= Solution
Substitute the <power series> $y=\sum_{n\geq0}a_nx^n$ and compare the coefficient of $x^n$. This gives
$$
(n+2)(n+1)a_{n+2}+(\lambda^2-n^2)a_n=0,
$$
or
$$
\boxed{a_{n+2}=\frac{n^2-\lambda^2}{(n+1)(n+2)}a_n}.
$$
The two arbitrary coefficients $a_0,a_1$ determine all solutions. The condition $y'(0)=0$ sets $a_1=0$, leaving the even series. It terminates precisely when $\lambda^2=(2k)^2$ for some nonnegative integer $k$. Therefore the permitted real values are $\lambda=0$ or $\lambda=\pm2k$ for $k\geq1$; these are the even <Chebyshev polynomials> up to normalization.
The degrees below six occur for $\lambda=0,\pm2,\pm4$. Taking $a_0=1$, the corresponding polynomials are
$$
1,\qquad 1-2x^2,\qquad 1-8x^2+8x^4.
$$
For the forced equation, try $y=a_0+a_2x^2+a_4x^4$. Comparing the coefficients of $x^4,x^2,1$ gives
$$
-15a_4=8,
\qquad 12a_4-3a_2=0,
\qquad 2a_2+a_0=-3.
$$
Thus one polynomial satisfying $y'(0)=0$ is
$$
\boxed{y(x)=\frac{19-32x^2-8x^4}{15}}.
$$
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