Math 201 Lecture 09: Undetermined Coefficient Cont.

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1 +B Math 201 Lecture 09: Undetermined Coefficient Cont. Jan. 27, 2012 Many examples here are taken from the textbook. The first number in () refers to the problem number in the UA Custom edition, the second number in () refers to the problem number in the 8th edition. To solve 0. Review ay +by +cy=f(t), (1) 1. Solve the homogeneous problem ay +by +cy=0 (2) to get y 1,y 2. Keep record of the roots r 1,2 of the characteristic equation. 2. Guess what y p should look like according to the following rules: If f(t)=ct m e rt, then If f(t)=ct m e αt cosβt or Ct m e αt sinβt, then 1 t 0 )e αt cosβt+t s (B m t m + 1 t 0 )e rt (3) 0 )e αt sinβt. (4) with s=the number of times the pair (α±βi) appears in the list of roots r 1,r The general solution is then given by 4. If initial conditions are given, use them to determine C 1,C 2. Quiz. Solve Solution. First solve the homogeneous equation Next observe 2e t =2t 0 e 1 t, guess y=c 1 y 1 +C 2 y 2 +y p. (5) y y=2e t. (6) y y=0 r 2 1=0 r 1,2 =1, 1;y 1,y 2 =e t,e t. (7) y p =t s Ae t. (8) To determine s, check how many times 1 appears in r 1,r 2 : We have s=1. Therefore our guess is Now substitute this y p into the equation to find A: Therefore So y p =te t. Putting everything together, we have y=ate t. (9) y p =Ate t +2Ae t y p y p =2Ae t. (10) 2Ae t =2e t A=1. (11) y=c 1 e t +C 2 e t +te t. (12) 1

2 +B 2 Math 201 Lecture 09: Undetermined Coefficient Cont. The Equation with more general f(t): 1. Basic Information ay +by +cy=f(t) (13) f(t)=p m (t)e rt or f(t)=p m (t)e αt cosβt+q n (t)e αt sinβt (14) where P m (t) stands for a polynomial of degree m, and Q n stands for a polynomial with degree n. How to get general solution Recall: General solution to reads ay +by +cy=f(t) (15) y=c 1 y 1 +C 2 y 2 +y p (16) where C 1 y 1 +C 2 y 2 is general solution to the homogeneous equation and y p is any one solution to ay +by +cy=0. (17) ay +by +cy=f(t). (18) Compare with the situation in the last lecture, we see that y 1,y 2 stays the same, all we need to do is to figure out y p in this more general case. Key: Linearity. (See Notes & Comments for more explanation). Procedure: If f(t)=cp m (t)e rt, then If f(t)=p m (t)e αt cosβt+q n (t)e αt sinβt, then y p =t s (A m t k + 1 t 0 )e αt cosβt+t s (B m t k + 1 t 0 )e rt (19) 0 )e αt sinβt. (20) with s=the number of times the pair (α± βi) appears in the list of roots r 1,r 2, and k=max{m,n} that is the bigger number of m,n. Example: (4.5 18; ) Solve Solution. First solve the homogeneous equation: y 2y 3y=3t 2 5. (21) y 2y 3y=0 r 2 2r 3=0 r 1,2 =3, 1 (22) so the general solution to the homogeneous equation is y=c 1 e 3t +C 2 e t. (23) Now notice the right hand side is of the form P m (t)e rt with m=2, r=0. So guess y p =t s (A 2 t 2 1 t 0 )e 0t. (24) To determine s, check how many times r=0 appears in the list of roots: r 1,2 =3, 1. 0 times. So s=0. Substitute y p =A 2 t 2 1 t 0 into the equation we reach 2A 2 2(2A 2 t 1 ) 3(A 2 t 2 1 t 0 )=3t 2 5 (25)

