Let's say I have this equation (where m, W, B, and k are known constants):
$$mv'=W-B-kv$$
and I want to solve it for v, like this: $$ \begin{align*} mv' &= W-B-kv \\ \int{\frac{dv}{W-B-kv}} &= \int{\frac{1}{m}dt} \\ -\int{\frac{dv}{kv-W+B}} &= \frac{t}{m}+C_0 \\ -\frac{1}{k}ln|kv-W+B| &= \frac{t}{m}+C_1 \\ ln|kv-W+B| &= C_2k-\frac{tk}{m} \\ kv-W+B &= e^{C_2k-\frac{tk}{m}} \\ v &= \frac{1}{k}(e^{C_2k-\frac{tk}{m}}+W-B) \tag{1}\\ \end{align*} $$
Given v(0)=0, the particular solution is: $$ \begin{align*} v=\frac{1}{k}(|B-W|e^{\frac{-tk}{m}}+W-B) \tag{2} \end{align*} $$
But, if I had simplified the constant in (1), like this: $$ \begin{align*} v &= \frac{1}{k}(C_3e^{-\frac{tk}{m}}+W-B) \end{align*} $$ and then found the particular solution, I would have: $$ \begin{align*} v &= \frac{1}{k}((B-W)e^{\frac{-tk}{m}}+W-B) \tag{3}\\ \end{align*} $$
Are my solutions different because I've made a mistake? If not, how do I explain the difference between (2) and (3) when I absorb k into the arbritrary constant?