Class 12 Differential Equations · Long Answer3 Marksseparable_differential_equations

Class 12 Maths · Differential Equations · Long Answer #102

Solve (x² − yx²)dy + (y² + x²y²)dx = 0, y(1) = 1.

✅ Step-by-Step Solution

Final Answer: ln y + 1/y = x − 1/x + 1, as an implicit integral curve; no differentiable y(x) solves the IVP through (1, 1).

For xy ≠ 0 separate (1 − y)dy/y² = −(1 + x²)dx/x².
Integrate −1/y − ln|y| = −x + 1/x + C, hence ln|y| + 1/y = x − 1/x + C₁.
The point (1, 1) gives C₁ = 1. But in the original graph equation at this point, x²(1 − y)y′ + y²(1 + x²) = 0 becomes 0·y′ + 2 = 0.
The displayed implicit curve passes through (1, 1) with vertical tangent; there is no differentiable solution graph y(x) through it. As x(y), the relation is valid locally.

[0.5 marks] For xy ≠ 0 separate (1 − y)dy/y² = −(1 + x²)dx/x².
[1 mark] Integrate −1/y − ln|y| = −x + 1/x + C, hence ln|y| + 1/y = x − 1/x + C₁.
[1 mark] The point (1, 1) gives C₁ = 1. But in the original graph equation at this point, x²(1 − y)y′ + y²(1 + x²) = 0 becomes 0·y′ + 2 = 0.
[0.5 marks] The displayed implicit curve passes through (1, 1) with vertical tangent; there is no differentiable solution graph y(x) through it. As x(y), the relation is valid locally.

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