y=x+Ae^x+Be^x .... eq 1
Differentiating both sides with respect to x,
dy/dx = 1 +Ae^x+Be^x .... eq 2
Differentiating again with respect to x,
dy^2/d^2x = Ae^x + Be^x ...eq 3
Substituting eq. 3 to eq. 1, we get
y = x +dy^2/d^2x
Thus, ans. is y = x + dy^2/d^2x
