 Solution

The Backward Euler method uses the new point:Writing $a=h(x_n+h)>0$ gives the quadratic equation $a{y}_{n+1}^2+y_{n+1}-y_n=0$. Starting from $y_n>0$, its two polynomial roots have opposite signs. The positive root, which continues the solution as $h\to 0$, isThe other root is negative and does not approach $y_n$ as $h\to 0$. Rationalizing the numerator gives an equivalent form that avoids cancellation error when the step size is small:The defining quadratic equation gives $0<y_{n+1}<y_n$ for this root, so positivity is maintained inductively from $y_0=1$. The branch choice is part of specifying the Backward Euler method here: the algebraic update alone has two solutions.