Does the constant $C$ in this solution to a differential equation equal infinity? The problem is $y' = -\frac{1}{t^2} - \frac{1}{t}y + y^2;\ y_p = \frac{1}{t}$.  My solution is 
$$\begin{align}
y = \frac{1}{t} + B &\implies y' = -\frac{1}{t^2} + B' \\
&\implies -\frac{1}{t^2} - \frac{1}{t}y + y^2 = -\frac{1}{t^2} + B' \\
&\implies -\frac{1}{t^2} - \frac{1}{t} \left(\frac{1}{t} + B \right) + \left(\frac{1}{t} + B\right)^2 = -\frac{1}{t^2} + B' \\
&\implies B' - \frac{1}{t}B = B^2 \\
&\implies L = B^{-1} \\
&\implies L' = -B^{-2} \left(B^2 + \frac{1}{t}B \right) \\
&\implies L' + \frac{1}{tB} = -1 \\
&\implies L' + \frac{1}{t}L = -1 \\
&\implies L_h = \frac{1}{t} \\
&\implies L = \frac{1}{t}\int\frac{-1}{\frac{1}{t}}dt \\
&\implies L = \frac{1}{t} \left(-\frac{1}{2}t^2 + C_{tentative} \right) \\
&\implies L = \frac{C - t^2}{2t} \\
&\implies B = \frac{2t}{C - t^2} \\
&\implies y = \frac{1}{t} + \frac{2t}{C - t^2}
\end{align}$$
Although this solution does not appear equivalent to the answer given in my book, one or two people in chat reviewed this work and could not see anything incorrect.  The issue is that the given particular solution, $y_p = \frac{1}{t}$, cannot be obtained by plugging any finite value into the $C$ in my general solution.  We could say that the general solution yields the given particular solution when $C =$ infinity, more specifically, when $C =$ some $\aleph$ expression which makes the term $\frac{2t}{C - t^2}$ disappear regardless of $t$, but this feels a bit outside the box for a textbook problem.  Is my general solution correct, and if so, how if at all can we derive the given particular solution from it?

For non-linear equations like this, it is valid to take $C\to\infty$. You can also write $c = \frac{1}{C}$ to get an alternate form
$$ y(t) = \frac{1}{t} + \frac{2ct}{1-ct^2} $$
