MA2051 - Ordinary Differential Equations
Sample Exam 2 Solutions - A96

Second Exam - Originally Given 1993 A Term




1
A

tex2html_wrap_inline70 dynes/cm; the model is 200 x'' + 24,500 x = 0, x(0) = 0, x'(0) = -6.

B
Use characteristic equations: tex2html_wrap_inline78 ; tex2html_wrap_inline80 . Hence, a general solution is tex2html_wrap_inline82 . The initial conditions force tex2html_wrap_inline84 , C2 = -.06/11.07 = -.00542 . Hence, tex2html_wrap_inline88 .
C
To find the period, solve tex2html_wrap_inline90 to find tex2html_wrap_inline92 s.
D
The maximum displacement occurs in tex2html_wrap_inline88 when the sine function is tex2html_wrap_inline96 . Hence, the maximum displacement (or amplitude of the motion) is 1.85 cm.

2
A
Use undetermined coefficients. First guess tex2html_wrap_inline98 . Because these functions are solutions of the homogeneous equation, the correct guess is tex2html_wrap_inline100 . Substitute and collect coefficients of sine and cosine to find A = 0, B = 1/4. Hence, tex2html_wrap_inline106 .
B
Using the particular solution just found and the given homogeneous solutions, tex2html_wrap_inline108 .
C
Because of the leading factor of t in tex2html_wrap_inline112 , the solution grows without bound. The system is being forced at its resonant frequency.

3
A
r = -4, 2
B
tex2html_wrap_inline116 .
C
Choose tex2html_wrap_inline118 , tex2html_wrap_inline120 .

4
A
Write tex2html_wrap_inline122 .

B
tex2html_wrap_inline124
C
Let tex2html_wrap_inline126 . Then tex2html_wrap_inline128 and

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  • tex2html_wrap_inline130 . Using partial fractions, we find

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    Then tex2html_wrap_inline132 , tex2html_wrap_inline134 , and

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    Hence, the solution is

    displaymath65

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    © 1996 by
    Will Brother. All rights Reserved. File last modified on December 6, 1996.