Applying an Initial Condition

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).