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Unformatted text preview: AMS 361: Applied Calculus IV by Prof. Y. Deng
Quiz 2 Wednesday (03/25/2009) at 8:05AM9:25AM
Open Book and Open Own Lecture Notes; Calculators Allowed
(1) Do Any Two of the Three Problems; Mark the Two Problems You Attempt; Will Only Grade the First
Two if You Don’t Mark the Problems You Attempted
(2) Each Problem Is Worth 7.5 Points
(3) No Points for Guessing Work and for Solutions without Appropriate Intermediate Steps; Partial
Credits are given only for Steps that are Relevant to the Solutions Q21 (7.5 Points): Find the general solution of the following nonhomogeneous equation
Solution:
The homogeneous characteristic equation is
cos
1 , , => 1 , 4 The particular solution 1 1 1
,
4 = cos Therefore, the general solution is sin where
Q22 (7.5 Points): For a given linear DE:
√ 1 and
are constants, find the general solution with the following two conditions:
(3.0 Points) and
(4.5 Points).
Solution:
The homogeneous general solution is
IF
Plug = cos suppose the particular solution: ,
, to , suppose the particular sol. IF
Plug , to sin . sin
and find , then
1 So, the general solution
cos sin 1 . and find , then find
1
2
cos So, the general solution 0, sin 1 . and Q23 (7.5 Points): For a 2nd order linear DE:
0 where
and
are given functions of independent variable . You are given a solution
,
for this equation as
(1) (5.0 Points) Find the general solution in terms of
,
, and
.
Solution:
0
: the given solution.
By reduction of order method,
Let’s determine is a second linearly independent solution of (*). which in (*). By substitution, 0 2 is a separable equation because we know
We can get . by integration. After we get , then is another solution.
. Therefore, the general solution is
2 (2) (2.5 Points) Find the general solution of 1
.
given 2 0 with Solution:
Let x = vt , x ' = tv '+ v, x ' ' = tv ' '+ v '+ v ' = v ' '+2v ' . Then the equation is
1
1
2
v' ' 4t 2 − 2
=
=
−
− .
2
v' t (1 − t ) 1 − t 1 + t t
ln v ' = −2 ln t − ln(t + 1) − ln(t − 1) + ln A. v" (t − t 3 ) + v' (2 − 4t 2 ) = 0. 1 A
1
1
1
(t + 1) 2 A
v' = 2
= −{ 2 −
+
} A, v = A ln
+ + B.
1
t (1 + t )(t − 1)
t
2(t + 1) 2(t − 1)
t
2
(t − 1)
1 Let A = 1, B = 0, we have an independent solution x = tv = t ln (t + 1) 2
(t − 1) Therefore, the general solution is x = c1t ln (t + 1)
(t − 1) 2 1
2
1
2 . + c1 + c2t. 1
2 . + 1. ...
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This note was uploaded on 05/11/2009 for the course AMS 361 taught by Professor Staff during the Spring '08 term at SUNY Stony Brook.
 Spring '08
 Staff

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