# Practice for test2 Fa2011 by ajizai

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```									                                                    Practice for test2 Fa2011
Find a steady state solution via the Method of Undetermined Coefficients for the equation y’’+3y’+y=cos3t                              d2y     dy
1.                                                                                                                                        1.      2
 3  y  cos 3t
a)      by using a guess Y=Acos3t+Bsin3t and                                                                                                   dt      dt
b)      by using the guess Y=Acos(3t+fi)

2. a)Find the general solution of the given differential equation, and b) compute the amplitude and the phase angle of the st eady-state
Solution y’’+4y’+3y=5sin2t

3. x’(t)= 3x(t) - 4y(t) + 1
y’(t)= 4x(t) - 7y(t) + 10t             Solve by systematic elimination.
4.Find the form of particular solution for each of the following equations. Do not solve.
(a) y ''  3t 3  t                       (b) y '' y ' 2 y  3te t  2 cos 4t
(c) y '' 2 y ' 5 y  t 2et sin 2t
5. Find a particular solution of y’’+4y=3 csc t .
Use variation of parameters

6. Solve Cauchy-Euler equation and use variation of parameters to f ind y particular. Review your hw f rom 4.7

7 Do problems f rom sect.5.1#1,3,5,7, 21,27,29,30,31,32,33. Also determine a steady -state solution, transient solution and amplitude and a phase shif t in each case.

8. Eight second-order equations and f our у(t)-graphs are given below. For each y(t)-graph, determine the second-order equation f or which y(t) is a solution,
and state brief ly how you know your choice is correct. You should do this exercise without using technology.
4.Solutions for (a),(b),(c)
(a) y ''  3t  t
3

Homo : y ''  0  y  c1  c2t and g (t )  3t 3  t is a third deg ree polynomial  Y (t )  t s [ A3t 3  A2t 2  A1t  A0 ]
We must take s  2 to ensure that none of the functions in the assumed form for Y (t ) appear in the fundamental set.
 Y (t )  A3t 5  A2t 4  A1t 3  A0t 2
(b) y '' y ' 2 y  3te t  2 cos 4t
Homo : y  c1e t  c2e 2t
Non hom o : Identify g1 (t )  3te t and g 2 (t )  2 cos 4t
Since g1 is the exp onential function e t multiplied by a first deg ree polynomial , we assume
Y1 (t )  t s [( A1t  A0 )e t ]. Since e t is solution of hom o, we must take s  1  Y1 (t )  ( A1t 2  A0t )e t
For Y2 (t )  t s [ B0 cos 4t  C0 sin 4t ]. Neither of solutions cos 4t or sin 4t are solutions of hom o, so s  0
Y2 (t )  B0 cos 4t  C0 sin 4t.
Substitute Y1 (t )  ( A1t 2  A0t )e t int o y '' y ' 2 y  3te t and Y2 (t )  B0 cos 4t  C0 sin 4t int o y '' y ' 2 y  2 cos 4t
The general solution for (b) is y  c1e t  c2e 2t  Y1 (t )  Y2 (t )
The method of undetermined coefficients
Is self-correcting in the sense that if one assumes
(c) y '' 2 y ' 5 y  t 2e t sin 2t                                                          too little for Y(t), then a contradiction is soon reached
t
Homo : y '' 2 y ' 5 y  0  y  c1e cos 2t  c2e sin 2t   t                                 that usually points the way to the modification
that is needed in the assumed form. On the other
g (t )  t 2e t sin 2t  (sec ond deg ree polynomial ) ( e t sin 2t )                        hand, if one assumes too many terms, then some
Since it is necessary to include both sin e and cos ine functions even if only one or anotherunnecessary work is done and some coefficients
is present , Y (t )  t s [( A2t 2  A1t  A0 )e t cos 2t  ( B2t 2  B1t  B0 )e t sin 2t ] turn out to be zero, but at least the correct answer
Is obtained.
s  1 (to insure that none of the terms in Y (t ) are solutions of hom ogeneous equation.
 Y (t )  ( A2t 3  A1t 2  A0t )e t cos 2t  ( B2t 3  B1t 2  B0t )e t sin 2t.
8      9
1. a )     A         B
145    145
1
b) A             fi  arctan(9 / 8)
145

8        1
2. (a) y  c1e 3t  c2e t  cos 2t  sin 2t
•     1. A                13       13                    y’’+4y’+3y=5sin2t
65
b) A           0.62   arctan(1/ 8)  0.1243549945
13
arctan(1/ 8)    3.265947648 in radians and  Ans *180 /  187     #3
1
3.General      solution      is    x(t )  C1e 5t  2C2et  8t  5
2
y(t )  c1e5t  c2et  6t  2
y (t )  C1e 5t  C2et  6t  2

5. y=u1cos2t+u2sin2t
#5
3csc t sin 2t
u '1                 3cos t  u1  3sin t  c1
2
3cos t cos 2t 3(1  2sin 2 t ) 3                     3
u '2                                csc t  3sin t  u2  ln | csc t  cot t | 3cos t  c2
sin 2t          2sin t      2                     2

3
y  3sin t cos 2t  ln | csc t  cot t | sin 2t  3cos t sin 2t  c1 cos 2t  c2 sin 2t 
2
3
 3sin t  ln | csc t  cot t | sin 2t  c1 cos 2t  c2 sin 2t
2
3
y p  3sin t  ln | csc t  cot t | sin 2t
2

#         8        a=vii,b=ii,c=v,d=viii

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