I know that if you think a book is wrong, most probably it is your own mistake. However, I can't understand the following Laplace transforms in K. A. Stroud's "Advanced Engineering Mathematics". In page 131 (Harmonic Oscillators part), it is written:
$$ af''(t)+bf(t)=0 \quad f(0)=\alpha,\;f'(0)=\beta $$
$$ L\{af''(t)+bf(t)\}=L\{0\} $$
$$ a[s^2F(s)-s\alpha-\beta]+bF(s)=0 $$
$$ (as^2+b)F(s)=s\alpha+\beta $$
$$ F(s)=\frac{s(\alpha/a)}{s^{2}+(b/a)}+\frac{\beta/a}{s^{2}+(b/a)} $$
giving:
$$ f(t)=\frac{\alpha}{a}Cos\sqrt{\frac{b}{a}}t+\frac{\beta}{a}Sin\sqrt{\frac{b}{a}}t $$
However, according to me it should be as below:
$$ (as^2+b)F(s)=as\alpha+a\beta $$
$$ F(s)=\frac{s\alpha}{s^{2}+(b/a)}+\frac{\beta}{s^{2}+(b/a)} $$
which should give:
$$ f(t)=\alpha Cos\sqrt{\frac{b}{a}}t+\beta\sqrt{\frac{a}{b}}Sin\sqrt{\frac{b}{a}}t $$
Is there a mistake in the book or am I terribly missing something?
A quick check would be to plug it back into the original differential equation and see if it works.
Also note that the book's result doesn't satisfy the initial conditions, since $f(0) = \alpha/a \neq \alpha$. Whereas your result does.
Taking your function: $$f(t) = \alpha \cos \sqrt{\frac{b}{a}} t + \beta \sqrt{\frac{a}{b}} \sin \sqrt{\frac{b}{a}} t$$
We wish to show that $af'' + bf = 0$.
$$a f''(t) = - \alpha b \cos \sqrt{\frac{b}{a}} t - \beta b \sqrt{\frac{a}{b}} \sin \sqrt{\frac{b}{a}} t = - b f(t)$$
So then we see that $af'' + bf = 0$, which means you are correct.
Also it is useful to solve the DE is multiple ways as a check. We can quickly see that two linearly independent solutions to $f'' = -b/a f$ is $\sin(\sqrt{b/a} t)$ and $\cos(\sqrt{b/a}t)$. Which means the solution to the differential equation is of the form $A\cos(\sqrt{b/a}t) + B\sin(\sqrt{b/a}t = f(t)$ then solve for $A$ and $B$ given the initial conditions.