Class 12 Differential Equations · Very Long Answer5 Markshomogeneous_differential_equations

Class 12 Maths · Differential Equations · Very Long Answer #114

For (x² + y²)dy = xy dx, y(1) = 1, find all real x₀ on the maximal solution with y(x₀) = e.

✅ Step-by-Step Solution

Final Answer: x₀ = ±√3 e.

Treat u = x² as dependent on y > 0: u′ = 2u/y + 2y.
Using IF y^(−2), integrate (u/y²)′ = 2/y to get x²/y² = 2ln y + C.
At (1, 1), C = 1.
At y = e, x₀²/e² = 3. The positive solution extends smoothly through x = 0 with y = e^(−1/2), so both signs belong to its maximal continuation.
Thus x₀ = ±√3 e. With an additional x₀ > 0 condition, only √3 e remains.

[1 mark] Treat u = x² as dependent on y > 0: u′ = 2u/y + 2y.
[1 mark] Using IF y^(−2), integrate (u/y²)′ = 2/y to get x²/y² = 2ln y + C.
[1 mark] At (1, 1), C = 1.
[1 mark] At y = e, x₀²/e² = 3. The positive solution extends smoothly through x = 0 with y = e^(−1/2), so both signs belong to its maximal continuation.
[1 mark] Thus x₀ = ±√3 e. With an additional x₀ > 0 condition, only √3 e remains.

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