Document 6539413

Sample Solutions of Assignment 7 for MAT3270A
a
Note: Any problems to the sample solutions, email Mr. Yao Xiao (yxiaomath.cuhk.edu.hk)
directly.
1. Compute the Wronskian of the following functions
a). et , tet , t2 et
b). 1, t, cos t, sin t
c). 1, t, e−t , te−t
Answer: a). W = 2e3t .
b). W = 1.
c). W = −5e−2t .
2. Use Abel’s formula to find the Wronskian of the following ODEs:
000
00
0
2
a). ty + 2y − t3 y + et y = 0
00
b). t2 y (4) + ty − 4y = 0
Answer: a). By Abel’s formula, W 0 = − 2t W , hence W =
C
.
t2
b). By Abel’s formula, W 0 = 0, hence W = C.
3. Show that if y1 is a solution of
000
00
0
y + p1 y + p2 y + p3 y = 0
0
then the substitution y = y1 (t)v(t) leads to the following second order equation for v :
0
00
0
y1
y1
y1
000
00
0
v + 3 + p1 v + 3 + 2p1 + p2 v = 0
y1
y1
y1
Answer: By direct computation, we have
y 0 = y10 v + v 0 y1
y 00 = y100 v + 2y10 v 0 + v 00 y1
(3)
y (3) = y1 v + 3y100 v 0 + 3y10 v 00 + v (3) y1
1
2
Then
000
00
0
y + p1 y + p2 y + p3 y
(3)
= y1 v + 3y100 v 0 + 3y10 v 00 + v (3) y1 + p1 (y100 v + 2y10 v 0 + v 00 y1 ) + p2 (y10 v + v 0 y1 ) + p3 y1 v
0
00
000
0
00
0
=
v y1 + 3y1 + p1 y1 v + 3y1 + 2p1 y1 + p2 y1 v
Hence
00
0
0
y1
y1
y1
00
0
v + 3 + p1 v + 3 + 2p1 + p2 v = 0
y1
y1
y1
000
4. Use Problem 3 to solve the following ODE:
000
00
0
(2 − t)y + (2t − 3)y − ty + y = 0,
t < 2;
y1 (t) = et
Answer: . The original ODE can be written as
2t − 3 00
t 0
1
000
y +
y −
y +
y=0
2−t
2−t
2−t
Let y = et v, according, the equation of v is
t − 3 00
v 000 +
v =0
t−2
Then v = C1 te−t + C2 t + C3 . Hence
y = et v = C1 t + C2 tet + C3 et
5. Solve the following equations
a). y (7) − 8y (5) + 16y (3) = 0(There are some typo in the previous paper we handed out,please
change it accordingly)
00
0
b). y (5) + y (4) + 2y (3) + 2y + y + y = 0
c). y (4) − y = 0,
y(0) = 1,
0
y (0) = 0,
00
y (0) = −1,
000
y (0) = 0
d). y (4) − y = 3t + cos t
000
00
0
e). y − 3y + 2y = t + et ,
000
0
f). y + y = tan t,
0<t<
y(0) = 1,
π
2
0
y (0) = − 41 ,
00
y (0) = − 32
3
000
00
0
g). t3 y − 3t2 y + 6ty − 6y = t4
(Hint: t, t2 and t3 are solutions of the homogeneous equation.)
Answer: . a). The characteristic equation is
r7 − 8r6 + 16r3 = r3 (r2 − 4)2 = 0.
We get r1 = 0, r2 = 0, r3 = 0, r4 = 2, r5 = 2, r6 = −2, r7 = −2.
The general solutions are
y(t) = c1 + c2 t + c3 t2 + c4 e2t + c5 te2t + c6 e−2t + c7 te−2t .
b). The characteristic equation is
r5 + r4 + 2r3 + 2r2 + r + 1 = 0
Thus the possible values of r are r1,2 = i, r3,4 = −i, r5 = −1, and the general solution of the
equation is
y(t) = (c1 + c2 t) cos t + (c3 + c4 t) sin t + c5 e−t
c).The characteristic equation is
r4 − 1 = 0
Thus the possible values of r are r1 = 1, r2 = −1, r3 = i, r4 = −i, and the general solution of the
equation is
y(t) = c1 et + c2 e−t + c3 cos t + c4 sin t.
From the initial data, we have
c1 + c2 + c3 = 1
c1 − c2 + c4 = 0
c1 + c2 − c3 = −1
c1 − c2 − c4 = 0
hence c1 = 0, c2 = 0, c3 = 1, c4 = 0. Then
y = cos t
4
d). It is easy to know that the general solution of the homogeneous equation is
Y (t) = c1 et + c2 e−t + c3 cos t + c4 sin t.
Since f = 3t + cos t, we can assume a particular solution is A0 t + A1 + t(C1 cos t + C2 sin t) then
−A0 t + A1 − 4C2 cos t − 4C1 sin t = 3t + cos t
hence A0 = −3, A1 = 0, C1 = 0, C2 = − 41 . Then the general solution of the ODE is
1
c1 et + c2 e−t + c3 cos t + c4 sin t − 3t − t sin t.
