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Question #70167

Difficulty: 8/10 | 🟢 Free

The Question

Find the equation of the orthogonal trajectories of the family of curves x=k∣y∣aeb∣y∣x = k |y|^a e^{b|y|}, where kk is an arbitrary non-zero constant, and a,ba, b are positive constants.

Options

  • A.

    b2x2+2b∣y∣−2aln⁡(a+b∣y∣)=Cb^2 x^2 + 2b|y| - 2a \ln(a+b|y|) = C

  • B.

    b2x2+2by−2aln⁡∣a+by∣=Cb^2 x^2 + 2by - 2a \ln|a+by| = C

  • C.

    b2x2+2by−2asgn(y)ln⁡(a+b∣y∣)=Cb^2 x^2 + 2by - 2a \text{sgn}(y) \ln(a+b|y|) = C

  • D.

    b2x2+2b∣y∣+2aln⁡(a+b∣y∣)=Cb^2 x^2 + 2b|y| + 2a \ln(a+b|y|) = C

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Orthogonal Trajectories|Tangent/Normal Slope Interplay|Modulus Sign Dependency|The Modulus Trap

Why This Question Is Tricky

Examiner Levers

Variable Manipulation

Algebraic solutions only (No numbers); tests pure symbolic logic.

Difficulty Boost : +4

Common Mistakes

This question tests the student's ability to handle orthogonal trajectories in the presence of a modulus function, specifically targeting the 'Modulus Trap' during integration.

Students often blindly differentiate and integrate without respecting the piecewise nature of ∣y∣|y|, leading them to drop the modulus (Option B) or incorrectly extract the signum function as a constant (Option C).

The mathematically robust path requires recognizing that ydyy dy elegantly transforms into uduu du under the substitution u=∣y∣u = |y|, thereby maintaining sign consistency across the y=0y=0 boundary and correctly reconstructing the orthogonal family.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterDifferential Equations - Class 12
ConceptOrthogonal Trajectories
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree