JEE Advanced Mathematics Question (2026) — Solution

Step-by-step solution

The given differential equation is dy dx = e^ 5x y^3 + y^3 e^x + e^x y^4 Factoring the numerator and denominator, we get dy dx = y^3 (e^ 5x + 1) e^x (1 + y^4) Separating the variables x and y, we obtain 1 + y^4 y^3 dy = e^ 5x + 1 e^x dx ( 1 y^3 + y ) dy = (e^ 4x + e^ -x ) dx Integrating both sides yields ( y^ -3 + y ) dy = (e^ 4x + e^ -x ) dx - 1 2y^2 + y^2 2 = e^ 4x 4 - e^ -x + C Multiplying by 2, we get y^2 - 1 y^2 = e^ 4x 2 - 2e^ -x + 2C Using the initial condition y(0) = 1 2 , we substitute x = 0 and y = 1 2 : ( 1 2 )^2 - 1 ( 1 2 )^2 = e^0 2 - 2e^0 + 2C 1 2 - 2 = 1 2 - 2 + 2C This gives 2C = 0, so C = 0. The equation becomes y^2 - 1 y^2 = e^ 4x 2 - 2e^ -x To find y( _e 2), we substitute x = _e 2 into the equation. We have e^ 4x = e^ 4 _e 2 = 16 and e^ -x = e^ - _e 2 = 1 2 . y^2 - 1 y^2 = 16 2 - 2 ( 1 2 ) y^2 - 1 y^2 = 8 - 1 = 7 Let y^2 = t. Since y (0, ), t > 0. t - 1 t = 7 t^2 - 7t - 1 = 0 Solving for t using the quadratic formula gives t = 7 49 - 4(1)(-1) 2 = 7 53 2 Since t = y^2 > 0, we must choose the positive root: y^2 = 7 + 53 2 Taking the positive square root since y > 0, we get y = 7 + 53 2