4
e). It is easy to know that the general solution of the homogeneous equation is
Y (t) = c1 + c2 e2t + c3 e3t .
Since f = t + et , we can assume a particular solution is A0 t + A1 + C1 tet then
2A0 + C1 et = t + et
hence A0 = 21 , A1 = 0, C1 = 1. Then the general solution of the ODE is
1
y(t) = c1 + c2 e2t + c3 e3t + t + tet .
2
From the initial data, we have
1
=1
2
1
2c2 + 3c3 + 1 = −
4
1
4c2 + 9c3 + 2 = −
2
c1 + c2 + c3 +
solve this equation, we get c1 = 89 , c2 = − 58 , c3 = 0. Then the solution is
9
8
+ 85 e2t + 12 t + tet .
f). It is easy to know that 1, cos t, sin t are the solutions of the homogeneous equation. By
direct computation,
1
W = 0
0
1
W2 = 0
0
0
cos t
sin t − sin t cos t = 1, W1 = 0
1
− cos t − sin t 1
0 sin t 0 cos t = − cos t, W3 = 0
0
1 − sin t cos t
sin t − sin t cos t = 1,
− cos t − sin t cos t 0 − sin t 0 = − sin t,
− cos t 1 5
Then the general solution of the ODE is
Z
Z
Z
tan t
tan t cos t
tan t sin t
y(t) = c1 + c2 cos t + c3 sin t +
+ cos t −
+ sin t −
1
1
1
2
2
= c1 + c2 cos t + c3 sin t − ln cos t + cos t − x sin t tan t.
g). It is known that t, t2 , t3 are the solutions of the homogeneous equation. Since f = t4 , we
can assume the a particular solution is ct4 and substitute it into the ODE, we get
−42ct4 = t4 , c = −
1
.
42
Hence the general solution of the ODE is
y = c1 t + c2 t2 + c3 t3 −
1 4
t.
42
6. Use the method of undetermined coefficients to find the solutions of the following inhomogeneous ODEs.
00
a). y (4) − 4y = t2 + 3et + cos t
00
b). y (4) + 2y + y = 3 + cos 2t + t sin 3t + e5t
Answer: . a). It is easy to know that the general solution of the homogeneous equation is
c1 + c2 t + c3 e2t + c4 e−2t .
Since g = t2 +3et +cos t, we can assume the particular solution is c1 t4 +c2 et +c3 cos t+c4 sin t.and
substitute it into the ODE, we get
−48c1 t2 − 3c2 et + 5c3 cos t + 5c4 sin t = t2 + 3et + cos t.
hence
1
1
, c2 = −1, c3 = , c4 = 0.
48
5
The general solution of the original ODE is
c1 = −
y(t) = c1 + c2 t + c3 e2t + c4 e−2t − t4 − et +
1
cos t.
5
b). It is easy to know that the general solution of the homogeneous equation is
(c1 + c2 t) cos t + (c3 + c4 t) sin t.
6
Since g = 3 + cos 2t + t sin 3t + e5t , we can assume the particular solution is
c1 + c2 cos 2t + c3 sin 2t + (c4 t + c5 ) cos 3t + (c6 t + c7 ) sin 2t + c8 e5t .
Substitute it into the ODE, we get
c1 + 9c2 cos 2t + 9c3 sin 2t + 64c4 t cos 3t + 64c6 sin 3t
+ (64c5 − 96c6 ) cos 3t + (64c7 + 96c1 ) sin 3t + 576c8 e5t
3 + cos 2t + t sin 3t + e5t .
=
Hence
1
3
1
1
c1 = 3, c2 = , c3 = 0, c4 = 0, c5 =
, c6 = , c7 = 0, c8 =
.
9
128
64
576
The general solution of the original ODE is
y(t) = (c1 + c2 t) cos t + (c3 + c4 t) sin t + 3 +
1
3
1
1 5t
cos 2t +
cos 3t + t sin 3t +
e .
9
128
64
576
7. Find the solution of the following IVB:
000
00
0
y − 3y + 2y = t + et ,
y(0) = 1,
0
y (0) = −2,
00
y (0) = −3
Answer: . It is easy to know that the general solution of the homogeneous is
c1 + c2 e2t + c3 e3t .
Since g = t + et , we can assume a particular solution is c1 t2 + c2 t + c3 et . Substitute it into the
ODE, we get
−3(2c1 + c3 ) + 2(2c1 t + c2 + c3 ) = t + et
Hence
3
1
c1 = , c2 = , c3 = −1.
4
4
Then the general solution of the original ODE is
1
3
c1 + c2 e2t + c3 e3t + t2 + t − et .
4
4
7
From the initial data, we have
c1 + c2 + c3 = 2
11
2c2 + 3c3 = −
4
7
4c2 + 9c3 = −
2
hence
c1 =
19
2
89
, c2 = − , c3 = .
