Integrating two equations that equal, what happens to the constant on one of the sides? In class, we were talking about Newton's 3rd law and how to integrate. 
$\int(g)dt = \int(y''(t))dt \implies g(t) + C = y'(t)$
I am confused about why the right hand side of the equation doesn't get a constant. After asking the professor, he said that it was because the two constants would cancel each other out. But I still don't understand why that should prevent us from writing a constant on the right side. 
 A: Recall, you can add constants together into a single constant (as was done with C). 
Also, you can define a $C_1$ and $C_2$ - one to each side. You can also show the constant as being on the RHS of the equation. All will produce the same result. The choice is typically one of convenience to make solving easiest. 
Example: Consider the separable equation:
$$y' = x y$$
After separation, we can integrate both sides as:
$$\int \dfrac{1}{y}~dy = \int x~dx$$
Approach 1: Single constant (we could have also put $C$ on the LHS - try this)
$$\ln y = \dfrac{x^2}{2} + C$$
We take the exponential of each side and have:
$$y = e^{x^2/2 + C} = e^{x^2/2}~e^C = w~e^{x^2/2}$$
Note: $w = e^C$, which is just some constant (totally arbitrary).
Approach 2: Constant on each side
$$\ln y + c_1 = \dfrac{x^2}{2} + c_2$$
Taking exponential of both sides:
$$e^{\ln y + c_1} = e^{x^2/2 + c_2}$$
The RHS is as above and the LHS, we have ($q$ is just any arbitrary constant):
$$e^{\ln y}~e^{c_1} = y~q$$
Now we write:
$$q y = c_2~e^{x^2/2}$$
Dividing constants and just calling it $w$, we have:
$$y(x) = w~e^{x^2/2}$$
In both approaches, we just get some constant $w$.
Now, if you were provided with an initial condition like $y(0) = 2$, you would plug in $x=0$ and see that $w = 2$.
