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(C) The differential equation is given by $\frac{dy}{dx} = \frac{xy^2 + y}{x} = y^2 + \frac{y}{x}$.Rearranging the terms,we get $\frac{dy}{dx} - \frac

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$-\frac{18}{19}$

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(C) The differential equation is given by $\frac{dy}{dx} = \frac{xy^2 + y}{x} = y^2 + \frac{y}{x}$.Rearranging the terms,we get $\frac{dy}{dx} - \frac{y}{x} = y^2$.Dividing by $y^2$,we have $y^{-2} \frac{dy}{dx} - \frac{1}{x} y^{-1} = 1$.Let $v = y^{-1}$,then $\frac{dv}{dx} = -y^{-2} \frac{dy}{dx}$,which implies $y^{-2} \frac{dy}{dx} = -\frac{dv}{dx}$.Substituting this into the equation,we get $-\frac{dv}{dx} - \frac{1}{x} v = 1$,or $\frac{dv}{dx} + \frac{1}{x} v = -1$.This is a linear differential equation with integrating factor $IF = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$.The solution is $v \cdot x = \int (-1) \cdot x dx + C = -\frac{x^2}{2} + C$.Since $v = \frac{1}{y}$,we have $\frac{x}{y} = -\frac{x^2}{2} + C$.Given the curve intersects $x + 2y = 4$ at $x = -2$,we find $y$ by substituting $x = -2$: $-2 + 2y = 4 \Rightarrow 2y = 6 \Rightarrow y = 3$.The curve passes through $(-2, 3)$,so $\frac{-2}{3} = -\frac{(-2)^2}{2} + C \Rightarrow -\frac{2}{3} = -2 + C \Rightarrow C = 2 - \frac{2}{3} = \frac{4}{3}$.Thus,$\frac{x}{y} = -\frac{x^2}{2} + \frac{4}{3}$.For the point $(3, y)$,we have $\frac{3}{y} = -\frac{3^2}{2} + \frac{4}{3} = -\frac{9}{2} + \frac{4}{3} = \frac{-27 + 8}{6} = -\frac{19}{6}$.Therefore,$y = 3 \cdot (-\frac{6}{19}) = -\frac{18}{19}$.

Let the slope of the tangent line to a curve at any point $P(x, y)$ be given by $\frac{xy^2 + y}{x}$. If the curve intersects the line $x + 2y = 4$ at $x = -2$,then the value of $y$,for which the point $(3, y)$ lies on the curve,is ..... .

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