Let y=a0+a1x+a2x^2+...+anx^n, then
y'=a1+2a2x+3a3x^2+...+nanx^(n-1) and
y''=2a2+6a3x+...+n(n-1)anx^(n-2).
y(0)=-1 means a0=-1; y'(0)=2 means a1=2, so y=-1+2x+a2x^2...
y''-2xy'+3y=0 becomes ("an" also shows as "a[n]" in the following):
2a2+6a3x+...+n(n-1)anx^(n-2)-2x(2+2a2x+3a3x^2+...+nanx^(n-1))+3(-1+2x+a2x^2+...+anx^n)=0.
2a2-3+x(6a3-4+6)+x^2(12a4-4a2+3a2)+x^3(20a5-6a3+3a3)+x^4(30a6-8a4+3a4)+...⇒
2a2-3+x(6a3+2)+x^2(12a4-a2)+x^3(20a5-3a3)+...+x^n((n+1)(n+2)a[n+2]-(2n-3)a[n])=0 for n>2.
The variable x is general, that is, the power series must be true for all values of x. Therefore the second order differential equation is zero when the coefficients are all zero. We can relate the a values. To find a2 we know that 2a2=3, making a2=3/2. To find a4, we have the x^2 coefficient 12a4-a2, which must also be zero, making a4=a2/12=1/8. The x coefficient gives us a3=-1/3 and x^3 coefficient gives us a5=3a3/20=-1/20. Plugging these values into the initial power series y=-1+2x+3x^2/2-x^3/3+x^4/8-x^5/20+x^6/48-...
The general expression for n>1 is (n+1)(n+2)a[n+2]=a[n](2n-3), or a[n+2]=a[n](2n-3)/((n+1)(n+2)).