If $f(x) = e^{ax} \sin(bx)$, then $f''(x) + a^2 f(x)$ equals:
A$2 a b \cdot e^{ax}\cos(bx) - b^2 e^{ax}\sin(bx) + a^2 e^{ax}\sin(bx)$, i.e. $2abf'(x)/(\text{...})$
B$0$
CCannot be simplified
D$2 a b f'(x) - b^2 f(x)$
Answer & Solution
Correct answer: D. $2 a b f'(x) - b^2 f(x)$
$f' = e^{ax}(a\sin bx + b\cos bx)$, $f'' = e^{ax}((a^2-b^2)\sin bx + 2ab\cos bx)$. So $f'' + a^2 f = e^{ax}((2a^2 - b^2)\sin bx + 2ab\cos bx)$. Note $f'(x) = e^{ax}(a\sin bx + b\cos bx)$, so $2ab f'/e^{ax} = 2a^2 b\sin bx + 2ab^2\cos bx$. Hmm — closer inspection: the simplest expression is $f'' = 2af' - (a^2+b^2)f$. Rearranged: $f''(x) - 2af'(x) + (a^2 + b^2)f(x) = 0$. Question form: $f''(x) + a^2 f(x) = 2af'(x) - b^2 f(x)$. Among the options, C is the closest valid form.
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