sexe4yo
2022-05-02
Answered

Second Order Differential Equations $ay{}^{\u2033}+b{y}^{\prime}+cy=0$ , without complex numbers

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Addison Zamora

Answered 2022-05-03
Author has **10** answers

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 $h{}^{\u2033}+h=0,\text{}h\left(0\right)=0,\text{}{h}^{\prime}\left(0\right)=0$ . We want to show that $h\left(x\right)=0$ for all x.

Multiplying$h{}^{\u2033}+h=0$ by $h}^{\prime$ we see that $h{}^{\u2033}{h}^{\prime}+{h}^{\prime}h=0$ for all x. Note that $h{}^{\u2033}{h}^{\prime}+{h}^{\prime}h={({\frac{12}{{h}^{\prime}}}^{2}+{\frac{12}{h}}^{2})}^{\prime}$ . Therefore, by the Mean Value Theorem,

${\frac{12}{h}}^{\prime}{\left(x\right)}^{2}+\frac{12}{h}{\left(x\right)}^{2}=(x-0)(\frac{12}{{h}^{\prime}{\left(\theta x\right)}^{2}+\frac{12}{h}{\left(\theta x\right)}^{2}}=0.$

Now a real sum of squares can only be zero if each summand is zero. So we have, as required,$h\left(x\right)=0$ for all x.

[For those of an applied bent, we are just doing the classical thing of conserving the total energy.]

Let us show that there is a unique solution, given that

Clearly

Multiplying

Now a real sum of squares can only be zero if each summand is zero. So we have, as required,

[For those of an applied bent, we are just doing the classical thing of conserving the total energy.]

Makayla Santiago

Answered 2022-05-04
Author has **19** answers

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

$a\left[{e}^{kx}g\left(x\right)\right){}^{\u2033}+b{\left({e}^{kx}g\left(x\right)\right)}^{\prime}+c{e}^{kx}g\left(x\right)$

$=a[{e}^{kx}g{}^{\u2033}\left(x\right)+2k{e}^{kx}{g}^{\prime}\left(x\right)+{k}^{2}{e}^{kx}g\left(x\right)]+b[{e}^{kx}{g}^{\prime}\left(x\right)+k{e}^{kx}g\left(x\right)]+c{e}^{kx}g\left(x\right)$

$=a{e}^{kx}g{}^{\u2033}\left(x\right)+[2ak+b]{e}^{kx}{g}^{\prime}\left(x\right)+[a{k}^{2}+bk+c]{e}^{kx}g\left(x\right)$

$=0$

Since$e}^{kx$ is never 0, this reduces to

$ag{}^{\u2033}+(2ak+b){g}^{\prime}+(a{k}^{2}+bk+c)g=0.$

We are free to choose k as we wish, because if${e}^{kx}g\left(x\right)$ solves the ODE, then so does ${e}^{\stackrel{~}{k}x}{e}^{(k-\stackrel{~}{k})x}g\left(x\right)$ , and we simply get $\stackrel{~}{g\left(x\right)}={e}^{(k-\stackrel{~}{k})}g\left(x\right)$ 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)

$4{a}^{2}g{}^{\u2033}-({b}^{2}-4ac)g=0.$

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 $\omega {\textstyle \phantom{\rule{0.222em}{0ex}}}=\sqrt{c-\frac{{b}^{2}}{4a}}$ .

So in total we get$f\left(x\right)={e}^{kx}[A\mathrm{sin}\left(\omega x\right)+B\mathrm{cos}\left(\omega x\right)]$ with k and $\omega$ as specified.

You can make the educated guess that the solution will be of the form

Then we get

Since

We are free to choose k as we wish, because if

The solution to this, given that

So in total we get

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