Question

Alguém resolve essa edo : dy/dx= x-y

Answer 1

Notemos que esta é uma EDO de primeira ordem, linear, não-homogênea, e pode ser escrita na forma

$\mathsf{\dfrac{dy}{dx}+p(x)\,y=q(x)~~~~~~\mathbf{(i)}}$

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$\mathsf{\dfrac{dy}{dx}=x-y}\\\\\\ \mathsf{\dfrac{dy}{dx}+y=x}\\\\\\ \mathsf{\dfrac{dy}{dx}+1y=x~~~~~~\mathbf{(ii)}}$


Comparando $\mathbf{(ii)}$ com $\mathbf{(i)},$ tiramos que

$\mathsf{p(x)=1~~e~~q(x)=x}$

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Fator integrante:

$\mathsf{\mu(x)=e^{\int{\!p(x)\,dx}}}\\\\ \mathsf{\mu(x)=e^{\int{1\,dx}}}\\\\ \mathsf{\mu(x)=e^{x}}$

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Multiplicando os dois lados da equação $\mathbf{(ii)}$ por $\mathsf{\mu(x)=e^{x}},$ obtemos

$\mathsf{\mu(x)\cdot \left( \dfrac{dy}{dx}+1y\right )=\mu(x)\cdot x}\\\\\\ \mathsf{e^{x}\cdot \left( \dfrac{dy}{dx}+1y\right )=e^{x}\cdot x}\\\\\\ \mathsf{\dfrac{dy}{dx}\,e^{x}+ye^{x}=xe^{x}}$


Enxergamos o lado esquerdo como a derivada de um produto:

$\mathsf{\dfrac{d}{dx}(ye^{x})=xe^{x}}$


Integrando os dois lados em x, obtemos

$\mathsf{\displaystyle\int\dfrac{d}{dx}(ye^{x})\,dx=\int xe^{x}\,dx}\\\\\\ \mathsf{ye^{x}=\underbrace{\mathsf{\int xe^{x}\,dx}}_{I_{1}}~~~~~~\mathbf{(iii)}}$


Resolvemos $\mathsf{I_{1}}$ pelo método de integração por partes:

$\begin{array}{lcl} \mathsf{u=x}&~\Rightarrow~&\mathsf{du=dx}\\\\ \mathsf{dv=e^{x}\,dx}&~\Leftarrow~&\mathsf{v=e^{x}}\\\\ \end{array}\\\\\\ \displaystyle\mathsf{\int{u\,dv}=uv-\int{v\,du}}\\\\\\ \mathsf{\int{xe^{x}\,dx}=xe^{x}-\int{e^{x}\,dx}}\\\\\\ \mathsf{\int{xe^{x}\,dx}=xe^{x}-e^{x}+C}\\\\\\ \mathsf{\int{xe^{x}\,dx}=(x-1)\,e^{x}+C}\\\\\\ \mathsf{\therefore~~I_{1}=(x-1)\,e^{x}+C}$


Substituindo em $\mathbf{(iii)},$ temos

$\mathsf{ye^{x}=(x-1)\,e^{x}+C}\\\\ \mathsf{y=\dfrac{1}{e^{x}}\left[(x-1)\,e^{x}+C \right]}\\\\\\ \mathsf{y=(x-1)+\dfrac{C}{e^{x}} }\\\\\\ \boxed{\begin{array}{c} \mathsf{y=x-1+C\cdot e^{-x}} \end{array}}\mathsf{\,,~~~C\in\mathbb{R}.}$