JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The given differential equation can be rewritten as: dy dx + ( x)y = ^3 x ( x)^ 1/2 This is a linear differential equation of the form dy dx + P(x)y = Q(x). Integrating Factor (IF) = e^ x \, dx = e^ ( x) = x Multiplying both sides by the IF and integrating, we get: y x = ^4 x ( x)^ 1/2 \, dx + C y x = (1 + ^2 x) ^2 x ( x)^ 1/2 \, dx + C Let x = t, then ^2 x \, dx = dt. y x = 1 + t^2 t^ 1/2 \, dt + C y x = (t^ -1/2 + t^ 3/2 ) \, dt + C y x = 2t^ 1/2 + 2 5 t^ 5/2 + C y x = 2( x)^ 1/2 + 2 5 ( x)^ 5/2 + C Given y ( 4 ) = 6 2 5 , substituting x = 4 : 6 2 5 2 = 2(1)^ 1/2 + 2 5 (1)^ 5/2 + C 12 5 = 2 + 2 5 + C C = 0 The solution is y x = 2( x)^ 1/2 + 2 5 ( x)^ 5/2 . Substituting x = 3 : y ( 3 ) 2 = 2( 3 )^ 1/2 + 2 5 ( 3 )^ 5/2 y ( 3 ) = 3^ 1/4 + 1 5 3^ 5/4 = 3^ 1/4 (1 + 3 5 ) = 8 5 3^ 1/4 Given y ( 3 ) = 4 5 , we have: 4 5 = 8 5 3^ 1/4 = 2 3^ 1/4 Therefore, ^4 = (2 3^ 1/4 )^4 = 16 3 = 48 Answer: 48