24
8
3
The solution of IVB is
y(t) =
89 19 2t 2 3t 1 2 3
− e + e + t + t − et
24
8
3
4
4
8. Use the method of variation of parameters to find general solutions of the following ODEs:
000
0
000
0
000
0
a). y + y = tan t,
b). y + y = sec t,
c). y − y = csc t,
0<t<π
−π/2 < t < π/2
y(π/2) = 2,
0
y (π/2) = 1,
00
y (π/2) = −1
Answer: . a). It is easy to know that 1, cos t, sin t are the solutions of the homogeneous equation.
By direct computation,
1 cos t
sin t
W = 0 − sin t cos t
0 − cos t − sin t
1 0 sin t
W2 = 0 0 cos t
0 1 − sin t
0 cos t
sin t
= 1, W1 = 0 − sin t cos t
1 − cos t − sin t
= 1,
1 cos t 0 = − cos t, W3 = 0 − sin t 0 = − sin t,
0 − cos t 1 Then the general solution of the ODE is
Z
Z
Z
sec t cos t
sec t sin t
sec t
+ cos t −
+ sin t −
y(t) = c1 + c2 cos t + c3 sin t +
1
1
1
1 1 + sin t
= c1 + c2 cos t + c3 sin t + ln
− t cos t − sin t ln cos t.
2 1 − sin t
8
b). It is easy to know that 1, cos t, sin t are the solutions of the homogeneous equation. By
direct computation,
1 cos t
0 cos t
sin
t
sin
t
W = 0 − sin t cos t = 1, W1 = 0 − sin t cos t = 1,
0 − cos t − sin t 1 − cos t − sin t 1 0 sin t 1 cos t 0 W2 = 0 0 cos t = − cos t, W3 = 0 − sin t 0 = − sin t,
0 1 − sin t 0 − cos t 1 Then the general solution of the ODE is
Z
Z
Z
tan t
tan t cos t
tan t sin t
y(t) = c1 + c2 cos t + c3 sin t +
+ cos t −
+ sin t −
1
1
1
2
2
= c1 + c2 cos t + c3 sin t − ln cos t + cos t − x sin t tan t.
c). It is easy to know that 1, et , e−t are the solutions of the homogeneous equation. By direct
computation,
0 et e−t 1 et e−t W = 0 et −e−t = 2, W1 = 0 et −e−t = −2,
1 et e−t 0 et e−t 1 et 0 1 0 e−t W2 = 0 0 −e−t = e−t , W3 = 0 et 0 = et ,
0 et 1 0 1 e−t Then the general solution of the ODE is
Z
Z −t
Z t
2 csc t
e csc t
e csc t
t
−t
t
−t
y(t) = c1 + c2 e + c3 e + −
+e
+e
2
2
2
Z −t
Z t
sin t
e csc t
e csc t
= c1 + c2 et + c3 e−t − ln
+ et
+ e−t
.
1 + cos t
2
2
From the initial data, we get
π
π
π
π
c1 + c2 e 2 + c3 e − 2 = 2
c2 e 2 − c3 e − 2 = 1
π
π
c2 e 2 + c3 e− 2 = −1
π
Solve this equation, we have c1 = 3, c2 = 0, c3 = −e 2 . Then the solution of the original IVB is
Z t −s
Zt s
π
sin t
e csc s
e csc s
−t
t
−t
y(t) = 3 − e 2 − ln
+e
ds + e
ds.
1 + cos t
2
2
π
2
π
2
9
Section 4.1
Determine the intervals in which solutions are sure to exists.
2. ty 000 + sin ty 00 + 3y = cos t
Answer: y 000 +
Since
sin t 00
y
t
cos t sin t 3
, t ,t
t
+ 3t y =
cos t
.
t
are not continuous at 0. Hence the interval is t > 0, or t < 0.
4.y 00 + ty 00 + t2 y 0 + t3 y = 2 ln t
Answer: Since t,t2 ,t3 ,2 ln t are continuous on (0, +∞),hence the interval is (0, +∞).
6.(x2 − 4)y (6) + x3 y 000 + 9y = 0
Answer: y (6) +
Since
x3
, 9
x2 −4 x2 −4
x3
y 000
x2 −4
+
9
y
x2 −4
= 0.
are not continuous at ±2.Hence the interval is (−∞, −2) and (2, +∞).
In each of Problem determine whether the given set of functions is linearly dependent or linearly
independent. If they are linearly dependent, find a linear relation among them.
8. f1 (t) = 2t − 3, f2 (t) = 4t2 + 2, f3 (t) = 3t2 + t
Answer: 8. By direct computation,
2t − 3 4t2 + 2 3t2 + t
8t
6t + 1
W (f1 , f2 , f3 ) = 2
0
8
6
=0
So f1 , f2 , f3 are linear dependent, f1 (t) + 3f2 (t) − 2f3 (t) = 0.
In each problems 11 throuth 16,verify that the given functions are solutions of the differential
equation,and determine their Wronskian.
