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.