You're looking for a solution of the form
[tex]\displaystyle y = \sum_{n=0}^\infty a_n x^n[/tex]
Differentiating twice yields
[tex]\displaystyle y' = \sum_{n=0}^\infty n a_n x^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n[/tex]
[tex]\displaystyle y'' = \sum_{n=0}^\infty n(n-1) a_n x^{n-2} = \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n[/tex]
Substitute these series into the DE:
[tex]\displaystyle (x-1) \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n - x \sum_{n=0}^\infty (n+1) a_{n+1} x^n + \sum_{n=0}^\infty a_n x^n = 0[/tex]
[tex]\displaystyle \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^{n+1} - \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n \\\\ \ldots \ldots \ldots - \sum_{n=0}^\infty (n+1) a_{n+1} x^{n+1} + \sum_{n=0}^\infty a_n x^n = 0[/tex]
[tex]\displaystyle \sum_{n=1}^\infty n(n+1) a_{n+1} x^n - \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n \\\\ \ldots \ldots \ldots - \sum_{n=1}^\infty n a_n x^n + \sum_{n=0}^\infty a_n x^n = 0[/tex]
Two of these series start with a linear term, while the other two start with a constant. Remove the constant terms of the latter two series, then condense the remaining series into one:
[tex]\displaystyle a_0-2a_2 + \sum_{n=1}^\infty \bigg(n(n+1)a_{n+1}-(n+1)(n+2)a_{n+2}-na_n+a_n\bigg) x^n = 0[/tex]
which indicates that the coefficients in the series solution are governed by the recurrence,
[tex]\begin{cases}y(0)=a_0 = -7\\y'(0)=a_1 = 3\\(n+1)(n+2)a_{n+2}-n(n+1)a_{n+1}+(n-1)a_n=0&\text{for }n\ge0\end{cases}[/tex]
Use the recurrence to get the first few coefficients:
[tex]\{a_n\}_{n\ge0} = \left\{-7,3,-\dfrac72,-\dfrac76,-\dfrac7{24},-\dfrac7{120},\ldots\right\}[/tex]
You might recognize that each coefficient in the n-th position of the list (starting at n = 0) involving a factor of -7 has a denominator resembling a factorial. Indeed,
-7 = -7/0!
-7/2 = -7/2!
-7/6 = -7/3!
and so on, with only the coefficient in the n = 1 position being the odd one out. So we have
[tex]\displaystyle y = \sum_{n=0}^\infty a_n x^n \\\\ y = -\frac7{0!} + 3x - \frac7{2!}x^2 - \frac7{3!}x^3 - \frac7{4!}x^4 + \cdots[/tex]
which looks a lot like the power series expansion for -7eˣ.
Fortunately, we can rewrite the linear term as
3x = 10x - 7x = 10x - 7/1! x
and in doing so, we can condense this solution to
[tex]\displaystyle y = 10x -\frac7{0!} - \frac7{1!}x - \frac7{2!}x^2 - \frac7{3!}x^3 - \frac7{4!}x^4 + \cdots \\\\ \boxed{y = 10x - 7e^x}[/tex]
Just to confirm this solution is valid: we have
y = 10x - 7eˣ ==> y (0) = 0 - 7 = -7
y' = 10 - 7eˣ ==> y' (0) = 10 - 7 = 3
y'' = -7eˣ
and substituting into the DE gives
-7eˣ (x - 1) - x (10 - 7eˣ ) + (10x - 7eˣ ) = 0
as required.