10
11.y 000 + 4y 0 = 0;1,cos 2t,sin 2t
Answer:
1
cos 2t
sin 2t
0
−2
sin
2t
2
cos 2t
W (1, cos 2t, sin 2t) = 0 −4 cos 2t −4 sin 2t
=8
13.y 000 − 3y 00 − y 0 + 3y = 0;et ,e−t ,e3t
Answer:
et e−t
e3t
W (et , e−t , e3t ) = et −e−t 3e3t
et e−t 9e3t
= −16e3t
15.xy 000 − y 00 = 0;1,x,x3
Answer:
1 x x3 W (1, x, x3 ) = 0 1 3x2 = 6x
0 0 6x 17. Show that W (5, cos2 t, cos 2t) = 0 for all t. Can you establish this result without direct
evaluation of Wronskian?
Answer: By direct computation, we have
5
cos2 t
cos 2t
2
W (5, cos t, cos 2t) = 0 −2 sin t cos t −2 sin 2t
0
−2 cos 2t
−4 cos 2t
It is easy to know that
cos2 t =
1
1
× 5 + cos 2t
10
2
i.e. 5, cos2 t, cos 2t are linearly dependent, so W (5, cos2 t, cos 2t) = 0.
18. Verify the differential operator defined by
L[y] = y (n) + p1 (t)y (n−1) + ... + pn (t)y
is a linear differential operator. That is show that
L[c1 y1 + c2 y2 ] = c1 L[y1 ] + c2 L[y2 ].
=0
11
where y1 and y2 are n times differentiable functions and c1 and c2 are arbitrary constants. Hence
show that if y1 , y2 , ..., yn are solutions of L[y] = 0, then the linear combination c1 y1 + ... + cn yn
is also a solution of L[y] = 0.
Answer: Direct computation easily gives the answer.
19. Let the linear differential operator L be defined by
L[y] = a0 y (n) + a1 y (n−1) + · · · + an y,
where a0 , a1 , · · · , an are real constants.
(a). Find L[tn ].
(b). Find L[ert ].
000
(c). Determine four solutions of the equation y (4) − 5y + 4y = 0. Do you think the four solutions
form a fundamental set of solutions? why?
Answer: (a) Since (tn )0 = ntn−1 , (tn )00 = n(n − 1)tn−2 ,· · · ,(tn )(n) = n!, so
L[tn ] = a0 n · (n − 1) · · · 2 · 1 + a1 n · (n − 1) · · · 2t + · · · + an tn .
(b) Since (ert )0 = rert , (ert )(n) = rn ert , so
L[ert ] = a0 rn ert + a1 rn−1 ert + · · · + an ert = (a0 rn + · · · + an )ert .
(c) The characteristic equation is r4 − 5r2 + 4 = 0, we have r1 = 1, r2 = −1, r3 = 2, r4 = −2.
Since W [et , e−t , e2t , e−2t ] 6= 0, −∞ < t < ∞, so the four solutions forms a fundamental set of
solutions.
20.In this problem we show how to generalize Theorem3.3.7(Abel’s theorem) to higher order
equations. We first outline the procedure for the third order equation
000
00
0
y + p1 (t)y + p2 (t)y + p3 (t)y = 0
Let y1 , y2 , and y3 be solutions of this equation on an interval I.
(a) If W = W (y1 , y2 , y3 ), show that
y1 y2 y3
0
0
0
0
W = y1 y2 y3
y 000 y 000 y 000
1
2
3
.
12
Hint: The derivative o fa 3-by-3 determinant is the sum of three 3-by-3 determinants obtained
by differentiating the first, second,and third rows, respectively.
000
000
000
(b) Substitute for y1 , y2 , andy3 from the differential equation; multiply the first row by p3 ,
multiply the second row by p2 , and add these to the last row to obtain
0
W = −p1 (t)W.
(c) Show that
Z
W (y1 , y2 , y3 ) = c exp[−
p1 (t)dt].
It follows that W is either always zero or nowhere zero on I. (d) Generalize this argument to
the nth order equation
y (n) + p1 (t)y (n−1) + · · · + pn (t)y = 0
with solutions y1 , . . . , yn . That is, establish Abel’s formula,
Z
W (y1 , . . . , yn )(t) = c exp[− p1 (t)dt],
for this case.
y1 y2 y3
y10 y20 y30
y1 y2 y3
y1 y2 y3
00
00
00
0
0
0
0
0
0
Answer: (a). W = y1 y2 y3 + y1 y2 y3 + y1 y2 y3 = y10 y20 y30
00
00
000
000
000
000
00
000
000
00
00
00
y1 y2 y3
y1 y2 y3
y1 y2 y3
y1 y2 y3
0
(b). According the equation, we have
000
00
0
000
00
0
000
00
0
y1 = −(p1 (t)y1 + p2 (t)y1 + p3 (t)y1 )
y2 = −(p1 (t)y2 + p2 (t)y2 + p3 (t)y2 )
y3 = −(p1 (t)y3 + p2 (t)y3 + p3 (t)y3 )
then,
y1
y2
y3
0
0
0
0
y
y
y3
W =
1
2
00
0
00
0
00
−(p1 (t)y + p2 (t)y + p3 (t)y1 ) −(p1 (t)y + p2 (t)y + p3 (t)y2 ) −(p1 (t)y + p2 (t)y 0 + p3 (t)y3 )
1
1
2
2
3
3
the first row × p3 + the second row × p2 + the last row =⇒
y
y
y
1
2
3
0
0
0
0
= −p1 (t)W
y1
y2
y3
W = −p1 (t)y 00 −p1 (t)y 00 −p1 (t)y 00 1
2
3
.
