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

Second Order Differential Equations $ay{}^{″}+b{y}^{\prime }+cy=0$, without complex numbers
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Step 1
Let us show that there is a unique solution, given that
$g\left(0\right)=A,$
${g}^{\prime }\left(0\right)=B$.
Clearly $A\mathrm{cos}x+B\mathrm{sin}x$ is a solution with these initial conditions, so look at $h\left(x\right):g\left(x\right)-A\mathrm{cos}x+B\mathrm{sin}x$; this satisfies . We want to show that $h\left(x\right)=0$ for all x.
Multiplying $h{}^{″}+h=0$ by ${h}^{\prime }$ we see that $h{}^{″}{h}^{\prime }+{h}^{\prime }h=0$ for all x. Note that $h{}^{″}{h}^{\prime }+{h}^{\prime }h={\left({\frac{12}{{h}^{\prime }}}^{2}+{\frac{12}{h}}^{2}\right)}^{\prime }$. Therefore, by the Mean Value Theorem,
${\frac{12}{h}}^{\prime }{\left(x\right)}^{2}+\frac{12}{h}{\left(x\right)}^{2}=\left(x-0\right)\left(\frac{12}{{h}^{\prime }{\left(\theta x\right)}^{2}+\frac{12}{h}{\left(\theta x\right)}^{2}}=0.$
Now a real sum of squares can only be zero if each summand is zero. So we have, as required, $h\left(x\right)=0$ for all x.
[For those of an applied bent, we are just doing the classical thing of conserving the total energy.]
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Step 1
You can make the educated guess that the solution will be of the form
$f\left(x\right)={e}^{kx}g\left(x\right)$.
Then we get
$a\left[{e}^{kx}g\left(x\right)\right){}^{″}+b{\left({e}^{kx}g\left(x\right)\right)}^{\prime }+c{e}^{kx}g\left(x\right)$
$=a\left[{e}^{kx}g{}^{″}\left(x\right)+2k{e}^{kx}{g}^{\prime }\left(x\right)+{k}^{2}{e}^{kx}g\left(x\right)\right]+b\left[{e}^{kx}{g}^{\prime }\left(x\right)+k{e}^{kx}g\left(x\right)\right]+c{e}^{kx}g\left(x\right)$
$=a{e}^{kx}g{}^{″}\left(x\right)+\left[2ak+b\right]{e}^{kx}{g}^{\prime }\left(x\right)+\left[a{k}^{2}+bk+c\right]{e}^{kx}g\left(x\right)$
$=0$
Since ${e}^{kx}$ is never 0, this reduces to
$ag{}^{″}+\left(2ak+b\right){g}^{\prime }+\left(a{k}^{2}+bk+c\right)g=0.$
We are free to choose k as we wish, because if ${e}^{kx}g\left(x\right)$ solves the ODE, then so does ${e}^{\stackrel{~}{k}x}{e}^{\left(k-\stackrel{~}{k}\right)x}g\left(x\right)$, and we simply get $\stackrel{~}{g\left(x\right)}={e}^{\left(k-\stackrel{~}{k}\right)}g\left(x\right)$ as the solution for g instead. So we choose k as simple as possible: such that $2ak+b=0$, that is, $k=-\frac{b}{2a}$. Then the equation reads (after multiplying by 4a)
$4{a}^{2}g{}^{″}-\left({b}^{2}-4ac\right)g=0.$
The solution to this, given that ${b}^{2}-4ac<0$, is known: it's of the form $g\left(x\right)=A\mathrm{sin}\left(\omega x\right)+B\mathrm{cos}\left(\omega x\right)$, where $\omega \phantom{\rule{0.222em}{0ex}}=\sqrt{c-\frac{{b}^{2}}{4a}}$.
So in total we get $f\left(x\right)={e}^{kx}\left[A\mathrm{sin}\left(\omega x\right)+B\mathrm{cos}\left(\omega x\right)\right]$ with k and $\omega$ as specified.