Class 12 Differential Equations · Long Answer3 Markshomogeneous_differential_equations

Class 12 Maths · Differential Equations · Long Answer #64

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

✅ Step-by-Step Solution

Final Answer: x₀ = ±√3 e.

Treat x² as a function u of y > 0: du/dy = 2x dx/dy = 2u/y + 2y.
Then (u/y²)′ = 2/y, giving x²/y² = 2ln y + C. At (1, 1), C = 1.
For y = e, x₀²/e² = 2 + 1 = 3, so x₀ = ±√3 e. The positive solution branch extends through x = 0 with y(0) = e^(−1/2).
Both x₀ = √3 e and x₀ = −√3 e occur on the maximal solution. If an additional x₀ > 0 restriction is imposed, only √3 e remains.

[0.5 marks] Treat x² as a function u of y > 0: du/dy = 2x dx/dy = 2u/y + 2y.
[1 mark] Then (u/y²)′ = 2/y, giving x²/y² = 2ln y + C. At (1, 1), C = 1.
[1 mark] For y = e, x₀²/e² = 2 + 1 = 3, so x₀ = ±√3 e. The positive solution branch extends through x = 0 with y(0) = e^(−1/2).
[0.5 marks] Both x₀ = √3 e and x₀ = −√3 e occur on the maximal solution. If an additional x₀ > 0 restriction is imposed, only √3 e remains.

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