13
(c).By (b),
dW/dt
W
R
R
= −p1 (t) ⇒ ln W = − p1 (t)dt + C ⇒ W (y1 , y2 , y3 )(t) = c exp [− p1 (t)dt]
(d).For the nth order equation
y1
y2
···
y1
y2
···
yn
0
0
0
.
.
..
..
..
y1
y2
···
yn
0
.
W =
, and W = (n−2) (n−2)
..
..
..
..
.
.
.
.
y1
y2
···
(n−1)
(n−1)
(n−1)
(n)
(n)
y1
y2
· · · yn
y1
y2
···
Substitute for y1n , · · · , ynn from the differential equation,
yn
..
.
(n−2)
yn
(n)
yn
.
(n)
y1 = −[p1 (t)y1n−1 + · · · + pn (t)y1 ]
..
.
(n)
yn = −[p1 (t)ynn−1 + · · · + pn (t)yn ]
like (a), row (1) ×pn (t) + · · · + row (n − 1) × p2 (t)+ row (n), then
y1
y2
···
yn
..
..
..
..
0
.
.
.
.
= −p1 (t)W
W =
(n−2)
(n−2)
(n−2)
y1
y2
···
yn
(n−1)
(n−1)
(n−1)
−p1 (t)y1
−p1 (t)y2 R · · · −p1 (t)yn
so, W (y1 , · · · , yn )(t) = c exp [− p1 (t)dt].
26. Show that if y1 is a solution of
y 000 + p1 (t)y 00 + p2 (t)y 0 + p3 (t)y = 0,
then the substitution y = y1 (t)v(t) leads to the following second order equation for v 0 :
y1 v 000 + (3y10 + p1 y1 )v 00 + (3y100 + 2p1 y10 + p2 y1 )v 0 = 0.
Proof: By direct computation and use the fact that y1000 + p1 (t)y100 + p2 (t)y10 + p3 (t)y1 = 0, it is
easy to obtain the result.
Section 4.2
In each of the problem express the given complex number in the form
R(cos θ + i sin θ) = Reiθ .
14
1. 2 + 2i
3. −4
√
5. − 3 − i
√
Answer: 1.The magnitude of 2 + 2i is 2 2 and the polar angle is π/4. Hence
√
2 + 2i = 2 2ei(π/4+2kπ) .
k∈Z
3.The magnitude of −4 is 4 and the polar angle is π. Hence
−4 = 4ei(π+2kπ)
k∈Z
√
5.The magnitude of − 3 − i is 2 and the polar angle is 67 π. Hence
√
7
− 3 − i = 2e( 6 π+2kπ)i
k∈Z
In each of the problem follow the procedure illustrated in Example 4 to determine the indicated
roots of the given complex number.
8. (1 + i)1/2
10. [2(cos 2π/3 + i sin 2π/3)]1/2
Answer: 8. In polar form 1 + i =
√
2ei(π/4+2mπ) , where m is any integer.
1
Thus (1+i) 2 = 21/4 ei(π/8+mπ) . Setting m=0,1 successively, we obtain the roots 21/4 eiπ/8 , 21/4 ei7π/8 .
10. In polar form 2(cos 2π/3 + i sin 2π/3) = 2ei(2π/3+2mπ) , where m is any integer.
√
Thus [2(cos 2π/3+i sin 2π/3)]1/2 = 2ei(π/3+mπ) . Setting m=0,1 successively, we obtain the roots
√
√
√
√
( 3 + i)/ 2, −( 3 + i)/ 2.
In each of the problem find the general solution of the given differential equation.
11. y 000 − 3y 00 + 4y = 0
13. 2y 000 − y 00 − 2y 0 + y = 0
15. y (6) + y = 0
24. y 000 + 5y 00 + 6y 0 + 2y = 0
15
Answer:
11. The characteristic equation is
r3 − 3r2 + 4 = (r − 2)2 (r + 1) = 0
The roots are r1 = r2 = 2 and r3 = −1. Thus the general solution is
y(t) = (c1 + c2 t)e2t + c3 e−t
13. The characteristic equation is
2r3 − r2 − 2r + 1 = (r − 1)(r + 1)(2r − 1)
The roots are r1 = 1, r2 = −1, and r3 = 21 . Thus the general solution is
1
y(t) = c1 et + c2 e−t + c3 e 2 t
15. The characteristic equation is
r6 + 1 = 0
√
3
2
The roots are r1 =
+ 12 i, r2 =
√
3
2
− 12 i, r3 = i, r4 = −i, r5 = −
√
3
2
+ 12 i, and r6 = −
√
3
2
− 12 i.
Thus the general solution is
√
y(t) = e
3t
2
(c1 cos
√
3t
t
t
t
t
+ c2 sin ) + c3 cos t + c4 sin t + e− 2 (c5 cos + c6 sin )
2
2
2
2
24. The characteristic equation is
r3 + 5r2 + 6r + 2 = (r + 1)(r2 + 4r + 2) = 0
√
√
The roots are r1 = −1, r2 = −2 + 2, and r3 = −2 − 2. Thus the general solution is
√
y(t) = c1 e−t + c2 e(−2+
2)t
√
+ c3 e(−2−
2)t
37. Show that the general solution of y 4 − y = 0can be written as
y = c1 cos t + c2 sin t + c3 cosh t + c4 sinh t.
0
00
000
Determine the solution satisfying the initial conditions y(0) = 0, y (0) = 0, y (0) = 1, y (0) = 1.
Why is it convenient to use the solutions cosh t and sinh t rather than et and e−t ?
16
Answer: The characteristic equation is
r4 − 1 = 0
the roots are r1 = i, r2 = −i, r3 = 1, r4 = −1. Since sinh t + cosh t = et and cosh t − sinh t = e−t ,
the general solution is
y(t) = c1 cos t + c2 sin t + c3 cosh t + c4 sinh t
From the initial data, we have
c1 + c3 = 0
c2 + c4 = 0
−c1 + c3 = 1
−c2 + c4 = 1
hence c1 = c2 = − 21 , c3 = c4 = 12 . Then
1
1
y(t) = (cosh t − cos t) + (sinh t − sin t)
2
2
0
0
0
We know, cosh t = sinh t, sinh t = cosh t, but (e−t ) = −e−t . Hence, using the solutions cosh t
0
00
000
and sinh t, it is convenient to compute the y (t), y (t), y (t) and easy to obtain c1 , c2 , c3 , c4
according the initial conditions.
38. Consider the equation y (4) − y = 0,
(a) use Able’s formula to find the Wronskian of a fundamental set of solutions of the given equation.
(b) Determine the Wronskian of the solutions et , e−t , cos t and sin t.
(c) Determine the Wronskian of the solutions cosh t, sinh t, cos t and sin t.
Answer: (a). Since P1 (t) = 0, =⇒ W = C, where C is a constant.
(b). W (et , e−t , cos t, sin t) = −8.
(c). W (cosh t, sinh t, cos t, sin t) = 4.
17
40. In this problem we outline one way to show that if r1 , ..., rn are all real and different, then
er1 t , ..., ern t are linearly independent on −∞ < t < ∞. To do this, we consider the relation
c1 er1 t + ... + cn ern t = 0, −∞ < t < ∞ (i)
and show all the constants are zero.
(a) Multiply Eq.(i) by er1 t and differentiate with respect to t, thereby obtaining
c2 (r2 − r1 )e(r2 −r1 )t + ... + cn (rn − r1 )e(rn −r1 )t = 0
(b) Multiply the result of part(a) by e−(r2 −r1 )t and differentiate with respect to t to obtain
c3 (r3 − r2 )(r3 − r1 )e(r3 −r2 )t + ... + cn (rn − r2 )(rn − r1 )e(rn −r2 )t = 0
(c) Continue the procedure from parts (a) and (b), eventually obtaining
cn (rn − rn−1 ) · · · (rn − r1 )e(rn −rn−1 )t = 0.
Hence cn = 0, and therefore
c1 er1 t + ... + cn−1 ern−1 t = 0
(d) Repeat the preceding argument to show that cn1 = 0. In a similar way it follows that
cn−2 = · · · = c1 = 0. Thus the functions er1 t , ..., ern t are linearly independent.
Answer: Follow the procedure, carrying out direct computations, we can show the functions
er1 t , ..., ern t are linearly independent.
Section 4.3
Determine the general solution of the given differential equation.
1.y 000 − y 00 − y 0 + y = 4e−t + 3
3.y 000 + y 00 + y 0 + y = 2e−t + 4t
4.y 000 − y 0 = 2 sin t
6.y (4) + 2y 00 + y = 4 + cos 2t
Answer: 1.The characteristic equation of homogeneous equation is
r3 − r2 − r + 1 = 0,
18
the roots are r1 = 1, r2 = 1, r3 = −1. So the general solution of homogeneous equation is
y(t) = c1 et + c2 e−t + c3 tet .
Let the particular solution is Y (t) = Ate−t + B, by computation, we have A = 1, B = 3. The
general solution is
y = c1 et + c2 e−t + c3 tet + te−t + 3.
Answer: 3.The characteristic equation of homogeneous equation is
r3 + r2 + r + 1 = 0,
the roots are r1 = −1, r2 = i, r3 = −i. So the general solution of homogeneous equation is
y(t) = c1 e−t + c2 cos t + c3 sin t.
