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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.

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    $\begingroup$ 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 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. $\endgroup$ – Amzoti Jan 16 '14 at 18:13
  • $\begingroup$ As an example, take $y' = x y$. Solve it with a constant on one side after separation and integration. Then, have a constant on each side. What do you notice when you solve for $y$ in both approaches? $y = c e^{x^2/2}$. $\endgroup$ – Amzoti Jan 16 '14 at 18:21
  • $\begingroup$ I think I understand, thank you! $\endgroup$ – user121860 Jan 16 '14 at 18:29
  • $\begingroup$ I haven't taken calculus any calculus courses for about two years now and I've forgotten some of the rules, such as combining the constants. The professor told me that when we integrated both constants would be C1, ie. the same, and as such would end up canceling each other out by way of subtraction if we wanted to integrate more. So he wrote C1 on the left hand side, but it seems to make sense if he actually wanted to write C (the inclusion of both constants). $\endgroup$ – user121860 Jan 16 '14 at 18:33
  • $\begingroup$ Though if you don't mind, examples are always helpful. Thank you for your time and help! $\endgroup$ – user121860 Jan 16 '14 at 18:33
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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 $w = 2$.

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