Differential of $x= (A\cos t +B\sin t)e^{-3t}$

Question

Just a quick question. I'm trying to find the solution of the following differential equations satisfying the given conditions:

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My general solution was: $x= (A\cos t +B\sin t)e^{-3t}$. I think I'm going wrong on the differentiation.

Is the differential $x'= -3(A\cos t +B\sin t)e^{-3t}.(-A\sin t+B\cos t)$ ?

I'm supposed to get an $A$ value=1 (which I got using the above calculation) and a $B$ value=3 (but I keep getting a $B$ value of 0)

Does anybody understand where I'm going wrong?

Many thanks

Your differential is wrong. $(f(x)e^{g(x)})'=f'(x)e^{g(x)}+f(x)g'(x)e^{g(x)}$ — Cortizol, Mar 04 '13 at 22:51

score 2 · Accepted Answer · answered Mar 04 '13 at 22:52

2

No, the differentiation of $x$ w.r.t. $t$ will not yield that result. You're applying the product rule wrongly.

$$ \dfrac{dx}{dt} = (A \cos{t} + B \sin{t}) (-3 e^{-3t}) + ( - A \sin{t} + B \cos{t})e^{-3t} $$

answered Mar 04 '13 at 22:52

hjpotter92

1 Answers1