Let the particular solution is Y (t) = At + B + Ce−t t, by computation, we have A = 4, B =
−4, C = 1. The general solution is
y = c1 e−t + c2 cos t + c3 sin t + 4(t − 1) + te−t .
Answer: 4.The characteristic equation of homogeneous equation is
r3 − r = 0,
the roots are r1 = 0, r2 = 1, r3 = −1. So the general solution of homogeneous equation is
y(t) = c1 + c2 et + c3 e−t .
Let the particular solution is Y (t) = A sin t + B cos t, by computation, we have A = 0, B = 1.
The general solution is
y = c1 + c2 et + c3 e−t + cos t.
Answer: 6.The characteristic equation of homogeneous equation is
r4 + 2r2 + 1 = 0,
19
the roots are r1 = i, r2 = −i, r3 = i, r4 = −i. So the general solution of homogeneous equation is
y(t) = c1 cos t + c2 sin t + c3 t cos t + c4 t sin t.
Let the particular solution is Y (t) = A cos 2t + B sin 2t + C, by computation, we have A = 91 , B =
0, C = 4. The general solution is
y = c1 cos t + c2 sin t + c3 t cos t + c4 t sin t +
1
cos 2t + 4.
9
In each of problem determine a suitable form for Y (t) if the method of undetermined coefficients
is to be used. Do not evaluate the constants.
13. y 000 − 2y 00 + y 0 = 3t3 + 2et
Answer: The characteristic equation of homogeneous equation is
r3 − 2r2 + r = 0
the roots are r1 = 0, r2 = r3 = 1. So the general solution of homogeneous equation is
Y (t) = c1 + c2 et + c3 tet .
Since g = t3 + 2et , we can assume a particular solution is
Y = t(c1 t3 + c2 t2 + c3 t + c4 ) + c5 t2 et
18.
y (4) + 2y 000 + 2y 00 = 3et + 2te−t + e−t sin t
Answer: The characteristic equation of homogeneous equation is
r4 + 2r3 + 2r2 = 0,
the roots are r1 = r2 = 0, r3 = −1 + i, r4 = −1 − i. So the general solution of homogeneous
equation is
y(t) = c1 + c2 t + c3 e−t cos t + c4 e−t sin t.
20
Hence we can assume the particular solution is
Y (t) = Aet + (B0 t + B1 )e−t + te−t (C cos t + D sin t).
Section 4.4
In each of problem use the method of variation of parameters to determine the general solution
of the given differential equation.
2. y 000 − y 0 = 4t
5. y 000 − y 00 + y 0 − y = 2e−t sin t
The characteristic equation of homogeneous equation is
r3 − r = 0
the roots are r1 = −1, r2 = 0, r3 = 1. So the general solution of homogeneous equation is
yc (t) = c1 e−t + c2 + c3 et .
By method of variation of parameters, a particular solution is
Y (t) =
3
X
m=1
−t
Zt
ym (t)
g(s)Wm (s)
ds
W (s)
t
Where y1 = 1, y2 = e , y3 = e . By direct computation,
1 e−t et −t
t e = −2 W1 = W = 0 −e
0 e−t et 1 0 et 1
W2 = 0 0 et = −et W3 = 0
0 1 et 0
0 e−t et 0 −e−t et = 2
1 e−t et e−t 0 −e−t 0 = −e−t
e−t 1 hence
Zt
Zt
1 t
1 −t
s
e sds + e
e−s sds
Y (t) = − sds + e
2
2
1
1
1
= − t2 + (t − 1) − (t + 1) = −2t2 + 2
2
2
2
Zt
21
Then the general solution is
y(t) = c1 + c2 et− + c3 et − 2t2
Answer: 5.The characteristic equation of homogeneous equation is
r3 − r2 + r − 1 = 0
the roots are r1 = 1, r2 = i, r3 = −i. So the general solution of homogeneous equation is
yc (t) = c1 et + c2 cos t + c3 sin t.
By method of variation of parameters, a particular solution is
Zt
3
X
g(s)Wm (s)
ds
Y (t) =
ym (t)
W (s)
m=1
Where y1 = et , y2 = cos t, y3 = sin t. By direct computation,
0 cos t
1 cos t
sin t W = et 1 − sin t cos t = 2et W1 = 0 − sin t
1 − cos t
1 − cos t − sin t 1 cos t
1 0 sin t t
t
t
W2 = e 1 0 cos t = −e (cos t − sin t) W3 = e 1 − sin t
1 − cos t
1 1 − sin t hence
2
Y (t) = − e−t cos t
5
Then the general solution is
2
y(t) = c1 et + c2 cos t + c3 sin t − e−t cos t
5
sin t cos t = 1
− sin t 0 0 = −et (sin t + cos t)
1 8. Find the general solution of the given differential equation. Leave your answer in terms of
one or more integrals.
y 000 − y 0 = csc t.
0 < t < π.
Answer: . The characteristic equation of homogeneous equation is
r3 − r = 0
22
the roots are r1 = 0, r2 = 1, r3 = −1. So the general solution of homogeneous equation is
yc (t) = c1 + c2 et + c3 e−t .
