= Solution
{c}
With $f(x,y)=-xy^2-y^5$, write $t=x_{n+1}=x_n+h$ and $z=y_{n+1}$. The <Backward Euler method> requires
$$
z=y_n-htz^2-hz^5,
$$
or the monic <quintic polynomial> equation
$$
p(z)=z^5+tz^2+h^{-1}z-h^{-1}y_n=0.
$$
Construct its $5\times5$ <companion matrix>
$$
\boxed{C_n=\begin{pmatrix}
0&0&0&0&y_n/h\\
1&0&0&0&-1/h\\
0&1&0&0&-t\\
0&0&1&0&0\\
0&0&0&1&0
\end{pmatrix}.}
$$
Its <characteristic polynomial> is $\det(zI-C_n)=p(z)$. Thus the <polynomial roots> needed for this update are exactly the <eigenvalues> of $C_n$.
At each step, assemble $C_n$, compute its <eigenvalues>, for example with the <QR algorithm>, and choose its unique positive real <eigenvalue> as $y_{n+1}$. To justify this selection, let
$$
F(z)=hz^5+htz^2+z-y_n.
$$
If $y_n>0$, then $F(0)=-y_n<0$, $F(y_n)=hy_n^5+hty_n^2>0$, and
$$
F'(z)=5hz^4+2htz+1>0\qquad(z\geq0).
$$
The <intermediate value theorem> and strict <monotonicity> prove there is exactly one positive <polynomial root>, located in $(0,y_n)$. Starting at $y_0=1$ therefore makes every update well defined, positive and decreasing. In numerical arithmetic, select the nearly real <eigenvalue> in this interval and check the <polynomial residual> $|F(z)|$; a final <Newton root-finding iteration> can refine it. This reduces each update to the requested <eigenvalue problem>.
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