= Solution
{c}
The <Backward Euler method> uses the new point:
$$
y_{n+1}=y_n-hx_{n+1}y_{n+1}^2,
\qquad x_{n+1}=x_n+h.
$$
Writing $a=h(x_n+h)>0$ gives the <quadratic equation> $ay_{n+1}^2+y_{n+1}-y_n=0$. Starting from $y_n>0$, its two <polynomial roots> have opposite signs. The positive root, which continues the solution as $h\to0$, is
$$
\boxed{y_{n+1}=\frac{-1+\sqrt{1+4h(x_n+h)y_n}}{2h(x_n+h)}.}
$$
The other root is negative and does not approach $y_n$ as $h\to0$. Rationalizing the numerator gives an equivalent form that avoids <cancellation error> when the <step size> is small:
$$
\boxed{y_{n+1}=\frac{2y_n}{1+\sqrt{1+4h(x_n+h)y_n}}.}
$$
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.
Back to article page
