Question

A solução do problema de valor inicial dy/dx=3x-2y-6+xy, com y(2)=-2, é uma função y(x). Então, pode-se afirmar q y(1) vale, aproximadamente?
- 3,36


1,64


- 1,35


3,15


1,00

Answer 1

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——————————

Resolver o problema de valor inicial:
 
 dy
——  =  3x – 2y – 6 + xy
 dx


com  y(2) = – 2.

 
Rearrumando a equação:

 dy
——  + 2y – xy =  3x – 6
 dx

 dy
——  + (2 – x) · y =  3x – 6
 dx

 dy
——  + (2 – x) · y =  – 6 + 3x
 dx

 dy
——  + (2 – x) · y =  – 3 · (2 – x)          (i)
 dx


Temos uma equação diferencial ordinária de 1ª ordem, linear, não-homogênea e a coeficientes não-constantes.


A equação  (i)  é na forma

 dy
——  +  p(x) · y  = q(x)
 dx


com  p(x) = 2 – x  e   q(x) = – 3 · (2 – x).


Fator integrante:

$\mathsf{\mu(x)=e^{\int\!p(x)\,dx}}\\\\\\ \mathsf{\mu(x)=e^{\int\!(2-x)\,dx}}\\\\\\ \mathsf{\mu(x)=e^{\footnotesize\begin{array}{l}\!\!\!\mathsf{2x-\frac{x^2}{2}} \end{array}}}$


Se multiplicarmos os dois lados pelo fator integrante, obtemos

$\mathsf{e^{\footnotesize\begin{array}{l}\!\!\!\mathsf{2x-\frac{x^2}{2}}\end{array}\!\!\!}\cdot \left[\dfrac{dy}{dx}+(2-x)\cdot y \right]=e^{\footnotesize\begin{array}{l}\!\!\!\mathsf{2x-\frac{x^2}{2}}\end{array}\!\!\!}\cdot (-3)\cdot (2-x)}\\\\\\ \mathsf{e^{\footnotesize\begin{array}{l}\!\!\!\mathsf{2x-\frac{x^2}{2}}\end{array}\!\!\!}\cdot \dfrac{dy}{dx}+\left[e^{\footnotesize\begin{array}{l}\!\!\!\mathsf{2x-\frac{x^2}{2}}\end{array}\!\!\!}\cdot (2-x)\right]\cdot y =e^{\footnotesize\begin{array}{l}\!\!\!\mathsf{2x-\frac{x^2}{2}}\end{array}\!\!\!}\cdot (-3)\cdot (2-x)}$


O lado esquerdo pode ser visto como a derivada de um produto:

$\mathsf{\dfrac{d}{dx}\!\left(e^{\footnotesize\begin{array}{l}\!\!\!\mathsf{2x-\frac{x^2}{2}}\end{array}\!\!\!}\cdot y\right )=-\,3\cdot e^{\footnotesize\begin{array}{l}\!\!\!\mathsf{2x-\frac{x^2}{2}}\end{array}\!\!\!}\cdot (2-x)}$


Integrando ambos os lados com respeito a  x:

$\mathsf{\displaystyle\int\!\frac{d}{dx}\!\left(e^{\footnotesize\begin{array}{l}\!\!\!\mathsf{2x-\frac{x^2}{2}}\end{array}\!\!\!}\cdot y\right )dx=-\,3\int\! e^{\footnotesize\begin{array}{l}\!\!\!\mathsf{2x-\frac{x^2}{2}}\end{array}\!\!\!}\cdot (2-x)\,dx}\\\\\\ \mathsf{e^{\footnotesize\begin{array}{l}\!\!\!\mathsf{2x-\frac{x^2}{2}}\end{array}\!\!\!}\cdot y=-\,3\displaystyle\int\! e^u\,du\qquad\qquad onde~~u=2x-\dfrac{x^2}{2}}\\\\\\ \mathsf{e^{\footnotesize\begin{array}{l}\!\!\!\mathsf{2x-\frac{x^2}{2}}\end{array}\!\!\!}\cdot y=-\,3e^u+C}\\\\\\ \mathsf{e^{\footnotesize\begin{array}{l}\!\!\!\mathsf{2x-\frac{x^2}{2}}\end{array}\!\!\!}\cdot y=-\,3e^{\footnotesize\begin{array}{l}\!\!\!\mathsf{2x-\frac{x^2}{2}}\end{array}\!\!\!}+C}$


