Math 234 Summer 2002 Assignment 11 - Solutions
Turn in the Problems with ! and show your work in detail.
Reading assignment: Section 3.1. Show your work in detail.
1. Page 113: 2, 4, 9, ! 10, 11, ! 12, 13, ! 14(b)(c)
2 y ! c 1 e 4x " c 2 e "x , " # # x # #, y!0" ! 1, y $ !0" ! 2
y $ ! 4c 1 e x " c 2 e "x , y!0" ! c 1 " c 2 ! 1, y $ !0" ! c 1 " c 2 ! 2. Solve c 1 and c 2 .
c1 " c2 ! 1
4c 1 " c 2 ! 2
#
c1
c2
!
1
"1
1
4 "1
1
!
2
3
5
The solution of the initial value problem: y !
e 4x "
2
5
3
5
2
5
,
e "x .
4 y ! c 1 " c 2 cos!x" " c 3 sin!x", , " # # x # #, y!!" ! 0, y $ !!" ! 2, y $$ !!" ! "1
y $ ! "c 2 sin!x" " c 3 cos!x", y $$ ! "c 2 cos!x" " c 3 sin!x"
y!!" ! c 1 " c 2 ! 0, y $ !!" ! "c 3 ! 2, y $$ !!" ! c 2 ! "1, c 3 ! "2, c 1 ! c 2 ! "1
The solution of the initial value problem: y ! "1 " cos!x" " 2 sin!x"
9 !x " 2"y $$ " 3y ! x, y!0" ! 0, y $ !0" ! 1
a 2 !x" ! x " 2, a 1 !x" ! 0, a 0 !x" ! 3, f!x" ! x
All these functions are continuous on "#, # . a 2 !x" ! x " 2 $ 0 when x $ 2.
Let I ! "1. 5, 1. 5 .
10 y $$ " tan!x"y ! e x , y!0" ! 1, y $ !0" ! 0
a 2 !x" ! 1, a 1 !x" ! 0, a 0 !x" ! tan!x", f!x" ! e x .
a 2 !x" $ 0, and a 2 !x", a 1 !x" and f!x" are continuous everywhere.
However, a 0 !x" ! tan!x" is continuous on " ! , ! .
2 2
So, the initial value problem has a unique solution on " ! , ! .
2 2
11. y $$ " y ! 0, y!0" ! 0, y!1" ! 1, y ! c 1 e x " c 2 e "x .
y!0" ! c 1 " c 2 ! 0
y!1" ! c 1 e " c 2 e "1 ! 1
,
1 1
c1
e e "1
c2
!
1
"1
e
"e
!
!
!
"1
1
e e
1
c2
e "1 " e
e ex "
e e "x
The solution of the initial value problem: y ! "
2
1"e
1 " e2
c1
1 1
"1
0
12 y ! c 1 " c 2 x 2 , " # # x # #, y!0" ! 1, y $ !1" ! 6
y $ ! 2c 2 x, y!0" ! c 1 ! 1, y $ !1" ! 2c 2 ! 6, c 2 ! 3
The solution of the initial value problem: y ! 1 " 3x 2
13 y ! c 1 e x cos x " c 2 e x sin x,
1
"
0
1
"
e
1 " e2
e
1 " e2
!a"
y!0" ! 1,
$
y !!" ! 0
; !b"
y!0" ! 1,
y!0" ! 1,
; !c"
y!!" ! "1
!
2
y
!1
; !d"
y!0" ! 0,
y!!" ! 0
.
y $ ! c 1 e x cos x " c 1 e x sin x " c 2 e x sin x " c 2 e x cos x ! c 1 e x !cos x " sin x" " c 2 e x !sin x " cos x"
!a"
!b"
!c"
!d"
y!0" ! c 1 ! 1,
, the solution y ! e x cos x " e x sin x
y $ !!" ! e ! !"1" " c 2 e ! !"1" ! 0, c 2 ! "1
y!0" ! c 1 ! 1,
y!!" ! e ! !"1" ! "1
, no solution since e ! $ 1.
y!0" ! c 1 ! 1,
y
!
2
! c 2 e !/2 ! 1, c 2 ! e "!/2
y!0" ! c 1 ! 0,
!
y!!" ! c 2 e !0" ! 0, c 2 is free.
, the solution y ! e x cos x " e "!/2 e x sin x
, the solution y ! c 2 e x sin x
14 y ! c 1 x 2 " c 2 x 4 " 3, " # # x # #
!b" y!0" ! 1, y!1" ! 2
!c" y!0" ! 3, y!1" ! 0
!b" y!0" ! 3 $ 1, we cannot find c 1 and c 2 so that y satisfies the boundary conditions. So, there is no
solution for this boundary value problem.
!c" y!0" ! 3, so it is true for any c 1 and c 2 . y!1" ! c 1 " c 2 " 3 ! 0, c 1 ! "3 " c 2
y ! !"3 " c 2 "x 2 " c 2 x 4 " 3, infinitely many solutions.
2. State the criterion for linear independent solutions of a homogeneous linear nth-order differential equation
on an interval I.
If y 1 , . . . , y n are linearly independent on I if and only W!y 1 , . . . , y n " $ 0 for all x in I.
Page 113: 15, 17, 18, ! 20, ! 21
15 f 1 !x" ! x, f 2 !x" ! x 2 , f 3 !x" ! 4x " 3x 2
Since f 3 ! 4f 1 " 3f 2 , f 1 , f 2 , f 2 are linearly dependent.
17 f 1 !x" ! 5, f 2 !x" ! cos 2 x, f 3 !x" ! sin 2 x
Since f 1 ! 5f 2 " 5f 1 , f 1 , f 2 , f 3 are linearly dependent
18 f 1 !x" ! cos!2x", f 2 !x" ! 1, f 3 !x" ! cos 2 x
Note that cos 2 !x" ! 12 " 12 cos!2x". Since f 3 !
1
f
2 2
"
1
f ,
2 1
f 1 , f 2 , f 2 are linearly dependent.
20. f 1 !x" ! 2 " x, f 2 !x" ! 2 " |x|, " # # x # #
Let c 1 f 1 !x" " c 2 f 2 !x" ! c 1 !2 " x" " c 2 !2 " |x|" ! 0. Now we solve c 1 and c 2 .
x % 0, c 1 !2 " x" " c 2 !2 " x" ! !2 " x"!c 1 " c 2 " ! 0 # c 1 " c 2 ! 0
x # 0, c 1 !2 " x" " c 2 !2 " x" ! 0 # 2!c 1 " c 2 " " x!c 1 " c 2 " ! 0 # c 1 " c 2 ! 0
c1
c2
!
1
1
1 "1
Since c 1 ! 0, c 2 ! 0, f 1 , f 2 are linearly independent.
2
"1
0
0
!
0
0
21 f 1 !x" ! 1 " x, f 2 !x" ! x, f 3 !x" ! x 2
Since f 1 !x" $ c 1 f 2 !x" " c 2 f 3 !x", f 2 !x" $ c 1 f 1 !x" " c 2 f 3 !x", f 3 !x" $ c 1 f 1 !x" " c 2 f 2 !x",
f 1 !x", f 2 !x", f 3 !x" are linearly independent.
3