Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation : dydx+(6x2+(3x2+2x3+4)e−2x(x3+2)(2+e−2x))y=2+e−2x\dfrac{dy}{dx}+ \left(\dfrac{6x^{2}+ (3x^{2}+2x^{3}+4)e^{-2x}}{(x^{3}+2)(2+e^{-2x})}\right) y = 2+e^{-2x}dxdy+((x3+2)(2+e−2x)6x2+(3x2+2x3+4)e−2x)y=2+e−2x, x∈(−1,2)x \in (-1,2)x∈(−1,2) , satisfying y(0)=32y(0)=\dfrac{3}{2}y(0)=23. If y(1)=α(2+e−2)y(1)=\alpha (2+e^{-2})y(1)=α(2+e−2), than α is equal to :
138\dfrac{13}{8}813
613\dfrac{6}{13}136
1213\dfrac{12}{13}1312
1312\dfrac{13}{12}1213