Consistency in finding electric field!

When considering electric field in a dielectric media,there are two forms of Gauss's law:[itex]
\oint \vec{D}\cdot\vec{d \sigma}=q_f
[/itex]and [itex] \epsilon_0 \oint \vec{E}\cdot\vec{d\sigma}=q_f+q_b [/itex]

So we can do the following in a linear dielectric:
1-Finding D field from the free charge distribution using the first form of Gauss's law.
2-Finding E field from the D field by dividing it by [itex] \epsilon [/itex].
3-Finding P field from the E field.
4-Finding surface and volume bound charge distributions from the P field.
5-Finding E field from the free and bound charge distributions using the second form of Gauss's law.
For consistency,the result of steps 2 and 5 should be equal.How is that guaranteed?

I'm asking this because of the following example:
There is a linear dielectric spherical shell with inner and outer radii a and b.It has a free charge distribution of [itex] \rho_f=\frac{k}{2r} [/itex] and it contains a charge q at its center.Using the first form of Gauss's law,one can find D and so E to be [itex] \vec{E}=\frac{\pi k (r^2-a^2)+q}{4 \pi \epsilon r^2} \hat{r}[/itex]. Using the relations [itex] \vec{P}=\epsilon_0 \chi \vec{E} [/itex] and [itex] \rho_b=-\vec{\nabla}\cdot\vec{P} [/itex] it is easy to obtain [itex]\rho_b=-\frac{\epsilon_0 \chi k}{2r} [/itex].
Now by using [itex] \rho_f [/itex] and [itex] \rho_b [/itex] to find E field from the second form of Gauss's law,one can find [itex] \vec{E}=\frac{\pi k (1-\epsilon_0 \chi)(r^2-a^2)+q}{4 \pi \epsilon_0 r^2} \hat{r} [/itex].
But looks like the two E fields are not equal!!!
What's wrong???
Thanks