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 que y(1) vale, aproximadamente:

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\(\[\begin{align} & Temos: \\ & ~\frac{dy}{dx}=~\text{ }3x\text{ }\text{ }2y\text{ }\text{ }6\text{ }+\text{ }xy~ \\ & \text{ }y\left( 2 \right)\text{ }=\text{ }\text{ }2. \\ & Logo: \\ & \frac{dy}{dx}+\text{ }2y\text{ }\text{ }xy\text{ }=~\text{ }3x\text{ }\text{ }6 \\ & \frac{dy}{dx}+\text{ }\left( 2\text{ }\text{ }x \right)\text{ }\cdot \text{ }y\text{ }=~\text{ }3x\text{ }\text{ }6 \\ & \frac{dy}{dx}~\text{ }+\text{ }\left( 2\text{ }\text{ }x \right)\text{ }\cdot \text{ }y\text{ }=~\text{ }\text{ }6\text{ }+\text{ }3x \\ & \frac{dy}{dx}+\text{ }\left( 2\text{ }\text{ }x \right)\text{ }\cdot \text{ }y\text{ }=~\text{ }\text{ }3\text{ }\cdot \text{ }\left( 2\text{ }\text{ }x \right)~~~~~~~~~\text{ }\left( i \right) \\ \end{align}\] \)

\(\[\begin{align} & Equacao(i) \\ & dy/dx+~\text{ }p\left( x \right)\text{ }\cdot \text{ }y~\text{ }=\text{ }q\left( x \right) \\ & p\left( x \right)\text{ }=\text{ }2\text{ }\text{ }x~\text{ }e~~\text{ }q\left( x \right)\text{ }=\text{ }\text{ }3\text{ }\cdot \text{ }\left( 2\text{ }\text{ }x \right). \\ & \text{ Fator integrante:} \\ & \mu (x)={{e}^{\int{}p(x)dx}}\mu (x)={{e}^{\int{}(2-x)dx}}\mu (x)={{e}^{\begin{array}{*{35}{l}} 2x-\frac{{{x}^{2}}}{2} \\ \end{array}}} \\ & \text{Multiplicando os dois lados:} \\ & {{e}^{\begin{array}{*{35}{l}} 2x-\frac{{{x}^{2}}}{2} \\ \end{array}}}\cdot \left[ \frac{dy}{dx}+(2-x)\cdot y \right]={{e}^{\begin{array}{*{35}{l}} 2x-\frac{{{x}^{2}}}{2} \\ \end{array}}}\cdot (-3)\cdot (2-x){{e}^{\begin{array}{*{35}{l}} 2x-\frac{{{x}^{2}}}{2} \\ \end{array}}}\cdot \frac{dy}{dx}+\left[ {{e}^{\begin{array}{*{35}{l}} 2x-\frac{{{x}^{2}}}{2} \\ \end{array}}}\cdot (2-x) \right]\cdot y={{e}^{\begin{array}{*{35}{l}} 2x-\frac{{{x}^{2}}}{2} \\ \end{array}}}\cdot (-3)\cdot (2-x) \\ & \text{Lado esquerdo:} \\ & \frac{d}{dx}\left( {{e}^{\begin{array}{*{35}{l}} 2x-\frac{{{x}^{2}}}{2} \\ \end{array}}}\cdot y \right)=-3\cdot {{e}^{\begin{array}{*{35}{l}} 2x-\frac{{{x}^{2}}}{2} \\ \end{array}}}\cdot (2-x) \\ & \int{}\frac{d}{dx}\left( {{e}^{\begin{array}{*{35}{l}} 2x-\frac{{{x}^{2}}}{2} \\ \end{array}}}\cdot y \right)dx=-3\int{}{{e}^{\begin{array}{*{35}{l}} 2x-\frac{{{x}^{2}}}{2} \\ \end{array}}}\cdot (2-x)dx{{e}^{\begin{array}{*{35}{l}} 2x-\frac{{{x}^{2}}}{2} \\ \end{array}}}\cdot y=-3\int{}{{e}^{u}}du\qquad \\ & onde~~u=2x-\frac{{{x}^{2}}}{2}{{e}^{\begin{array}{*{35}{l}} 2x-\frac{{{x}^{2}}}{2} \\ \end{array}}}\cdot y=-3{{e}^{u}}+C{{e}^{\begin{array}{*{35}{l}} 2x-\frac{{{x}^{2}}}{2} \\ \end{array}}}\cdot y=-3{{e}^{\begin{array}{*{35}{l}} 2x-\frac{{{x}^{2}}}{2} \\ \end{array}}}+C \\ & \\ \end{align}\] \)

\(\[\begin{align} & y=\frac{-3{{e}^{\begin{array}{*{35}{l}} 2x-\frac{{{x}^{2}}}{2} \\ \end{array}}}+C}{{{e}^{\begin{array}{*{35}{l}} 2x-\frac{{{x}^{2}}}{2} \\ \end{array}}}}y=\frac{-3{{e}^{\begin{array}{*{35}{l}} 2x-\frac{{{x}^{2}}}{2} \\ \end{array}}}}{{{e}^{\begin{array}{*{35}{l}} 2x-\frac{{{x}^{2}}}{2} \\ \end{array}}}}+\frac{C}{{{e}^{\begin{array}{*{35}{l}} 2x-\frac{{{x}^{2}}}{2} \\ \end{array}}}}y=-3+C{{e}^{\begin{array}{*{35}{l}} -\left( 2x-\frac{{{x}^{2}}}{2} \right) \\ \end{array}}}y=-3+C{{e}^{\begin{array}{*{35}{l}} \frac{{{x}^{2}}}{2}-2x \\ \end{array}}} \\ & \text{Para encontrar C:} \\ & y\text{ }=\text{ }\text{ }2~\text{ }para~\text{ }x\text{ }=\text{ }2: \\ & y(2)=-3+C{{e}^{\begin{array}{*{35}{l}} \frac{{{2}^{2}}}{2}-2\cdot 2 \\ \end{array}}}-2=-3+C{{e}^{2-4}}-2=-3+C{{e}^{-2}}C{{e}^{-2}}=-2+3C{{e}^{-2}}=1C={{e}^{2}} \\ & \text{funcao:} \\ & y=-3+{{e}^{2}}\cdot {{e}^{\begin{array}{*{35}{l}} \frac{{{x}^{2}}}{2}-2x \\ \end{array}}}y=-3+{{e}^{\begin{array}{*{35}{l}} \frac{{{x}^{2}}}{2}-2x+2 \\ \end{array}}} \\ & Logo: \\ & y(1)=-3+{{e}^{\text{ }\!\!\backslash\!\!\text{ }\begin{array}{*{35}{l}} \frac{{{1}^{2}}}{2}-2\cdot 1+2 \\ \end{array}}}y(1)=-3+{{e}^{\begin{array}{*{35}{l}} \frac{1}{2}-2+2 \\ \end{array}}}y(1)=-3+{{e}^{\begin{array}{*{35}{l}} \frac{1}{2} \\ \end{array}}}y(1)=-3+\sqrt{e}y(1)\approx -3+1,6487\ldots \\ & y(1)=-1,3512\ldots ? \\ \end{align}\] \)