In Boas it says that if the roots of the auxiliary equation are $\alpha \pm i\beta$, then for the solution we get $$y=Ae^{(\alpha + i\beta)x} + Be^{(\alpha - i\beta)x} = e^{\alpha x}(Ae^{i\beta x}+Be^{-i\beta x})$$
This is fine so far. Next it says that if we substitute $e^{\pm i\beta}=\cos(\beta x) \pm i\sin(\beta x)$, then we get
$$y = e^{\alpha x}(c_{1}\sin(\beta x)+c_{2}\cos(\beta x))$$ which seems to disregard the imaginary part. Is the imaginary part just absorbed into the constants $c_{1}$ and $c_{2}$? Does this somehow not affect the solutions?
You write $$ y_1=\cos(\beta x)+i\sin(\beta x) $$ and $$ y_2=\cos(\beta x)-i\sin(\beta x). $$ Then, let $$ Y_1=\frac{1}{2}(y_1+y_2)=\cos(\beta x), $$ and $$ Y_2=\frac{1}{2i}(y_1-y_2)=\sin(\beta x). $$
Now, one can show that for any constants $c_1$ and $c_2$, there are corresponding constants $C_1$ and $C_2$ (which may be complex) so that $$ c_1y_1+c_2y_2=C_1Y_1+C_2Y_2. $$ In other words, $y_1$ and $y_2$ generate the same solutions as $C_1$ and $C_2$.
Moreover, since $$ y_1=Y_1+iY_2 $$ and $$ y_2=Y_1-iY_2, $$ if we look at $c_1y_1+c_2y_2$, we get $$ c_1y_1+c_2y_2=c_1(Y_1+iY_2)+c_2(Y_1-iY_2)=(c_1+c_2)Y_1+(ic_1-ic_2)Y_2. $$ Therefore, $C_1=c_1+c_2$ and $C_2=i(c_1-c_2)$, i.e., the imaginary part has been absorbed into the constant.