Pretty much the same way you would any such equation.
Let
[tex]y= \Sigma_{n=0}^\infty a_nx^n[/tex]
then
[tex]y'= \Sigma_{n=1}^\infty na_nx^{n-1}[/tex]
(we can start at n= 1 because if n= 0 the term is 0) and
[tex]y"= \Sigma_{n= 2}^\infty n(n-1)a_nx^{n-2}[/tex]
(we can start at n= 2 because both n=0 and n= 1 make n(n-1) 0)
Now put those into the equation:
y``-3y`+2y=0
[tex]\Sigma_{n= 2}^\infty n(n-1)a_nx^{n-2}-\Sigma_{n=1}^\infty 3na_nx^{n-1}+\Sigma_{n=0}^\infty2a_nx^n= 0[/tex]
Since n in each sum is a "dummy index", we can change indices to match powers. In the first sum let j= n-2, in the second sum, j= n-1, and in the last, j= n. The sums become
[tex]\Sigma_{j= 0}^\infty (j+2)(j+1)a_{j+2}x^j-\Sigma_{n=0}^\infty 3(j+1)a_{j+1}x^j+\Sigma_{j=0}^\infty2a_jx^j= 0[/tex]
or
[tex]\Sigma_{j= 0}^\infty\left( (j+2)(j+1)a_{j+2}- 3(j+1)a_{j+1}+2a_j\right)x^j= 0[/tex]
For that to be true for all x, we must have
[tex](j+2)(j+1)a_{j+2}- 3(j+1)a_{j+1}+2a_j= 0[/tex]
for all j. That's a recursive equation that you can solve for a_{j} in terms of a_{0} and a_{1}, the two constants in the general solution.
Another way to get a power series is to find Taylor's series for y:
[tex]y= \Sigma_{n=0}^\infty \frac{y^{(n)}}{n!}x^n[/tex]
Take y(0) and y'(0) as given initial values (again, the two constants in the general solution. The y"(0)= 3y'(0)- 2y(0), y^{(3)}(0)= 3y"(0)- 2y'(0), y^{(4)}(0)= 3y^{(3)}- 2y"(0), etc.
Of course, there is no need to do all that (unless some evil teacher is requiring it!) since this is a simple linear, homogeneous, differential equation with constant coefficients. It's characteristic equation is r^{2}- 3r+ 2= 0 which has roots r= -1 and r= -2. The general solution to the differential equation is y= C_{1}e^{-x}+ C_{2}e^{-2x}. You can use that to check your series solution.
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