Are you considering $d$ to be an arbitrary positive integer?
Then the answer is yes. In fact, one can show the following is true:
For the homogeneous equation with real coefficients:
$$\tag{1}\def\sss{}
a_{\sss n} y^{\sss(\!n\!)} +a_{\sss n\!-\!1} y^{\sss(\!n\!-\!1\!
)} +\cdots+a_{\sss1} y' +a_{\sss0} y = 0,\quad a_n\ne0
$$
The characteristic polynomial (c.p.) is
$$
\tag{2}a_{\sss n}x^n+a_{\sss n\!-\!1}x^{n-1} +\cdots+a_{\sss1}x +a_{\sss0} .
$$
Find the roots and their multiplicities of (2).
If $c$ is a real root of (2) with multiplicity $k $, then $k$-independent solutions of (1)
are
$$
e^{ct}, xe^{ct}, x^2 e^{ct}, \ldots, x^{k-1}e^{ct}
$$
If $a+bi$ is a complex root of (2) with multiplicity $k$, then $2k$-independent solutions of (1)
are
$$
e^{at}\sin (bt), x e^{at}\sin (bt), \ldots, x^{k-1} e^{at}\sin (bt)
$$
$$
e^{at}\cos (bt), xe^{at}\cos (bt) , \ldots, x^{k-1}e^{at}\cos (bt)
$$
Moreover, the set of all solutions found from the above will be independent (for instance, one can compute the Wronskian) and there will be $n$ of them (this follows from the Fundamental Theorem of Algebra).
The general solution to (1) is
$$y_c=c_{\sss 1}y_{\sss1}+c_{\sss2}y_{\sss2}+\cdots+c_{\sss n}y_{\sss n}$$
where $y_1$, $y_2$, $\ldots\,$, $y_n$ are the $n$-solutions found above.
There's three recurrence relations that help here:
$$P_{n+1}^{'} -P_{n-1}^{'} = (2n+1)P_n$$
$$(n+1)P_{n+1} = (2n+1)xP_n -nP_{n-1}$$
$$P_{n+1}-P_{n-1} = (x^2-1)\cdot \frac{2n+1}{n(n+1)}\cdot P_n^{'}$$
The induction step then looks as follows:
$\newcommand{\partial}[1]{\left[#1\right]}$
$\newcommand{\bracket}[1]{\left(#1\right)}$
\begin{equation}
\begin{split}
[(1-x^2)P_{n+1}^{'}]^{'} &=[(1-x^2)(P_{n-1}^{'}+(2n+1)P_n)]^{'}
\\ &=[(1-x^2)P_{n-1}^{'}]^{'}+(2n+1)\bracket{-2xP_n + (1-x^2)P_n^{'}}
\\ &=-(n-1)nP_{n-1}-2\{(n+1)P_{n+1}+nP_{n-1}\} + n(n+1)(P_{n-1}-P_{n+1})
\\ &=\{-(n-1)n-2n+ n(n+1)\}P_{n-1}+\{-2(n+1)-n(n+1)\}P_{n+1}
\\ &=n\{-(n-1)-2+(n+1)\}P_{n-1}-(n+1)(2+n)P_{n+1}
\\ &=-(n+1)(n+2)P_{n+1}
\end{split}
\end{equation}
Q.E.D.
Best Answer
Variation of parameters will do it. You can assume a form of solution $y=fP_{n}$ and substitute into the equation. The equation in expanded form is $$ (1-x^{2})P_{n}''-2xP_{n}'+n(n+1)P_{n}=0. $$ Now substitute $y=fP_{n}$ into the same equation $$ (1-x^{2})(fP_{n}''+2f'P_{n}'+f''P_{n})-2x(fP_{n}'+f'P_{n})+n(n+1)fP_{n}=0. $$ Subtracting $$ (1-x^{2})fP_{n}''-2xfP_{n}'+n(n+1)fP_{n}=0 $$ from the previous equation gives $$ (1-x^{2})(2f'P_{n}'+f''P_{n})-2xf'P_{n}=0. $$ This is a first order equation in $f'$ $$ (1-x^{2})P_{n}f''+\{2(1-x^{2})P_{n}'-2xP_{n}\}f'=0,\\ \frac{f''}{f'} = -2\frac{P_{n}'}{P_{n}}+\frac{2x}{1-x^{2}},\\ %% \ln|f'| = -2\ln|P_{n}|-\ln(1-x^{2})+C\\ %% f' = \frac{K}{(1-x^{2})P_{n}^{2}} \\ %% f = K\int\frac{1}{(1-x^{2})P_{n}^{2}}\,dx+L $$ A second solution has the following form for constant $K\ne 0$ and $L$: $$ fP_{n} = K\left(\int \frac{1}{(1-x^{2})P_{n}^{2}}\,dx\right)P_{n}+LP_{n}. $$