Let y=y(x) be the solution of the differential equation 4 x^ 3/4 (1+ x ) dy dx + y = x^ 1/4 e^ - ⁻¹(x^ 1/4 ) , x > 0 If y(1) = 1 2 2 e^ - /4 , then the value of y(16) e^ ⁻¹(2) is equal to :

Step-by-step solution

The given differential equation can be rewritten as: dy dx + 1 4 x^ 3/4 (1+ x ) y = x^ 1/4 e^ - ⁻¹(x^ 1/4 ) 4 x^ 3/4 (1+ x ) dy dx + 1 4 x^ 3/4 (1+ x ) y = e^ - ⁻¹(x^ 1/4 ) 4 x (1+ x ) This is a linear differential equation of the form dy dx + P(x)y = Q(x) . Integrating Factor (IF) = e^ P(x) dx = e^ 1 4 x^ 3/4 (1+ x ) dx Let x = t^4 dx = 4t^3 dt . P(x) dx = 4t^3 4t^3 (1+t^2) dt = 1 1+t^2 dt = ⁻¹(t) = ⁻¹(x^ 1/4 ) So, IF = e^ ⁻¹(x^ 1/4 ) Multiplying the differential equation by the IF and integrating: y e^ ⁻¹(x^ 1/4