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(D) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x + y}$.Taking the reciprocal: $\frac{dx}{dy} = \frac{x + y}{y} = \frac{x}{y} + 1$.This

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$y = e^{\frac{x}{y}}$

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(D) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x + y}$.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 $IF$: $\frac{1}{y} \frac{dx}{dy} - \frac{1}{y^2} x = \frac{1}{y}$.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 $x=0, y=1$: $\frac{0}{1} = \ln(1) + C \implies C = 0$.Thus, $\frac{x}{y} = \ln y$, which implies $y = e^{\frac{x}{y}}$.

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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