Verified Solution • 2025–26 Session moderate

QJEEM22Ma2a7f: JEE Main 2022 Mathematics - Class 12 JEEM22Ma2a7f Solution: If , then :

If {\cos ^{ - 1}}\left( {{y \over 2}} \right) = {\log _e}{\left( {{x \over 5}} \right)^5},\,|y| < 2 , then :

Multiple Choice Options

  • A x2y′′+xy′−25y=0{x^2}y'' + xy' - 25y = 0
  • B x2y′′−xy′−25y=0{x^2}y'' - xy' - 25y = 0
  • C x2y′′−xy′+25y=0{x^2}y'' - xy' + 25y = 0
  • D x2y′′+xy′+25y=0{x^2}y'' + xy' + 25y = 0

Step-by-Step Verified Solution

{\cos ^{ - 1}}\left( {{y \over 2}} \right) = {\log _e}{\left( {{x \over 5}} \right)^5}\,\,\,\,\,\,\,\,\,|y| < 2

Differentiating on both side

−11−(y2)2×y′2=5x5×15- {1 \over {\sqrt {1 - {{\left( {{y \over 2}} \right)}^2}} }} \times {{y'} \over 2} = {5 \over {{x \over 5}}} \times {1 \over 5}

−xy′2=51−(y2)2{{ - xy'} \over 2} = 5\sqrt {1 - {{\left( {{y \over 2}} \right)}^2}}

Square on both side

x2y′24=25(4−y24){{{x^2}y{'^2}} \over 4} = 25\left( {{{4 - {y^2}} \over 4}} \right)

Diff on both side

2xy′2+2y′y′′x2=−25×2yy′2xy{'^2} + 2y'y''{x^2} = - 25 \times 2yy'

xy′+y′′x2+25y=0xy' + y''{x^2} + 25y = 0

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