For the particular solution I have xp(t) = A cos(t) + Bsin(t). When substituting
ID: 2939748 • Letter: F
Question
For the particular solution I have xp(t) = A cos(t) + Bsin(t). When substituting xp(t) into x'' - 9x = cos(t) - 4 Iend up with -10Acos(t) - 10Bsin(t) = cos(t) - 4.
Are we allowed to assume that t equals some value, say 0, sothat we have 2 uquations and 2 unknowns to solve for A andB?? This would make -10Acos(0) - 10Bsin(0) = cos(0) - 4 =-10A = 1-4 = -3. So, A = 3/10?? The value of B could beanything though since sin(0) is zero.
Please let me know if I'm on the right track with this problemand how to finish solving it.
Thank you.
Explanation / Answer
For the particular solution I have xp(t) = A cos(t) + B sin(t)........OK
When substituting xp(t) into x'' - 9x = cos(t) - 4 I end up with-10Acos(t) - 10Bsin(t) = cos(t) - 4.
Are we allowed to assume that t equals some value, say 0, sothat we have 2 uquations and 2 unknowns to solve for A andB?? This would make -10Acos(0) - 10Bsin(0) = cos(0) - 4 =-10A = 1-4 = -3. So, A = 3/10?? The value of B could beanything though since sin(0) is zero.
Please let me know if I'm on the right track with this problemand how to finish solving it.
Thank you....
NO.......
EQUATE COEFFICIENT OF COS(T) TO ZERO AND LIKE WISE
EQUATE COEFFICIENT OF SIN (T) TO ZERO
TO GET 2 SIMULTANEOUS EQNS. IN A AND B TO SOLVE
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