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.
Related questions
An air pollution index is $p=x^2+2xy+4xy^2$. At the point $(10,5)$, the partial derivativeUsing the second derivative test on $y=3xe^{-x}$, the stationary point is:If $y=\tan^{-1}\left(\dfrac{ qrt{1+x}- qrt{1-x}}{ qrt{1+x}+ qrt{1-x}}\right)$, then $\dfra$\dfrac{d}{dx}\left[\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)\right]$ for $x > 0$ equals:If $f(x) = x^3 - 3x$, the values of $x$ where $f'(x) = 0$ are:If $y_n$ denotes the $n$-th derivative of $y = in x$, then $y_n = $:If $y = \tan^{-1}\left(\dfrac{ qrt{1+x^2} - 1}{x}\right)$, then $\dfrac{dy}{dx}$ equals:If $f(x) = |x - 1| + |x - 3|$, then $f'(2)$ equals: