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(A) Given the slope of the tangent $\frac{dy}{dx} = \frac{y}{x + y}$.Taking the reciprocal, $\frac{dx}{dy} = \frac{x + y}{y} = \frac{x}{y} + 1$.This i

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$x = y \ln(y)$

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(A) Given the slope of the tangent $\frac{dy}{dx} = \frac{y}{x + y}$.Taking the reciprocal, $\frac{dx}{dy} = \frac{x + y}{y} = \frac{x}{y} + 1$.This is a linear differential equation of the form $\frac{dx}{dy} - \frac{1}{y}x = 1$.The integrating factor $IF = e^{\int -\frac{1}{y} dy} = e^{-\ln y} = \frac{1}{y}$.Multiplying by $IF$, we get $\frac{1}{y} \frac{dx}{dy} - \frac{1}{y^2} x = \frac{1}{y}$.Integrating both sides with respect to $y$, $\frac{x}{y} = \int \frac{1}{y} dy = \ln|y| + C$.Since the curve passes through $(0, 1)$, substitute $x=0$ and $y=1$: $\frac{0}{1} = \ln(1) + C \implies 0 = 0 + C \implies C = 0$.Thus, $\frac{x}{y} = \ln y$, which gives $x = y \ln y$.

The equation of the curve passing through the point $(0, 1)$ and having a slope of the tangent at any point $(x, y)$ equal to $\frac{y}{x + y}$ is:

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