Class 12 Differential Equations · Very Long Answer5 Marksseparable_differential_equations
Class 12 Maths · Differential Equations · Very Long Answer #110
Solve xy y′ = (x + 2)(y + 2), y(1) = −1.
✅ Step-by-Step Solution
Final Answer: y − 2ln|y + 2| = x + 2ln x − 2, on the branch −2 < y < 0 through (1, −1).
- Near the initial point, divide by x(y + 2): y dy/(y + 2) = (1 + 2/x)dx.
- Use y/(y + 2) = 1 − 2/(y + 2).
- Integrate y − 2ln|y + 2| = x + 2ln|x| + C.
- At (1, −1), −1 = 1 + C, so C = −2.
- Thus y − 2ln(y + 2) = x + 2ln x − 2 on the connected branch through (1, −1), with x > 0 and −2 < y < 0. The branch ends before y reaches 0, where the original graph equation cannot hold.
📋 CBSE Stepwise Marking Scheme
- [1 mark] Near the initial point, divide by x(y + 2): y dy/(y + 2) = (1 + 2/x)dx.
- [1 mark] Use y/(y + 2) = 1 − 2/(y + 2).
- [1 mark] Integrate y − 2ln|y + 2| = x + 2ln|x| + C.
- [1 mark] At (1, −1), −1 = 1 + C, so C = −2.
- [1 mark] Thus y − 2ln(y + 2) = x + 2ln x − 2 on the connected branch through (1, −1), with x > 0 and −2 < y < 0. The branch ends before y reaches 0, where the original graph equation cannot hold.
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