Given differential equation:
$$x(1 - x \ln y)\frac{dy}{dx} + y = 0$$
Multiply by $$\frac{dx}{dy}$$ to treat $$x$$ as the dependent variable:
$$x(1 - x \ln y) + y\frac{dx}{dy} = 0$$ $$x - x^2 \ln y + y\frac{dx}{dy} = 0$$
Rearrange terms:
$$y\frac{dx}{dy} + x = x^2 \ln y$$
Divide through by $$y$$:
$$\frac{dx}{dy} + \frac{1}{y}x = \frac{\ln y}{y}x^2$$
Divide the equation by $$x^2$$:
$$x^{-2}\frac{dx}{dy} + \frac{1}{y}x^{-1} = \frac{\ln y}{y}$$
Let $$v = x^{-1} = \frac{1}{x}$$. Differentiating with respect to $$y$$:
$$\frac{dv}{dy} = -x^{-2}\frac{dx}{dy} \implies x^{-2}\frac{dx}{dy} = -\frac{dv}{dy}$$
Substitute these into the equation:
$$-\frac{dv}{dy} + \frac{1}{y}v = \frac{\ln y}{y}$$
Multiply by $$-1$$ to get the standard linear form:
$$\frac{dv}{dy} - \frac{1}{y}v = -\frac{\ln y}{y}$$
$$P(y) = -\frac{1}{y}$$$$\text{I.F.} = e^{\int -\frac{1}{y} \, dy} = e^{-\ln y} = \frac{1}{y}$$
$$v \cdot (\text{I.F.}) = \int Q(y) \cdot (\text{I.F.}) \, dy + C$$ $$\frac{v}{y} = \int \left(-\frac{\ln y}{y}\right) \cdot \frac{1}{y} \, dy + C$$ $$\frac{v}{y} = -\int \frac{\ln y}{y^2} \, dy + C$$
Using integration by parts ($$\int u \, dw = uw - \int w \, du$$) for $$\int \frac{\ln y}{y^2} \, dy$$:
- $$u = \ln y \implies du = \frac{1}{y} \, dy$$
- $$dw = y^{-2} \, dy \implies w = -\frac{1}{y}$$
$$\int \frac{\ln y}{y^2} \, dy = -\frac{\ln y}{y} - \int \left(-\frac{1}{y}\right)\left(\frac{1}{y}\right) \, dy = -\frac{\ln y}{y} - \frac{1}{y} = -\frac{1}{y}(\ln y + 1)$$
Substitute this back:
$$\frac{v}{y} = -\left[-\frac{1}{y}(\ln y + 1)\right] + C$$ $$\frac{v}{y} = \frac{\ln y + 1}{y} + C$$
Multiply by $$y$$ and substitute $$v = \frac{1}{x}$$:
$$\frac{1}{x} = (\ln y + 1) + Cy$$
The curve passes through $$\left(e, \frac{1}{e}\right)$$, so substitute $$x = e$$ and $$y = \frac{1}{e}$$:
$$\frac{1}{e} = \left(\ln\left(\frac{1}{e}\right) + 1\right) + C\left(\frac{1}{e}\right)$$
Since $$\ln\left(\frac{1}{e}\right) = -1$$:
$$\frac{1}{e} = (-1 + 1) + \frac{C}{e}$$$$\frac{1}{e} = \frac{C}{e} \implies C = 1$$
Substitute $$C = 1$$ back into the solution:
$$\frac{1}{x} = (\ln y + 1) + y$$
Cross-multiplying by $$x$$:
$$\{(\ln y + 1) + y\}x = 1$$
This perfectly matches Option (d).
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