Isolando  y,

$\mathsf{y=\dfrac{-\,3e^{\footnotesize\begin{array}{l}\!\!\!\mathsf{2x-\frac{x^2}{2}}\end{array}\!\!\!}+C}{e^{\footnotesize\begin{array}{l}\!\!\!\mathsf{2x-\frac{x^2}{2}}\end{array}\!\!\!}}}\\\\\\ \mathsf{y=\dfrac{-\,3e^{\footnotesize\begin{array}{l}\!\!\!\mathsf{2x-\frac{x^2}{2}}\end{array}\!\!\!}}{e^{\footnotesize\begin{array}{l}\!\!\!\mathsf{2x-\frac{x^2}{2}}\end{array}\!\!\!}}+\dfrac{C}{e^{\footnotesize\begin{array}{l}\!\!\!\mathsf{2x-\frac{x^2}{2}}\end{array}\!\!\!}}}\\\\\\ \mathsf{y=-\,3+Ce^{\footnotesize\begin{array}{l}\!\!\!\mathsf{-\!\left(2x-\frac{x^2}{2} \right )}\end{array}\!\!\!}}\\\\\\ \mathsf{y=-\,3+Ce^{\footnotesize\begin{array}{l}\!\!\!\mathsf{\frac{x^2}{2}-2x}\end{array}\!\!\!}}$


Para encontrar a constante  C, aplicamos o valor inicial:

y = – 2  para  x = 2:

$\mathsf{y(2)=-\,3+Ce^{\footnotesize\begin{array}{l}\!\!\!\mathsf{\frac{2^2}{2}-2\cdot 2}\end{array}\!\!\!}}\\\\ \mathsf{-2=-\,3+Ce^{2-4}}\\\\ \mathsf{-2=-\,3+Ce^{-2}}\\\\ \mathsf{Ce^{-2}=-2+3}\\\\ \mathsf{Ce^{-2}=1}\\\\ \mathsf{C=e^2}$


Portanto, a função em questão é

$\mathsf{y=-\,3+e^2\cdot e^{\footnotesize\begin{array}{l}\!\!\!\mathsf{\frac{x^2}{2}-2x}\end{array}\!\!\!}}$

$\mathsf{y=-\,3+e^{\footnotesize\begin{array}{l}\!\!\!\mathsf{\frac{x^2}{2}-2x+2}\end{array}\!\!\!}}$          ✔


O valor que a função assume para  x = 1  é

$\mathsf{y(1)=-\,3+e^{\footnotesize\begin{array}{l}\!\!\!\mathsf{\frac{1^2}{2}-2\cdot 1+2}\end{array}\!\!\!}}\\\\ \mathsf{y(1)=-\,3+e^{\footnotesize\begin{array}{l}\!\!\!\mathsf{\frac{1}{2}-2+2}\end{array}\!\!\!}}\\\\ \mathsf{y(1)=-\,3+e^{\footnotesize\begin{array}{l}\!\!\!\mathsf{\frac{1}{2}}\end{array}\!\!\!}}\\\\ \mathsf{y(1)=-\,3+\sqrt{e}}\\\\ \mathsf{y(1)\approx -\,3+1,\!6487\ldots}$


$\mathsf{y(1)\approx -1,\!3512\ldots}$          ✔


Resposta:  terceira alternativa:  – 1,35.


Bons estudos! :-)