Can there be different particular solutions for a differential equation since the constant MAY change? Say, we had the following differential equation:
$dy/dx = 2(y^2)$
Now, when we solve it, we may get one of the following equations, depending on IF we decide to multiply something with the constant OR leave it as it is:
$y = 1/-2x-2c$ (if we decide to multiply -2 with the constant)
$y = 1/-2x+c$ (if we decide to leave the constant alone)
If we're asked to find the particular solution for when x=2 and y=1, obviously, the value of the constant and the particular solution in each case will be different. Now, I'm a bit confused here. Some of the teachers like to leave the constant alone, explaining that it has a particular value which can not change and hence, we leave it as it is. Other teachers like to multiply stuff with the constant as well.
What's the right way to approach this, and why?
EDIT: The value of C maybe different, but the constant we get at the end, after solving for the particular solution and plugging the value of the constant into the general solution, will exactly be the same! Thanks, Joe!
 A: \begin{align}
\frac{dy}{dx} &= 2y^2 \\
\int 1 \, dx &= \int \frac{1}{2}y^{-2} \, dy \\
x + C &= -\frac{1}{2}y^{-1} \\
y(x+C) &= -\frac{1}{2} \\
y &= -\frac{1}{2(x+C)}
\end{align}
What the solution to this differential equation shows is that any function $y$ of the form
$$
y(x)=-\frac{1}{2(x+\text{some arbitrary constant})}
$$
has the derivative $2y^2$. It doesn't matter how you label that constant. For example, you could use $C$, you could use $K$, you could even use $17C-398$; it's all the same. It is very common to not even bother multiplying the constant as you are solving the equation, because we just need some letter that represents a member of the real numbers. For example, you might see
\begin{align}
2y &= 5x + C \\
y &= 5x + C
\end{align}
More formally, this could be justified in the following way:
\begin{align}
2y &= 5x + C \\
y &= \frac{5}{2}x + C/2
\end{align}
Let $K=C/2$. Then we have $y=\frac{5}{2}x+K$. However, since the value of $K$ is arbitrary, we may relabel it as $C$. (This is despite the fact that we are using the letter $C$ to represent something that earlier we would denote $C/2$.) Again, we just need to use a letter to emphasise that it is a family of solutions to the differential equation that we have found. The constant $C$ is just an arbitrary member of the real numbers.
