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The given equation is $y e^{y} dx = (y^{3} + 2x e^{y}) dy$. We can rewrite this as a linear differential equation in terms of $x$ by dividing by $dy$ and $y e^{y}$:
$$\frac{dx}{dy} = \frac{y^{3} + 2x e^{y}}{y e^{y}}$$ $$\frac{dx}{dy} = \frac{y^{2}}{e^{y}} + \frac{2x}{y}$$Rearranging the terms to fit the form $\frac{dx}{dy} + P(y)x = Q(y)$:
$$\frac{dx}{dy} - \left(\frac{2}{y}\right)x = y^{2} e^{-y}$$Here, $P(y) = -\frac{2}{y}$. The integrating factor is:
$$IF = e^{\int P(y) dy} = e^{\int -\frac{2}{y} dy} = e^{-2 \ln|y|} = e^{\ln(y^{-2})} = \frac{1}{y^{2}}$$Multiply the linear equation by the IF and integrate:
$$\frac{1}{y^{2}} \frac{dx}{dy} - \frac{2}{y^{3}} x = e^{-y}$$ $$\frac{d}{dy} \left( \frac{x}{y^{2}} \right) = e^{-y}$$ $$\frac{x}{y^{2}} = \int e^{-y} dy = -e^{-y} + C$$ $$x = y^{2} (C - e^{-y})$$Given $y(0) = 1$, substitute $x=0$ and $y=1$:
$$0 = 1^{2} (C - e^{-1}) \implies C = e^{-1} = \frac{1}{e}$$ $$x = y^{2} \left( \frac{1}{e} - e^{-y} \right)$$Final Answer: x = y^{2} (e^{-1} - e^{-y})
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