Verified Solution • 2025–26 Session moderate

QJEEM02Ma2b6f: JEE Main 2002 Mathematics - Class 12 JEEM02Ma2b6f Solution: The order and degree of the differential equation are

The order and degree of the differential equation
 (1+3dydx)2/3=4d3ydx3\,{\left( {1 + 3{{dy} \over {dx}}} \right)^{2/3}} = 4{{{d^3}y} \over {d{x^3}}} are

Multiple Choice Options

  • A (1,23)\left( {1,{2 \over 3}} \right)
  • B (3,1)(3, 1)
  • C (3,3)(3,3)
  • D (1,2)(1,2)

Step-by-Step Verified Solution

(1+3d yd x)2={\left( {1 + 3{{d\,y} \over {d\,x}}} \right)^2} = (4)3 (d3yd x3)3{\left( {{{{d^3}y} \over {d\,{x^3}}}} \right)^3}

⇒(1+3d yd x)2=16(d3yd x3)3\Rightarrow {\left( {1 + 3{{d\,y} \over {d\,x}}} \right)^2} = 16{\left( {{{{d^3}y} \over {d\,{x^3}}}} \right)^3}

∴\therefore Order = 3 and Degree = 3

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