Class 12 Differential Equations · Long Answer3 Marksfirst_order_differential_equations

Class 12 Maths · Differential Equations · Long Answer #67

Find the particular solution of (1 − y²)(1 + ln x)dx + 2xy dy = 0 with y(1) = 0.

✅ Step-by-Step Solution

Final Answer: No differentiable solution graph y(x) satisfies the printed IVP.

For a differentiable graph, the original equation is (1 − y²)(1 + ln x) + 2xyy′ = 0, x > 0.
At x = 1 and y = 0, its left-hand side is 1 + 0·y′(1) = 1, which cannot equal zero for any finite derivative.
Formally integrating away from y = 0 gives ln|1 − y²| = ln x + ½(ln x)² + C; the condition would give C = 0. The resulting implicit curve has a vertical tangent at (1, 0).
Therefore the implicit relation is not a differentiable solution graph through the prescribed point; the printed IVP has no classical solution.

[0.5 marks] For a differentiable graph, the original equation is (1 − y²)(1 + ln x) + 2xyy′ = 0, x > 0.
[1 mark] At x = 1 and y = 0, its left-hand side is 1 + 0·y′(1) = 1, which cannot equal zero for any finite derivative.
[1 mark] Formally integrating away from y = 0 gives ln|1 − y²| = ln x + ½(ln x)² + C; the condition would give C = 0. The resulting implicit curve has a vertical tangent at (1, 0).
[0.5 marks] Therefore the implicit relation is not a differentiable solution graph through the prescribed point; the printed IVP has no classical solution.

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