Answer
See below
Work Step by Step
Obtain:
$y_1(x)=a_0[1-\frac{\alpha (\alpha +1)}{2}x^2+\frac{\alpha(\alpha -2)(\alpha +1)(\alpha +3)}{4!}x^4+\frac{\alpha (\alpha +1)(\alpha -2)(\alpha +3)(\alpha -4)(\alpha +5)}{6!}x^6+...]$
and then $y_1(x)=a_1[x-\frac{(\alpha -1)(\alpha +2)}{3!}x^3+\frac{(\alpha -1)(\alpha +2)(\alpha -3)(\alpha +4)}{5!}x^5+...]$
Subsituting $\alpha =3$ we have:
$y_1(x)=a_0(1-6x^2+3x^4-\frac{4x^6}{5}...)\\
y_2(x)=a_1(x-\frac{5}{3}x^6)$
We can notice that $y_2(x)$ is a degree of 3.
Set $\alpha_0=0$, we have:
$P_3(1)=1\\
\rightarrow a_1=-\frac{3}{2}(x-\frac{5}{3}x^6)$
Subsituting $\alpha =4$ we have:
$y_1(x)=a_0(1-10x^2+\frac{35x^4}{3}...)\\
y_2(x)=a_1(x+\frac{6}{5}x^5-3x^6+...)$
Set $\alpha_1=0$, we have:
$P_4(1)=1\\
\rightarrow a_0=\frac{3}{8}\\
=\frac{3}{8}(1-10x^2+\frac{35}{3}x^4)$
