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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.
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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?
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
.
.
.
A=10/13, B= 2/13
.
.
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?