Solving for a general solution in $y' = y^2$ My book gives me $y' = y^2$ and then asks me to find the general solution. 
I am getting $\displaystyle\frac{-1}{(c+x)}$ as my answer.
However, both the book's answer key AND Wolfram report to me that $\displaystyle\frac{1}{(c-x)}$ is the answer which I am NOT getting.
 A: As noted in the comments, the answers are equivalent. Generally we assume $c$ to be a positive number, just so we have less negative terms to deal with. Note that $-c \equiv c$ since we just consider $c$ to be some constant (it could be negative or positive). Based on this, one can simply multiply through by $-1$ to obtain the answer that Wolfram got.
A: The difference between your answer and the Wolfram answer occurs because you can choose to multiply by $-1$ at any point. Note that $\int y^{-2}y'dx = \int y^{-2}dy$ so we have $$\int y^{-2}dy = \int 1 dx.$$ As $\frac{d}{dy}(y^{-1}) = -y^{-2}$, we see that $y^{-1}$ is an antiderivative for $-y^{-2}$. Multiplying the previous equation by $-1$ we have 
\begin{align*}
\int -y^{-2} dy =& \int -1 dx\\
y^{-1} &= -x + c_1
\end{align*}
so $y = \frac{1}{c_1 - x}$. Alternatively, from $\frac{d}{dy}(y^{-1}) = -y^{-2}$ we can multiply by $-1$ to obtain $\frac{d}{dy}(-y^{-1}) = y^{-2}$; that is, $-y^{-1}$ is an antiderivative for $y^{-2}$. Therefore we have
\begin{align*}
\int y^{-2} dy &= \int 1 dx\\
-y^{-1} &= x + c_2
\end{align*}
so $y = \frac{-1}{c_2 + x}$. This is why you get seemingly different answers. Note, as Daniel Fischer pointed out, the arbitrary constants in the two expressions are not the same, i.e. $c_1 \neq c_2$. In fact, $c_2 = - c_1$.
