# Thread: Initial Value Problem - Differential Equation!!!

1. ## Initial Value Problem - Differential Equation!!!

Hi, For the following problem I am having hard time trying to find the phase angle (alpha).

How to find the phase angle,

isnt that alpha = arctan(B/A) if A,B>0 ?

Problem:

Solve the initial value problem (where primes indicate derivatives with respect to t). Express your solution as the sum of two oscillations,

x
(t) = (49/27)*cos(6t - ) + (5/27)cos(3t)

where = (Recall that must be between 0 and 2 .)

But the phase angle is not right.

Many Thanks.

2. There is no phase angle here as the coefficient of sin (6t) = 0

x= Acos(6t)+Bsin(6t) + 5/27 cos(3t)

x(0) = A + 5/27 = 0 A = -5/27

x ' (0) = 6B = 0 B = 0

x(t) = -5/27 cos(6t) +5/27cos(3t)

3. Thanks a lot.

Actually, I misinterpreted A as 5/27 which made me write the angle as arctan((49/27)/(5/27)).

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