Question 9

**Solving the differential equation** The equation x² dy/dx + xy = x² + y² is homogeneous. Put y = vx, so dy/dx = v + x dv/dx: x²(v + x v') + x²v = x² + x²v² ⟹ x v' = 1 − 2v + v² = (1 − v)² Separating variables: dv/(1 − v)² = dx/x ⟹ 1/(1 − v) = ln x + C Using y(1) = 0 ⟹ v = 0 at x = 1, so C = 1: 1/(1 − v) = 1 + ln x ⟹ v = ln x/(1 + ln x) Hence y(x) = x ln x / (1 + ln x), valid for x > 1/e (where 1 + ln x > 0). **Evaluating** - y(e) = e · 1/2 = e/2 - y(e²) = e² · 2/3 = 2e²/3 Therefore 2·(y(e))² / y(e²) = 2·(e²/4) / (2e²/3) = (e²/2)·(3/(2e²)) = 3/4 = 0.75

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