Note that the factorization below
suggests a the simpler equation,
i.e. translating $(-2,-3)$ to the origin
$$ y'=(x+2)(y+3)-6 $$
and the initial condition $(0,-1)$ to $(2,2)$.
This, however, leaves $(y+3)'=y'$ unchanged:
$$
\eqalign{
y' &= xy - 6 \\
y' - xy &= - 6 \\
(y' - xy) \, e^{-x^2/2} &= - 6\,e^{-x^2/2} \\
(y \, e^{-x^2/2})' &= - 6\,e^{-x^2/2} \\
y \, e^{-x^2/2} &= - 6 \int e^{-x^2/2} dx \\
y \, e^{-x^2/2} &= c-6 \sqrt{\frac\pi2}\,
\text{erf}\frac{x}{\sqrt{2}}
\qquad \implies \quad c=y_0 \\
y &= e^{x^2/2} \left( y_0 - 3 \sqrt{2\pi}\,
\text{erf}\frac{x}{\sqrt{2}}\right) .
}
$$
Translating back, we get
$$
y = e^{(x+2)^2/2} \left( c - 3 \sqrt{2\pi}\,
\text{erf}\frac{x+2}{\sqrt{2}}\right)-3
$$
for the same constant $c$ which, however,
is no longer equal to $y_0$, but rather to
$$
\eqalign{
c &= (y_0+3) e^{-\frac12(x_0+2)^2}
+ 3\sqrt{2\pi}\,\text{erf}\frac{x_0+2}{\sqrt{2}}
\\&=2e^{-2}+3\sqrt{2\pi}\,\text{erf}\sqrt{2}
\approx 7.44839864640532
}
$$
which would give
$$
\eqalign{
y(0.1)
&= e^{1/200} \left( c - 3 \sqrt{2\pi}\,
\text{erf}\frac{2.1}{\sqrt{2}}\right)-3
\\&\approx -1.21143171766941
\,.
}
$$
If we had just let $u=x+2$ and $v=y+3$ to begin with,
we could have rewritten our differential equation as
$v'=uv-6$
which is easier to differentiate by hand anyway:
$$
\eqalign{
\\v^{(1)}&=\left(u\right)\cdot&v+\left(-6\right)
\\v^{(2)}&=\left(u^{2}+1\right)\cdot&v+\left(-6 \, u\right)
\\v^{(3)}&=\left(u^{3}+3\,u\right)\cdot&v+\left(-6\,u^{2}-12\right)
\\v^{(4)}&=\left(u^{4}+6\,u^{2}+3\right)\cdot&v+\left(-6\,u^{3}-30\,u\right)
\\v^{(5)}&=\left(u^{5}+10\,u^{3}+15\,u\right)\cdot&v
+\left(-6\,u^{4}-54\,u^{2}-48\right)
\\v^{(6)}&=\left(u^{6}+15\,u^{4}+45\,u^{2}+15\right)\cdot&v
+\left(-6\,u^{5}-84\,u^{3}-198\,u\right)
\\v^{(7)}&=\left(u^{7}+21\,u^{5}+105\,u^{3}+105\,u\right)\cdot&v
+\left(-6\,u^{6}-120\,u^{4}-522\,u^{2}-288\right)
\\v^{(8)}&=\left(u^{8}+28\,u^{6}+210\,u^{4}+420\,u^{2}+105\right)\cdot&v
+\left(-6\,u^{7}-162\,u^{5}-1110\,u^{3}-1674\,u\right)
\\v^{(9)}&=\left(u^{9}+36\,u^{7}+378\,u^{5}+1260\,u^{3}+945\,u\right)\cdot&v
+\left(-6\,u^{8}-210\,u^{6}-2070\,u^{4}-5850\,u^{2}-2304\right)
}
$$
especially if one notices the recursion
$$
\left.\matrix{
v^{(n)}=a_nv+b_n\\\\
a_1=u,~b_1=-6}\right\}
\quad\implies\quad
\left\{\matrix{
a_{n+1}=a_n'+u\,a_n\\\\
b_{n+1}=b_n'-6\,a_n\,,}\right.
$$
which I managed to calculate in sage:
var('a,b,u,v,z')
a(u) = u
b(u) = -6
for n in range(1,10):
# print n, a, b #Note: a & b are functions of u
# print n, a(u).simplify_full(), b(u).simplify_full()
print '\\\\v^{(%d)}&=\\left(%s\\right)\cdot v+\\left(%s\\right)' \
% (n, latex(a(u).simplify_full()), latex(b(u).simplify_full()))
# print 'v^{(%d)}=%d' % (n, 2*a(2)+b(2))
z(u) = a
a(u) = (diff(a,u) + u*z).simplify_full()
b(u) = (diff(b,u) - 6*z).simplify_full()
Using the initial condition $(u,v)=(2,2)$
corresponding to $(x,y)=(0,-1)$, we get
$$
v^{(1)}=-2\\
v^{(2)}=-2\\
v^{(3)}=-8\\
v^{(4)}=-22\\
v^{(5)}=-76\\
v^{(6)}=-262\\
v^{(7)}=-980\\
v^{(8)}=-3794\\
v^{(9)}=-15428
$$
so that, factoring out the $(-1)$,
I computed the following terms and partial sums:
t = [1, 2, 2, 8, 22, 76, 262, 980, 3794, 15428]
x = 0.1
y = 0
for n in range(10):
tn = t[n]*x^n/factorial(n)
y = y + tn
print '%-4d %-14g %-16.12g' % (n, tn, y)
0 1 1
1 0.2 1.2
2 0.01 1.21
3 0.00133333 1.21133333333
4 9.16667e-05 1.211425
5 6.33333e-06 1.21143133333
6 3.63889e-07 1.21143169722
7 1.94444e-08 1.21143171667
8 9.40972e-10 1.21143171761
9 4.25154e-11 1.21143171765
So the analytic and Taylor series solutions agree.
Best Answer
In general the equation $y''-qy=0$ with $0<q(x)\to\infty$ for $x\to\infty$ has the WKB approximation (see wikipedia, this is a standard example) of basis solutions $$ y(x)=q(x)^{-\frac14}\exp\left(\pm\int\sqrt{q(x)}dx\right) $$ where the first factor is the second order term in the expansion. So indeed, in first order you would get $y(x)=\exp(\pm\frac{x^2}2)$, while in second order, there would be additionally a factor $\frac1{\sqrt{x}}$.
The approach is to set $y=\exp(S)$ with an expansion $S=S_0+S_1+S_2+...$ where $S_0\gg S_1\gg S_2\gg...$ for $x\to\infty$. This scale also translates to the derivatives. Set for shortness $S'=s$ then isolating the components of equal scale in $s'+s^2-q=0$ gives $$ s_0^2=q\\ s_0'+2s_0s_1=0\\\vdots $$ which implies $s_0=\pm\sqrt q$ and $s_1=-\frac{s_0'}{2s_0}\implies S_1=-\frac12\ln|s_0|=-\frac14\ln(q)$.