Class 12 Differential Equations · Short Answer2 Marksseparable_differential_equations

Class 12 Maths · Differential Equations · Short Answer #42

Solve ln(y′) = ax + by, for real constants a and b.

✅ Step-by-Step Solution

Final Answer: If ab ≠ 0: e^(ax)/a + e^(−by)/b = C. If a = 0, b ≠ 0: −e^(−by)/b = x + C. If b = 0, a ≠ 0: y = e^(ax)/a + C. If a = b = 0: y = x + C.

Exponentiate to y′ = e^(ax+by) > 0 and separate e^(−by)dy = e^(ax)dx.
For ab ≠ 0, integrate to −e^(−by)/b = e^(ax)/a + C₁. If a or b vanishes, integrate the corresponding constant factor directly.
For ab ≠ 0, e^(ax)/a + e^(−by)/b = C. For a = 0: −e^(−by)/b = x + C; for b = 0: y = e^(ax)/a + C; for both zero: y = x + C. Use real branches with positive exponential arguments.

📋 CBSE Stepwise Marking Scheme

[0.5 marks] Exponentiate to y′ = e^(ax+by) > 0 and separate e^(−by)dy = e^(ax)dx.
[1 mark] For ab ≠ 0, integrate to −e^(−by)/b = e^(ax)/a + C₁. If a or b vanishes, integrate the corresponding constant factor directly.
[0.5 marks] For ab ≠ 0, e^(ax)/a + e^(−by)/b = C. For a = 0: −e^(−by)/b = x + C; for b = 0: y = e^(ax)/a + C; for both zero: y = x + C. Use real branches with positive exponential arguments.

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Class 12 Maths · Differential Equations · Short Answer #42

✅ Step-by-Step Solution

📋 CBSE Stepwise Marking Scheme

Master Chapter 9: Differential Equations

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