Question 74Asked in BPSC TRE 1.0 (2023) Mathematics Class 11-12

The general solution of the differential equation sec²y dy/dx + 2x tan y = x³ isअवकल समीकरण sec²y dy/dx + 2x tan y = x³ का व्यापक हल है

अवकल समीकरण sec²y dy/dx + 2x tan y = x³ का व्यापक हल है
  1. Atan y = ce^(−x²) + (1/2)(x² + 1)tan y = ce^(−x²) + (1/2)(x² + 1)
  2. Btan y = ce^(−x²) + (1/2)(x² − 1)tan y = ce^(−x²) + (1/2)(x² − 1)
  3. Ctan y = ce^(x²) + (1/2)(x² − 1)tan y = ce^(x²) + (1/2)(x² − 1)
  4. DMore than one of the aboveउपर्युक्त में से एक से अधिक
  5. ENone of the aboveउपर्युक्त में से कोई नहीं
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Correct answer: (B) tan y = ce^(−x²) + (1/2)(x² − 1)tan y = ce^(−x²) + (1/2)(x² − 1)

Explanation

Put v = tan y, so dv/dx + 2xv = x³. The integrating factor is e^(x²), giving v·e^(x²) = ∫x³e^(x²)dx = (1/2)e^(x²)(x² − 1) + c. Hence tan y = (1/2)(x² − 1) + ce^(−x²).v = tan y रखें, तब dv/dx + 2xv = x³। समाकलन गुणक e^(x²) है, जिससे v·e^(x²) = ∫x³e^(x²)dx = (1/2)e^(x²)(x² − 1) + c। अतः tan y = (1/2)(x² − 1) + ce^(−x²)।

From BPSC TRE 1.0 (2023) Mathematics Class 11-12 — Previous Year Paper

This is question 74 of BPSC TRE 1.0 (2023) Mathematics Class 11-12 — Previous Year Paper. See all questions of this paper with answers, or attempt the paper as a timed online test to get your score and all-India rank.