Magdalena Norton

2022-04-30

Question on a differential equation

Let $x:\mathbb{R}\Rightarrow \mathbb{R}$ a solution of the differenzial equation:

$5x{}^{\u2033}\left(t\right)+10{x}^{\prime}\left(t\right)+6x\left(t\right)=0$

Proof that the function:

$f:\mathrm{\mathbb{R}}\Rightarrow \mathrm{\mathbb{R}}$

Kendal Kelley

Beginner2022-05-01Added 16 answers

Step 1

Let$y:\mathbb{R}\Rightarrow \mathbb{R}$ be defined as follows:

$y\left(t\right)=\frac{{t}^{2}}{1+{t}^{4}}$

Consider that x exists;$x\left(t\right)={e}^{-t}(A\mathrm{cos}\left(\frac{t}{\sqrt{5}}\right)+B\mathrm{cos}\left(\frac{t}{\sqrt{5}}\right))$ fulfills the differential equation. Consider also that

$f\left(t\right)=y\left(x\left(t\right)\right)$

Now calculate the first and second derivatives of y:

$y}^{\prime}\left(t\right)=\frac{2t(1+{t}^{4})-{t}^{2}\left(4{t}^{3}\right)}{{(1+{t}^{4})}^{2}$

$y}^{\prime}\left(t\right)=\frac{2t(1-{t}^{4})}{{(1+{t}^{4})}^{2}}=\frac{2t(1-{t}^{2})(1+{t}^{2})}{{(1+{t}^{4})}^{2}$

and

$y{}^{\u2033}\left(t\right)=\frac{(2-10{t}^{4})(1+2{t}^{4}+{t}^{8})-(2t-2{t}^{5})(8{t}^{7}+8{t}^{3})}{{(1+{t}^{4})}^{4}}$

$y{}^{\u2033}\left(t\right)=\frac{2+4{t}^{4}+2{t}^{8}-10{t}^{4}-20{t}^{8}-10{t}^{12}+16{t}^{12}+16{t}^{8}-16{t}^{8}-16{t}^{4}}{{(1+{t}^{4})}^{4}}$

$y{}^{\u2033}\left(t\right)=\frac{2-22{t}^{4}-18{t}^{8}+6{t}^{12}}{{(1+{t}^{4})}^{4}}$

$y{}^{\u2033}\left(t\right)=\frac{2(1+{t}^{4})(1-12{t}^{4}+3{t}^{8})}{{(1+{t}^{4})}^{4}}$

$y{}^{\u2033}\left(t\right)=\frac{2(1-12{t}^{4}+3{t}^{8})}{{(1+{t}^{4})}^{3}}$

Consider that the stationary points of y (points where${y}^{\prime}\left(t\right)=0$ ) are $-1,0,1$ and of those points, only -1 and 1 yield negative values for the second derivative. Therefore the points making y maximal are -1 and 1.

Now it only remains to prove that x(t) reaches the value -1 or 1 somewhere on$\mathbb{R}$ , which should be fairly easy (with the intermediate value theorem). After that is proven, you have a maximal value of f, which is $\frac{1}{2}$ .

Exception: if$A=B=0$ , then x(t) does not reach -1 or 1, but f becomes constant and thus has a maximum.

Let

Consider that x exists;

Now calculate the first and second derivatives of y:

and

Consider that the stationary points of y (points where

Now it only remains to prove that x(t) reaches the value -1 or 1 somewhere on

Exception: if

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