Temos:

\(\[\begin{align} & \text{Reescrevendo:} \\ & y'(x)+p(x).y=q(x) \\ & p(x)=-1,q(x)=ex \\ & \frac{d}{dx}(y)-1.y=ex \\ & \text{Fator integrante:} \\ & \frac{d}{dx}\left( \text{ }\!\!\mu\!\!\text{ }\left( x \right) \right)=\text{ }\!\!\mu\!\!\text{ }\left( x \right)\left( -1 \right) \\ & \frac{\frac{d}{dx}\left( \text{ }\!\!\mu\!\!\text{ }\left( x \right) \right)}{\text{ }\!\!\mu\!\!\text{ }\left( x \right)}=\frac{\text{ }\!\!\mu\!\!\text{ }\left( x \right)\left( -1 \right)}{\text{ }\!\!\mu\!\!\text{ }\left( x \right)}\to \frac{\frac{d}{dx}\left( \text{ }\!\!\mu\!\!\text{ }\left( x \right) \right)}{\text{ }\!\!\mu\!\!\text{ }\left( x \right)}=-1 \\ \end{align}\] \)

\(\[\begin{align} & \frac{d}{dx}\left( \ln \left( \text{ }\!\!\mu\!\!\text{ }\left( x \right) \right) \right)=\frac{\frac{d}{dx}\left( \text{ }\!\!\mu\!\!\text{ }\left( x \right) \right)}{\text{ }\!\!\mu\!\!\text{ }\left( x \right)} \\ & \frac{d}{dx}\left( \ln \left( \text{ }\!\!\mu\!\!\text{ }\left( x \right) \right) \right)=-1 \\ & \text{Resolvendo:} \\ & \frac{d}{dx}\left( \ln \left( \text{ }\!\!\mu\!\!\text{ }\left( x \right) \right) \right)=-1:\quad \text{ }\!\!\mu\!\!\text{ }\left( x \right)={{e}^{-x+{{c}_{1}}}} \\ & \text{ }\!\!\mu\!\!\text{ }\left( x \right)={{e}^{-x+{{c}_{1}}}} \\ & \text{Multiplicando por }\!\!\mu\!\!\text{ }\left( x \right)={{e}^{-x}},temos: \\ & \frac{d}{dx}\left( {{e}^{-x}}y \right)={{e}^{1-x}}x \\ \end{align}\] \)

\(\[\begin{align} & y=\frac{-x{{e}^{1-x}}-{{e}^{1-x}}+{{c}_{1}}}{{{e}^{-x}}} \\ & {{e}^{-x}}y=-{{e}^{1-x}}x-{{e}^{1-x}}+{{c}_{1}} \\ & \frac{{{e}^{-x}}y}{{{e}^{-x}}}=-\frac{{{e}^{1-x}}x}{{{e}^{-x}}}-\frac{{{e}^{1-x}}}{{{e}^{-x}}}+\frac{{{c}_{1}}}{{{e}^{-x}}} \\ & \text{simplificando o resultado }\!\!\acute{\mathrm{e}}\!\!\text{ }: \\ & y=\frac{-x{{e}^{1-x}}-{{e}^{1-x}}+{{c}_{1}}}{{{e}^{-x}}} \\ & \\ \end{align}\] \)