3 +B Jan. 27, which simplifies to 3A 2 t 2 +( 4A 2 3A 1 )t+(2a 2 2A 1 3A 0 )=3t 2 5 (26) 3A 2 = 3; (27) 4A 2 3A 1 = 0 (28) 2A 2 2A 1 3A 0 = 5. (29) Solving it we have So Check solution: A 2 = 1;A 1 =4/3;A 0 =1/9. (30) y p = t t (31) e 3t,e t solves the homogeneous problem; y p solves the original problem. So the final answer is y=c 1 e 3t +C 2 e t t t (32) How to solve initial value problem (IVP) Nothing new here. Solve y 2y 3y=3t 2 5, y(0)=0,y (0)=0. (33) Solution. We have already found the general solution To apply initial conditions, prepare: y=c 1 e 3t +C 2 e t t t (34) Now y =3C 1 e 3t C 2 e t 2t (35) y(0)=0 C 1 +C =0 (36) 9 y (0)=0 3C 1 C =0. (37) 3 Solving it (Add two equation to get C 1, then get C 2 using either one equation) we get So the solution to the initial value problem is C 1 = ; C 2= 1 4. (38) How to check solutions y= e3t e t t t (39) Nothing new here. 2. Things to be Careful/Tricky Issues See Common Mistakes for examples. Keep in mind that in the 2nd case, y p =t s (A m t k + 1 t 0 )e αt cosβt+t s (B m t k + 0 )e αt sinβt. (40)

4 4 Math 201 Lecture 09: Undetermined Coefficient Cont. It s t k in both terms. When solving initial value problems, initial conditions should only be used when the general solution is found, that is when y 1, y 2, y p are all figured out. Using initial condition on C 1 y 1 + C 2 y 2 before finding y p is not correct, as adding y p will change the values of y(0),y (0). When the calculation is complicated, it is easy to forget the t s term when writing the final answer. 3. More Examples Sometimes the right hand side is not of the form P m (t)e rt or P m (t)e αt cosβt+q n (t)e αt sinβt, but a sum of terms of these forms. In this case we have to break y p into y p1, y p2, and find each one separately. Example 1. (4.5 28; ) Solve y +y 12y=e t +e 2t 1; y(0)=1,y (0)=3. (41) Solution. First solve the homogeneous equation: y +y 12y=0 r 2 +r 12 r 1,2 = 4,3 C 1 e 4t +C 2 e 3t. (42) Now to find y p, we have to break it up to y p =y p1 +y p2 +y p3 with the latter solving and It is easy to find So The general solution is y p1 +y p1 12y p1 =e t (43) y p2 +y p2 12y p2 =e 2t (44) y p3 +y p3 12y p3 = 1. (45) y p1 = 1 10 et ; y p2 = 1 6 e2t ; y p3 = (46) y p = 1 10 et 1 6 e2t (47) y=c 1 e 4t +C 2 e 3t 1 10 et 1 6 e2t (48) To apply initial conditions, prepare: y = 4C 1 e 4t +3C 2 e 3t 1 10 et 1 3 e2t. (49) Now use initial conditions: y(0)=1 C 1 +C =1; (50) 12 The solutions are The final answer is then y (0)=3 4C 1 +3C =3. (51) 3 C 1 = 1 60,C 2= 7 6. (52) y= 1 60 e 4t e3t 1 10 et 1 6 e2t (53) The method also applies to higher order equations. Example 2. Find a particular solution for y (4) +5y +4y= 10 cost 20 sint. (54)

5 This simplifies to 1 6Asint+6B cost=10 cost 20 sint. (63) Jan. 27, Solution. First solve the homogeneous equation (or more precisely, find the roots to the characteristic equation): y (4) +5y +4y=0 r 4 +5r 2 +4=0 (r 2 +1)(r 2 +4)=0 r 1,2,3,4 =±i,±2i. (55) Note that it s a good habit to organize complex roots in pairs. Now the right hand side is of the form f(t)=p m (t)e αt cosβt+ Q n (t)e αt sinβt with m=n=0, α=0,β=1. So guess y p =t s (A cost+b sint). (56) To determine s, take α ±iβ =±i, and check how many times this pair appears in the list of roots r 1,2,3,4 =±i,±2i. Once. So s=1 and Now substitute this guess into the equation: Compute So the equation becomes y p =t(a cost+b sint). (57) y p =Acost+B sint Atsint+Btcost; (58) y p = 2Asint+2B cost Atcost Btsint; (59) y p = 3Acost 3B sinttsint Btcost; (60) y p (4) =4Asint 4B costtcost+btsint. (61) [4 A sin t 4 B cos t + A t cos t + B t sin t] + 5 [ 2 A sin t + 2 B cos t A t cos t B t sin t] + 4t(A cost+b sint)=10 cost 20 sint. (62) So We have A= 10 3, B= 5 3. (64) ( 10 y p =t 3 cost+ 5 ) 3 sint. (65) If time allows, substitute it back into the equation to check. How are the formulas derived. Start from If f(t)=ct m e rt, then 4. Notes and Comments we will try to derive If f(t)=cp m (t)e rt, then 1 t 0 )e rt (66) 1 t 0 )e rt (67) 1. If at this stage you find that the t cost and t sint terms do not totally cancel out, go back and check your calculation. When s 0, it is very easy to mess up.

