Second Order Differential Equations ay''+by'+cy=0, without complex numbers

Step 1
Let us show that there is a unique solution, given that
${\displaystyle g\left(0\right)=A,}$
${\displaystyle g^{\prime }\left(0\right)=B}$.
Clearly ${\displaystyle A\cos x+B\sin x}$ is a solution with these initial conditions, so look at ${\displaystyle h\left(x\right):g\left(x\right)-A\cos x+B\sin x}$; this satisfies ${\displaystyle h{}^{″}+h=0,h\left(0\right)=0,h^{\prime }\left(0\right)=0}$. We want to show that ${\displaystyle h\left(x\right)=0}$ for all x.
Multiplying ${\displaystyle h{}^{″}+h=0}$ by ${\displaystyle h^{\prime }}$ we see that ${\displaystyle h{}^{″}{h}^{\prime }+h^{\prime }h=0}$ for all x. Note that ${\displaystyle h{}^{″}{h}^{\prime }+h^{\prime }h={({\frac{12}{h^{\prime }}}^2+{\frac{12}{h}}^2)}^{\prime }}$. Therefore, by the Mean Value Theorem,
${\displaystyle {\frac{12}{h}}^{\prime }{\left(x\right)}^2+\frac{12}{h}{\left(x\right)}^2=(x-0)(\frac{12}{h^{\prime }{\left(\theta x\right)}^2+\frac{12}{h}{\left(\theta x\right)}^2}=0.}$
Now a real sum of squares can only be zero if each summand is zero. So we have, as required, ${\displaystyle h\left(x\right)=0}$ for all x.
[For those of an applied bent, we are just doing the classical thing of conserving the total energy.]

Step 1
You can make the educated guess that the solution will be of the form
${\displaystyle f\left(x\right)=e^{kx}g\left(x\right)}$.
Then we get
${\displaystyle a\left[e^{kx}g\left(x\right)\right){}^{″}+b{\left(e^{kx}g\left(x\right)\right)}^{\prime }+c{e}^{kx}g\left(x\right)}$
${\displaystyle =a[e^{kx}g{}^{″}\left(x\right)+2k{e}^{kx}{g}^{\prime }\left(x\right)+k^2{e}^{kx}g\left(x\right)]+b[e^{kx}{g}^{\prime }\left(x\right)+k{e}^{kx}g\left(x\right)]+c{e}^{kx}g\left(x\right)}$
${\displaystyle =a{e}^{kx}g{}^{″}\left(x\right)+[2ak+b]e^{kx}{g}^{\prime }\left(x\right)+[a{k}^2+bk+c]e^{kx}g\left(x\right)}$
${\displaystyle =0}$
Since ${\displaystyle e^{kx}}$ is never 0, this reduces to
${\displaystyle ag{}^{″}+(2ak+b)g^{\prime }+(a{k}^2+bk+c)g=0.}$
We are free to choose k as we wish, because if ${\displaystyle e^{kx}g\left(x\right)}$ solves the ODE, then so does ${\displaystyle e^{\tilde{k}x}{e}^{(k-\tilde{k})x}g\left(x\right)}$, and we simply get ${\displaystyle \widetilde{g\left(x\right)}=e^{(k-\tilde{k})}g\left(x\right)}$ as the solution for g instead. So we choose k as simple as possible: such that ${\displaystyle 2ak+b=0}$, that is, ${\displaystyle k=-\frac{b}{2a}}$. Then the equation reads (after multiplying by 4a)
${\displaystyle 4a^{2}g{}^{″}-(b^2-4ac)g=0.}$
The solution to this, given that ${\displaystyle b^2-4ac<0}$, is known: it's of the form ${\displaystyle g\left(x\right)=A\sin \left(\omega x\right)+B\cos \left(\omega x\right)}$, where ${\displaystyle \omega =\sqrt{c-\frac{b^2}{4a}}}$.
So in total we get ${\displaystyle f\left(x\right)=e^{kx}[A\sin \left(\omega x\right)+B\cos \left(\omega x\right)]}$ with k and ${\displaystyle \omega }$ as specified.