By method of variation of parameters, a particular solution is
Y (t) =
3
X
Zt
ym (t)
m=1
−t
t
Where y1 = 1, y2 = e , y3 = e
1
W = 0
0
1
W2 = 0
0
g(s)Wm (s)
ds
W (s)
. By direct computation,
0
et e−t t
−t e −e = 2 W1 = 0
1
et e−t 0 e−t −t
−t 0 −e = −te
W3 = 1 e−t et e−t
et −e−t
et e−t
= −2
1 e−t 0 0 −e−t 0 = et
0 e−t 1 hence
1
Y (t) = − ln t + ln(cos t + 1) + et
2
Z
t
t0
1
e−s
ds + e−t
sin s
2
Z
t
t0
es
ds.
sin s
Then the general solution is
t
y(t) = c1 + c2 e + c3 e
−t
1
− ln t + ln(cos t + 1) + et
2
Z
t
t0
e−s
1
ds + e−t
sin s
2
Z
t
t0
es
ds.
sin s
13. Given that x, x2 , and 1/x are solutions of the homogeneous equation corresponding to
x3 y 000 + x2 y 00 − 2xy 0 + 2y = 2x4 ,
x>0
determine a particular solution.
Answer: The original ODE can be written as
1
2
2
y 000 + y 00 − 2 y 0 + 3 y = 2x
x
x
x
By method of variation of parameters, a particular solution is
Y (t) =
3
X
m=1
Zt
ym (t)
g(s)Wm (s)
ds
W (s)
23
Where y1 = x, y2 = x2 , y3 = x1 . By direct computation,
1
1
x x2
0 x2
x
x
6
1 1 W = 1 2x − x2 =
W1 = 0 2x − x2 = −3
2
2
0 2
x
1 2
x3
x3
1
x x2 0 x 0
x
2
W3 = 1 2x 0 = x2
W2 = 1 0 − x12 =
2
x
0 2 1 0 1
x3
hence
Zx
Zx
Zx
1
1
1
x3
1
Y (x) = x 2s(− s)ds + x2 2s( )ds +
2s( )ds = x4
2
3
x
6
15
Then a particular solution is
1
Y (t) = x4 .
15
14. Find a formula involving integrals for a particular solution of the differential equation
000
00
0
y − y + y − y = g(t).
Answer: The characteristic equation of homogeneous equation is
r3 − r2 + r − 1 = 0
the roots are r1 = 1, r2,3 = ±i. So the general solution of homogeneous equation is
yc (t) = c1 et + c2 cos t + c3 sin t
By method of variation of parameters, a particular solution is
Z t
3
X
g(s)Wm (s)
Y (t) =
ym (t)
ds
W (s)
m=1
Where y1 = et , y2 = cos t, y3 = sin t. By direct computation,
et cos t
0 cos t
sin
t
sin t
t
t
W = e − sin t cos t = 2e W1 = 0 − sin t cos t
et − cos t − sin t 1 − cos t − sin t
et 0 sin t
W2 = et 0 cos t
et 1 − sin t
=1
et cos t 0 = −et cos t + et sin t W3 = et − sin t 0 = −et cos t − et sin t
t
e − cos t 1 24
Hence
1
Y (t) =
2
Z
t
t−s
e
Z
t
Z
−(cos s − sin s)g(s)ds + sin t
g(s)ds + cos t
t
−(sin s + cos s)g(s)ds
Then a particular solution is
1
Y (t) =
2
Z
t
[et−s − sin (t − s) − cos (t − s)]g(s)ds
16. Find a formula involving integrals for a particular solutions of the differential equation.
y 000 − 3y 00 + 3y 0 − y = g(t),
if g(t) = t−2 et , determine Y (t).
Answer: The characteristic equation of homogeneous equation is
r3 − 3r2 + 3r − 1 = 0
the roots are r1 = r2,3 = 1. So the general solution of homogeneous equation is
yc (t) = c1 et + c2 tet + c3 t2 et .
By method of variation of parameters, a particular solution is
Z t
3
X
g(s)Wm (s)
Y (t) =
ym (t)
ds
W (s)
m=1
Where y1 = et , y2 = tet , y3 = t2 et . By direct computation,
t
2 t
0
et
te
t
e
tet
t2 et
t
3t
t
2
t
2
t
= 2e W1 = 0 (t + 1)e
(t + 2t)e
(t + 2t)et
W = e (t + 1)e
1 (t + 2)et (t2 + 4t + 2)et
et (t + 2)et (t2 + 4t + 2)et et 0
et
t2 et
tet
0 t
(t2 + 2t)et = −2e2t t W3 = et (t + 1)et 0 = e2t
W2 = e 0
et 1 (t2 + 4t + 2)et et (t + 2)et 1 Hence
1
Y (t) =
2
Z
t
et−s (t − s)2 g(s)ds.
t0
If g(t) = t−2 et , Then a particular solution is
Y (t) = −tet ln |t|.
= t2 et
25