6 +y +y 6 Math 201 Lecture 09: Undetermined Coefficient Cont. using linearity. The other case can be dealt with similarly so left as exercise. 2 Note that the value of s is determined by the roots of the characteristic equation together with the value of r. So should stay the same for this more general case. Let s first look at an example. Say f(t)=(t+1)e rt. We need to find y p solving By linearity we know that if we find y p1 solving and y p2 solving then y p is simply y p1 +y p2. For y p1, y p2 we know that it looks like 3 ay +by +cy=(t+1)e rt =te rt +e rt. (68) ay +by +cy=te rt (69) ay +by +cy=e rt (70) y p1 =t s (A 1 t 0 )e rt, y p2 =t s (B 0 )e rt (71) We can simply substitute them into the two equations and find out A 1, A 0, B 0. However there is a more efficient way: Observe y p =y p1 +y p2 =t s (A 1 t+(a 0 +B 0 ))e rt. (72) We see that it is not necessary to figure out A 0 and B 0, all we need to know is A 0 +B 0. Therefore it is more efficient to straightforwardly assume y p =t s (A 1 t 0 )e rt, (73) substitute into the original equation, and figure out A 1,A 0. Now back to the general case: Let P m (t)=a m t m +a m 1 t m 1 + where y pi solves So the correct forms for y pi are Sum them up, we see that y p =y pm +y pm 1 + +a 1 t+a 0. We know that p1 +y p0 (74) ay +by +cy=a i t i e rt. (75) y pi =t s (A ii t i ii 1 t i 1 + i1 t i0 )e rt. (76) where y p =y pm +y pm 1 + p1 +y p0 =t s (A m t m + A i =A mi m 1,i + 1 t 0 )e rt (77) ii. (78) The logic of undetermined coefficients. Warning. Undetermined coefficients is the result of trial and error and cannot be completely derived deductively. So the purpose of what s below is just to understand why the formulas look the way they are. Also we ignore the issue of s in what follows to make things less involved. The basic idea of undetermined coefficients is to make a guess of the form of y p based on the equation, and the right hand side f(t). Let s see what is required for a good guess. First, it s a good idea to include f(t) in the guess. Second, if y involves f(t), then y, y involves f and f. If there is any new type of functions involved in these derivatives, but is not in y, then we may be in trouble as there is no way to balance (See below for an example). So we better include terms arising from f,f in the original guess of y. 2. Some thinking is needed to understand why it s t k in both terms! 3. Make sure you understand why the number s stays the same for y p1 and y p2!

7 Jan. 27, Finally, once such terms are included, y,y will involve terms from f,f (4) so we should include them too in y. Repeating this, we see that a good guess of y must include all terms from all possible differentiations of f(t). Example 3. Let s look at f(t)=t 2 e rt. Taking derivatives we see that f gives a new type of terms: te rt, f gives another new typee rt. But further differentiation does not pop up anything new anymore. So a good guess is Example 4. Let s look at f(t)=tcost+sint. y=(a 2 t 2 1 t 0 )e rt. (79) f cost,tsint; (80) Further differentiation does not give anything new. So the good guess should be Example 5. Let s look at f(t)=t 1. y p =(A 1 t 0 ) cost+(b 1 t+b 0 ) sint. (81) f t 2 ; f t 3 ; f t 4, (82) It clearly never ends: Every time we take one more derivative, a new type of function pops up. So if we want to include them all, we have to set y p =A 1 t 1 2 t 2 + n t n +. (83) This is in fact a legitimate approach and may work in some cases. But it s much more involved in both theory and calculation and is not undetermined coefficients anymore.