auxiliary equation dx dx • ODE Result • Rearrange to get homogenous ODE equal zero with solutions y1 and y2  d 2y  d 2 y dy dy u  21  p 1  qy1   v  22  p 2  qy2  Homogenous dx dx  ODEs = zero  dx   dx  32 du dy1 dv dy2  r dx dx dx dx • Multiply first equation by y1, second equation by dy1/dx and subtract dy1  dv dv  dy2  y1 dx  y2 dx  dx  W dx  y1r 33 Variation of Parameters IX Variation of Parameters X • Repeat last chart with change to get equation for du/dx instead of dv/dx • Integrate du/dx and dv/dx equations u   dy1 du dy2 dv  r • ODE Result dx dx dx dx du dv • Auxiliary equation y1  y2 0 dx dx y2 r ( x ) dx W ( x) v y1r ( x) dx W ( x) • Plug into original solution: yP = y1u + y2v y P  y1u  y2 v   y1  • Multiply first equation by y2, second equation by dy2/dx and subtract first from second dy1  du du  dy2  y1 dx  y2 dx   dx  W dx   y2 r 34 y2 r ( x ) y r ( x) dx  y2  1 dx W ( x) W ( x) • Get y = yH + yP and evaluate constants in yH solution from initial conditions 35 36 Page 6 Nonhomogeneous, Linear, Second-order, Differential Equations Nonhomogenous Summary Exercise Solution I • Undetermined coefficients is simpler approach but is limited – Constant coefficient equations – Limited set of functions • Variation of parameters is more complex, but handles more cases • In reality, there are no general methods to get homogenous solution to linear, second-order ODE without constant coefficients • For right-hand-side 2cos(5x) use particular solution yP = Acos(5x) + Bsin(5x) • dyp/dx = -5Asin(5x) + 5Bcos(5x) • d2yp/dx = -25Acos(5x) - 25Bsin(5x) • Use ODE for yP: y’’ + 3y’ + 2y = 2cos(5x) • -25Acos(5x) - 25Bsin(5x) + 3[-5Asin(5x) + 5Bcos(5x)] + 2[Acos(5x) + Bsin(5x)] = 2cos(5x) and set coefficients of sine and cosine terms to zero 37 Exercise Solution II • • • • • • • • 38 Exercise Solution III Equate coefficients of like terms (-25A +15B +2A)cos(5x) = 2cos(5x) (-25B - 15A +2B)sin(5x) = 0 First equation gives -23A +15B = 2 Second equation gives A = -23B/15 Combined equations 529B/15+15B = 2 Solving gives 529B + 225B = 30 B = 30/754=15/377; A= -23B/15 =-23/377 • yP = Acos(5x) + Bsin(5x) = -23cos(5x)/377 + 15sin(5x)/377 • y = yH + yP = C1e-2x + C2e-x -23cos(5x)/377 + 15sin(5x)/377 • y(0) = y0 = C1 + C2 – 23/377 • y’ = -2C1e-2x – C2e-x + 115sin(5x)/377 + 75cos(5x)/377 • y’(0) = v0 = -2C1 – C2 + 75/377 • y0 + v0 = -C1 + 52/377 (Added red boxes) 39 Exercise Solution IV 40 Check Initial Conditions I • C1 = 52/377 – y0 – v0 C1 • Find C2 from first red box • C2 = y0 – C1 + 23/377 = y0 – [52/377 – y0 – v0] + 23/377 = 2y0 + v0 – 29/377 • y = yH + yP = C1e-2x + C2e-x – 23cos(5x)/377 + 15sin(5x)/377 • y = [52/377 – y0 – v0]e-2x + [2y0 + v0 – 29/377]e-x – 23cos(5x)/377 + 15sin(5x)/377 • y = [52/377 – y0 – v0]e-2x + [2y0 + v0 – 29/377]e-x – 23cos(5x)/377 + 15sin(5x)/377 • y(0) = [52/377 – y0 – v0]e-0 + [2y0 + v0 – 29/377]e-0 + 23cos(0)/377 + 15sin(0)/377 = (52 – 29 – 23)/377 – y0 – v0 + 2y0 + v0 = y0 Correct y(0)! • y’ = -2[52/377 – y0 – v0]e-2x – [2y0 + v0 – 29/377]e-x + 115cos(5x)/377 + 75sin(5x)/377 41 42 Page 7 Nonhomogeneous, Linear, Second-order, Differential Equations Check Initial Conditions II • y’ = -2[52/377 – y0 – v0]e-2x – [2y0 + v0 – 29/377]e-x + 115sin(5x)/377 + 75cos(5x)/377 • y’(0) = -2[52/377 – y0 – v0]e-0 – [2y0 + v0 – 29/377]e-0 + 115sin(0)/377 + 75cos(0)/377 • y’(0) = -2[52/377 – y0 – v0] – [2y0 + v0 – 29/377] + 75/377 = [-104 + 29 +75]/377 + 2y0 + 2v0 -2 y0 –v0 = v0 Correct y’(0)! 43 Page 8