JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The given differential equation is a linear differential equation of the form dy dx + P(x)y = Q(x), where P(x) = 6x^2 + (3x^2 + 2x^3 + 4)e^ -2x (x^3+2)(2+e^ -2x ) and Q(x) = 2 + e^ -2x . We can rewrite P(x) as: P(x) = 3x^2(2+e^ -2x ) + 2e^ -2x (x^3+2) (x^3+2)(2+e^ -2x ) = 3x^2 x^3+2 + 2e^ -2x 2+e^ -2x The integrating factor (IF) is given by: IF = e^ P(x) dx = e^ ( 3x^2 x^3+2 + 2e^ -2x 2+e^ -2x ) dx IF = e^ (x^3+2) - (2+e^ -2x ) = x^3+2 2+e^ -2x The general solution of the differential equation is: y (IF) = Q(x) (IF) dx + C Substituting the values, we get: y ( x^3+2 2+e^ -2x ) = (2+e^ -2x ) ( x^3+2 2+e^ -2x ) dx + C y ( x^3+2 2+e^ -2x ) = (x^3+2) dx + C y ( x^3+2 2+e^ -2x ) = x^4 4 + 2x + C Given y(0) = 3 2 , substituting x = 0: 3 2 ( 0+2 2+e^0 ) = 0 + 0 + C 3 2 ( 2 3 ) = C C = 1 Thus, the particular solution is: y ( x^3+2 2+e^ -2x ) = x^4 4 + 2x + 1 To find y(1), substitute x = 1: y(1) ( 1+2 2+e^ -2 ) = 1 4 + 2 + 1 y(1) ( 3 2+e^ -2 ) = 13 4 y(1) = 13 12 (2+e^ -2 ) Comparing this with y(1) = (2+e^ -2 ), we get: = 13 12