Question:

<div class='solution'><p><strong>Step 1:</strong> Tangent at $(x_0,y_0)$: $Y - y_0 = y'(X - x_0)$. At $Y=0$: $X = x_0 - y_0/y'$.</p> <p>Given $X = y_0$: $x_0 - y_0/y' = y_0 \implies y' = \dfrac{y_0}{x_0 - y_0}$, i.e. $\dfrac{dy}{dx} = \dfrac{y}{x-y}$.</p> <p><strong>Step 2:</strong> Homogeneous ODE. Let $y = vx$: $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$.</p> <p>$v + xv' = \dfrac{vx}{x-vx} = \dfrac{v}{1-v} \implies xv' = \dfrac{v^2}{1-v}$.</p> <p><strong>Step 3:</strong> Separate: $\dfrac{1-v}{v^2}\,dv = \dfrac{dx}{x}$.</p> <p>$-\dfrac{1}{v} - \ln v = \ln x + C \implies -\dfrac{x}{y} - \ln\dfrac{y}{x} = \ln x + C$.</p> <p>Simplify: $-x/y = \ln y + C$.</p> <p><strong>Step 4:</strong> At $(1,1)$: $-1 = 0 + C \implies C = -1$.</p> <p>$-x/y = \ln y - 1 \implies 1 - x/y = \ln y \implies y = e^{1-x/y} \implies ye^{x/y} = e$.</p> <p><strong>Answer: (A)</strong> $ye^{x/y} = e$.</p> <p class='key-concept'>🔑 Key Concept: Set up the tangent equation first, find x-intercept algebraically, then form the ODE. This is a standard geometric ODE setup.</p> <p class='trap-warning'>⚠️ Trap: Setting x-intercept = x (the abscissa) instead of = y (the ordinate). Read the problem statement very carefully.</p></div>