Answer
See below
Work Step by Step
Starting with $P_0(x)=1, P_1(x)=x$
$(n+1)P_{n+1}+nP_{n-1}=(2n+1)xP_n$
Substituting $n=1$, we get:
$P_0+2P_2=3P_1x \rightarrow P_2=\frac{1}{2}(3P_1x-P_0)=\frac{1}{2}(3x^2-1)$
Substituting $n=2$, we get:
$2P_1+3P_3=5P_2x \rightarrow P_3=\frac{1}{2}(5P_2x-2P_1)=\frac{1}{2}x(5x^2-3)$
Substituting $n=3$, we get:
$3P_2+4P_4=7P_3x \rightarrow P_4=\frac{1}{4}(7P_3x-3P_2)$
Consequently,
$P_4=\frac{1}{8}(35x^4-30x^2+3)$
