Assume that $u(x)$ is the classical solution solving $$a_{ij}(x)\partial_{ij}u(x)+b_i(x)\partial_iu(x)+c(x)u(x)=f(x)$$ on $\mathbb{R}^n$ for some smooth enough coefficients and uniformly elliptic $a_{ij}$. I am looking for the gradient bound of $u$ explicitly on the behavior of the coefficients.
I found that in Gilbarg and Trudinger's PDE book, Theorem 8.32 states that $$ |u|_{C^{1,\alpha}(B_1(x_0))}\leq C(n,K(x_0),\lambda_a)(|u|_{C^0(B_2(x_0))}+|f|_{C^{0}(B_2(x_0))}) $$ for some constant $C(n,K(x_0),\lambda_a)$ where $\lambda_a$ is the least eigenvalue of $a_{ij}$ and
$$\max \left\{1,|a_{ij}|_{C^{0,\alpha}(B_2(x_0))},|b_{i}-\partial_k a_{ij}|_{C^0(B_2(x_0))},|c|_{C^0(B_2(x_0))} \right\} = K(x_0). $$ Then I consider the equation, $$\dfrac{a_{ij}(x)}{K(x_0)}\partial_{ij}u(x)+\dfrac{b_i(x)}{K(x_0)}\partial_iu(x) +\dfrac{c(x)}{K(x_0)}u(x)=\dfrac{f(x)}{K(x_0)}$$ on $B_2(x_0)$. Therefore, the new $K$ in this situation should be smaller than 1.
My question is: using this rescaling, can I conclude that $$ |u|_{C^{1,\alpha}(B_1(x_0))}\leq C\left(|u|_{C^0(B_2(x_0))}+\left|\dfrac{f(x)}{K(x_0)}\right|_{C^{0}(B_2(x_0))}\right)$$ with $C$ only depending on $n$ and $\lambda_a$?