Undetermined Coeff's & solve eqn: y"-3y'-4y = -8e^t

shivers20

Junior Member
Joined
Mar 3, 2006
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68
I dont understand why we used the product rule when taking the derivatives of Y(t)'s when we didnt have to use it for other examples. Does it have something to do with this = -8e^t cos2t?

y"-3y'-4y = -8e^t cos2t

Y(t) = Ae^t cos2t + Be^t sin2t

Y(t)' =Ae^t cos2t - 2Ae^t sin2t + Be^t sin2t + 2Be^t cos2t <-Product rule & why?
=(A+2B)e^t cos2t + (-2A+B)e^t sin2t

Y(t)" = (A+2B)e^t cos2t - 2(A+2B)e^t sin2t + (-2A+B)e^t sin2t + 2(-2A+B)e^t cost
= (-3A+4B)e^t cos2t + (-4A-3B)e^t sin2t

Substitute
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A=10/13, B= 2/13
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Answer

2. I should feel ashamed but I am having a difficult time solving basic equations when it comes to this,

-12A+9B=2 How would I solve this?
 
The product rule was applied in picking Y(t).

The derivative of -8e^t*cos(2t) is -8e^t*cos(2t) + 16e^t*sin(2t). If you differentiate the latter, the result is still in terms of only e^t*cos(2t) and e^t*sin(2t).

The equations to solve after substituting Y''(t), Y'(t) and Y(t) into the DE are -10A - 2B = -8 (from equating the coefficients of cos(2t) terms) and 2A - 10B = 0 (from equating the coefficients of sin(2t) terms).
 
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