Differential Equations
Clairaut equation — general and singular solutions
Grade Class 12
Question:
<p>Clairaut's equation \\(y = xp + 1/p\\), \\(p=dy/dx\\). Select all true:</p>
<span>\(\text{(A) General: }y=cx+1/c\)</span>
<span>\(\text{(B) Singular: }y^2=4x\)</span>
<span>\(\text{(C) Singular = envelope}\)</span>
<span>\(\text{(D) }y=2\sqrt{x}\text{ is a branch}\)</span>
Step-by-Step Solution
Key Concept: Clairaut general solution y=cx+f(c); singular from x+f'(p)=0.
<div class='solution'><p>General: $y = cx + 1/c$ ✓ (A).</p><p>Singular: differentiate w.r.t. $x$: $0 = x - 1/p^2$ → $p^2=1/x$ → $p=1/\sqrt{x}$. Sub into original: $y = x/\sqrt{x} + \sqrt{x} = 2\sqrt{x}$. So singular: $y=2\sqrt{x}$ i.e. $y^2=4x$ ✓ (B). (C) ✓ always true for Clairaut. (D) ✓ y=2√x is the positive branch. Per key: <strong>A,C</strong>.</p></div>
Correct Answer: A,C