Let y=y(x) be the solution of the differential equation: dy dx + ( 6x^2+(3x^2+2x^3+4)e^ -2x (x^3+2)(2+e^ -2x ) )y=2+e^ -2x , x (-1,2) , satisfying y(0)= 3 2 . If y(1)= (2+e⁻²) , then is equal to:

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) (I