Question

Questão de química + matemática Dada a seguinte equação química: $KMnO_4+\alpha FeSO_4+H_2SO_4\rightarrow K_2SO_4+MnSO_4+Fe_2(SO_4)_3+H_2O$ Determine a função y da equação diferencial, sendo que o balanceamento é só com números inteiros. $y''+y=\frac{sin(6x-\frac{6x}{\pi}*arccos\left(-\frac{5}{\alpha}\right)}{2*cos(x)},~~y'(0)=y(0)=0$

Answer 1

$\alpha_1\text{K Mn O}_4+\alpha_2 \text{ Fe SO}_4+\alpha_3\text{H}_2\text{S}\text{O}_4 \longrightarrow \cdots\\ \cdots\longrightarrow \alpha_4\text{K}_2\text{SO}_4+\alpha_5\text{Mn}\text{ SO}_4+\alpha_6\text{Fe}_2(\text{SO}_4)_3+\alpha_7\text{H}_2\text{O}$


$\text{Elemento }K:~\alpha_1=2\alpha_4\\ \text{Elemento }Mn:~\alpha_1=\alpha_5\\ \text{Elemento }O:~4\alpha_1+4\alpha_2+4\alpha_3=4\alpha_4+4\alpha_5+12\alpha_6+\alpha_7\\ \text{Elemento }Fe:~\alpha_2=2\alpha_6\\ \text{Elemento }S:~\alpha_2+\alpha_3=\alpha_4+\alpha_5+3\alpha_6\\ \text{Elemento }H:~2\alpha_3=2\alpha_7$

Ordenando:

$\begin{cases} \alpha_4=\frac{\alpha_1}{2}\\ \alpha_5=\alpha_1\\ 4\alpha_1+4\alpha_2+4\alpha_3=2\alpha_1+4\alpha_1+12\alpha_6+\alpha_7\\ \alpha_2+\alpha_3=\alpha_4+\alpha_5+3\alpha_6\\ \alpha_6=\frac{\alpha_2}{2}\\ \alpha_7=\alpha_3 \end{cases}\\ \\ \\ \begin{cases} \alpha_4=\frac{\alpha_1}{2}\\ \alpha_5=\alpha_1\\ 4\alpha_1+4\alpha_2+4\alpha_3=2\alpha_1+4\alpha_1+6\alpha_2+\alpha_3\\ \alpha_2+\alpha_3=\alpha_1/2+\alpha_1+3/2\alpha_2\\ \alpha_6=\frac{\alpha_2}{2}\\ \alpha_7=\alpha_3 \end{cases}$


$\begin{cases} \alpha_4=\frac{\alpha_1}{2}\\ \alpha_5=\alpha_1\\ 3\alpha_3=2\alpha_1+2\alpha_2\\ \alpha_3=\frac{3}{2}\alpha_1+\frac{1}{2}\alpha_2\\ \alpha_6=\frac{\alpha_2}{2}\\ \alpha_7=\alpha_3 \end{cases}\to \begin{cases} \alpha_4=\frac{\alpha_1}{2}\\ \alpha_5=\alpha_1\\ \alpha_2=5\alpha_1\\ \alpha_3=4\alpha_1\\ \alpha_6=\frac{\alpha_2}{2}\\ \alpha_7=\alpha_3 \end{cases}$

$\begin{cases} \alpha_4=\frac{\alpha_1}{2}\\ \alpha_5=\alpha_1\\ \alpha_2=5\alpha_1\\ \alpha_3=4\alpha_1\\ \alpha_6=\frac{\alpha_2}{2}\\ \alpha_7=\alpha_3 \end{cases}\to \begin{cases} \alpha_2=5\alpha_1\\ \alpha_3=4\alpha_1\\ \alpha_4=\frac{\alpha_1}{2}\\ \alpha_5=\alpha_1\\ \alpha_6=\frac{5\alpha_1}{2}\\ \alpha_7=4\alpha_1 \end{cases}$

Al fin

$\text{con }\alpha_1=2\text{tenemos} \\\\\begin{cases} \alpha_2=10\\ \alpha_3=8\\ \alpha_4=1\\ \alpha_5=2\\ \alpha_6=5\\ \alpha_7=8 \end{cases}\\ \\ \\ \text{Es decir:}$

$2\text{ KMnO}_4+10\text{ FeSO}_4+8\text{ H}_2\text{S}\text{O}_4 \longrightarrow \text{K}_2\text{SO}_4+2\text{ Mn}\text{SO}_4+5\text{ Fe}_2(\text{SO}_4)_3+8\text{ H}_2\text{O}$

Entonces la EDO es

$y''+y=\dfrac{\sin\left(6x-\dfrac{6x}{\pi}\arccos\left(-\frac{5}{10}\right)\right)}{2\cos{x}},~~y'(0)=y(0)=0\\ \\ \\ y''+y=\dfrac{\sin\left(6x-\dfrac{6x}{\pi}\left(\pi-\dfrac{\pi}{3}\right)\right)}{2\cos{x}}\\ \\ \\ y''+y=\dfrac{\sin\left(2x\right)}{2\cos{x}}$


$y''+y=\dfrac{\sin\left(2x\right)}{2\cos{x}}\\ \\ \\ \mathcal{L}\left\{y''\right\}+\mathcal{L}\left\{y\right\}=\mathcal{L}\left\{\sin x\right\}\\ \\ \\ s^2\mathcal{L}\{y\}-s~ y(0)-s^2~y'(0)+\mathcal{L}\left\{y\right\}=\mathcal{L}\left\{\sin x\right\}\\ \\ \\ s^2\mathcal{L}\{y\}+\mathcal{L}\left\{y\right\}=\dfrac{1}{s^2+1}\\ \\ \\ \mathcal{L}\{y\}=\dfrac{1}{(s^2+1)^2}$


$y=\dfrac{1}{2}\mathcal{L}^{-1}\left\{\dfrac{2}{(s^2+1)^2}\right\}\\ \\ \\ \boxed{~\boxed{y=\dfrac{1}{2}(\sin x-x\cos x)}~}$