What happens to dt? I'm trying to understand a simple differential equation solution. 
$$\frac{dy}{dt} = t^3 - 3t^2 + t $$
To
$$ dy = t^3 dt - 3t^2 dt + t  dt $$
To
$$ y = \frac{t^4}{4} - t^3 + \frac{t^2}{2} + c $$
The polynomial is integrated like a normal polynomial. What happens to dt after multiplying it with the function?
Shouldn't $dt$ become $t_1 - t_0$ since $y_0 - t_0$ becomes the constant $C$?
 A: Integrate from $t_0$ (a constant) to $t$ (a variable).  Then just lump the constants from both sides into one and call it $C$.
A: The differentials are integrated / summed up on each side. However this is more a formal intermediate step, an application of Leibniz's Differential Kalkül.
A better description of what is going on is that one integrates both sides and applies the fundamental theorem of calculus:
$$
\int\limits_{t_0}^t \left.\frac{dy}{dt}\right\vert_{t=\tau} d\tau = 
\int\limits_{t_0}^t \left( \tau^3 - 3\tau^2 + \tau \right) \, d\tau \iff \\
y(t) - y(t_0) = \frac{1}{4} t^4 - t^3 + \frac{1}{2} t^2 - \left( \frac{1}{4} t_0^4 - t_0^3 + \frac{1}{2} t_0^2 \right)
$$
so we have
\begin{align}
y(t) &= \underbrace{\left[ \frac{1}{4} t^4 - t^3 + \frac{1}{2} t^2 \right]}_{p(t)} + 
\underbrace{\left[ y(t_0) - \left( \frac{1}{4} t_0^4 - t_0^3 + \frac{1}{2} t_0^2 \right) \right]}_{\text{const.}} \iff \\
y(t) &= p(t) + C
\end{align}
where "const." means this is a constant expression, thus having no dependency on the variable $t$.
A: Since $$\int\,dy=y+C_1$$ and $$\int(t^3-3t^2+t)\,dt=\frac14t^4-t^3+\frac12t^2+C_2$$ for some constants $C_1$ and $C_2,$ then putting $C=C_2-C_1$ yields the result.
On the other hand, taking an arbitrary constant $C$ and letting $$y=\frac14t^4-t^3+\frac12t^2+C,$$ we have $$\frac{dy}{dt}=t^3-3t^2+t,$$ as desired. Thus, the given result is the family of solutions to the ODE.
A: There is no reason to separate variables using differential notation.  
The solution to $y'(t) = f(t) \quad $ is $\quad y = \int f(t) dt$, up to adding a constant.
If you did separate variables then the constants can be combined as people have written.  But there is no reason to reach a point where there are two integrations, one in $y$ and one in $t$.
