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.
Remember that an expression with $n$ arbitrary constants will yield a differential equation of order $n$. So to get the $n^{th}$ order derivative you'll have to differentiate the expression $n$ times, and in that process you'll obtain $n$ more relations so that now you have a total of $n+1$ relations from which you can eliminate the $n$ arbitrary constants to obtain the differential equation.
Most of the times though the constants more or less dissappear by themselves. For example,consider
$y=ax^2+bx+c$.
There are 3 arbitrary constants $a$,$b$ and $c$ so just differentiate 3 times to obtain the DE $y'''=0$
Now consider $y^2=4ax$. Since there is only one constant $a$, differentiate once to get $2yy'=4a$. Now eliminate $4a$ to obtain the DE $2xy'=y$
I think with that in mind you can find the DE for a given solution.
Best Answer
One method is reduction of order. I describe it in the case of differential equation of order 2.
Reduction of order
Assume that you have the differential equation $$ y''+py'+qy=0, $$ and that you have one solution $y_1$. Then, try to find a solution $y$ in the form $$ y=y_1\int u\,dx,\tag{*} $$ where $u$ is a function to be determined. Differentiating, you will find $$ y_1''\int u\,dx+2y_1u+y_1u'+p\bigl[y_1'\int u\,dx+y_1u\bigr]+qy_1\int u\,dx=0, $$ which reduces to $$ y_1u'+(2y_1'+py_1)u=0.\tag{**} $$ This is a linear differential equation of order one, which in principle can be solved using the method with integrating factor.
Your example
For your example, this is not the easiest method, but you have that $y_1=1$ solves $$ y''-2xy'=0. $$ Here, $p=-2x$, $q=0$, and so the new differential equation $(**)$ becomes $$ u'-2xu=0. $$ An integrating factor is $e^{-x^2}$, so $$ (ue^{-x^2})'=0. $$ Integration gives $$ u=Ce^{x^2}. $$ Inserting this into $(*)$ gives your second solution $$ y=C\int e^{x^2}\,dx. $$