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(D) Given the differential equation $x dy + (y + y^2 x) dx = 0$. Dividing by $x dx$,we get $\frac{dy}{dx} + \frac{y}{x} + y^2 = 0$. This is a Bernoull

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$y = \frac{1}{x(1 + \log x)}$

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(D) Given the differential equation $x dy + (y + y^2 x) dx = 0$. Dividing by $x dx$,we get $\frac{dy}{dx} + \frac{y}{x} + y^2 = 0$. This is a Bernoulli differential equation. Divide by $y^2$: $y^{-2} \frac{dy}{dx} + \frac{1}{x} y^{-1} = -1$. Let $v = y^{-1}$,then $\frac{dv}{dx} = -y^{-2} \frac{dy}{dx}$,so $- \frac{dv}{dx} + \frac{v}{x} = -1$,which simplifies to $\frac{dv}{dx} - \frac{v}{x} = 1$. This is a linear differential equation with Integrating Factor $I.F. = e^{\int -\frac{1}{x} dx} = e^{-\log x} = \frac{1}{x}$. The solution is $v \cdot \frac{1}{x} = \int 1 \cdot \frac{1}{x} dx = \log x + C$. Substituting $v = \frac{1}{y}$,we get $\frac{1}{xy} = \log x + C$. Given $y = 1$ at $x = 1$,we have $\frac{1}{1 \cdot 1} = \log 1 + C \Rightarrow 1 = 0 + C \Rightarrow C = 1$. Thus,$\frac{1}{xy} = \log x + 1$,which gives $y = \frac{1}{x(1 + \log x)}$.

If $x dy + (y + y^2 x) dx = 0$ and $y = 1$ at $x = 1$,then

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