Differential Equations
Differential Equations
star_batch_jee_advanced_2025
Grade 12

Question:

The singular solution of the differential equation given in previous problem is :
y = -x + 1
y = x + 1
y + 1 = log x
y + 1 = -log(-x)

Step-by-Step Solution

Key Concept: Singular solutions arise from setting the coefficient of $dp/dx$ equal to zero in the differentiated equation.
Differentiating the given equation yields $\left(x + \frac{1}{p}\right)\frac{dp}{dx} = 0$. For a singular solution, $x + \frac{1}{p} = 0$, so $p = -\frac{1}{x}$. Substituting back into the original equation gives $y = -1 + \log(-x)$, or $y + 1 = -\log(-x)$.
Correct Answer: 4

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