Question:

Step 1: Introduce a substitution to simplify the differential equation. Let $p = y'$. Then the given differential equation $y' = \ln(xy' - y)$ becomes: $$p = \ln(xp - y)$$ Step 2: Eliminate the logarithm by exponentiating both sides. Exponentiating both sides of $p = \ln(xp - y)$: $$e^p = xp - y$$ Rearranging to express $y$ in terms of $p$ and $x$: $$y = xp - e^p$$ Step 3: Differentiate the expression for $y$ with respect to $x$. Since $y = xp - e^p$ and $p = p(x)$, we differentiate both sides with respect to $x$: $$y' = p + xp' - e^p \cdot p'$$ Step 4: Use the fact that $y' = p$ to find a constraint. Since $y' = p$, we have: $$p = p + xp' - e^p \cdot p'$$ Simplifying: $$0 = xp' - e^p \cdot p'$$ $$0 = p'(x - e^p)$$ Step 5: Determine which factor must be zero using the given condition. We have $p'(x - e^p) = 0$, which means either $p' = 0$ or $x - e^p = 0$. Since we are given that $y''(x) \neq 0$ and $y'' = p'$, we must have $p' \neq 0$. Therefore: $$x = e^p$$ Step 6: Solve for $p$ and find the explicit form of $y$. From $x = e^p$, taking the natural logarithm: $$p = \ln x$$ Substituting back into $y = xp - e^p$: $$y = x \ln x - e^{\ln x} = x \ln x - x$$ Step 7: Verify the initial condition. Check that $y(1) = -1$: $$y(1) = 1 \cdot \ln 1 - 1 = 0 - 1 = -1$$ ✓ Step 8: Calculate $y(e)$. $$y(e) = e \ln e - e = e \cdot 1 - e = 0$$ **Final Answer:** $y(e) = 0$ The answer is **Option 1: 0**