I'm having a physics exercise, but the question is more of math.
Assuming I have the following constants: $m_1, m_2, \alpha, V_0$ and two variables: $v, t$. (v as velocity).
I reach the following equation, and I wish to extract $v(t)$ out of it: $$ (m_1+m_2-\alpha\cdot{t})dv=(\alpha\cdot{V_0})dt$$
Usually the lecturer does the following trick in order to get $v(t)$: $$dv = \frac{\alpha\cdot{V_0}}{{m_1+m_2-\alpha\cdot{t}}}dt$$
$$\int_{0}^{v}{dv} = \int_{0}^{t}{\frac{\alpha\cdot{V_0}}{{m_1+m_2-\alpha\cdot{t}}}dt}$$
$$v = V_0 \cdot ln(\frac{m_1+m_2}{m_1+m_2-\alpha\cdot{t}})$$
And that's the correct result.
But why when I decide not to divide before integration gives me a total different result?
$$ (m_1+m_2-\alpha\cdot{t})dv=(\alpha\cdot{V_0})dt$$
$$\int_{0}^{v}(m_1+m_2-\alpha\cdot{t})dv = \int_{0}^{t}(\alpha\cdot{V_0})dt$$
$$(m_1+m_2-\alpha\cdot{t}) v = (\alpha\cdot{V_0})t$$
$$v = \frac{\alpha\cdot{V_0}}{m_1+m_2-\alpha\cdot{t}}$$
It's completely different.
How do this difference occur, and how can I know what the rules are?
Thank you.
You're solving a differential equation by separating the variables, but that technique is only valid if you actually separate the variables, and get all the $v$s on one side and all the $t$s on another side.
You're not doing that; instead you create the integral $$ \int_0^v (m_1 + m_2 - \alpha t)\,dv $$ which contains both a $t$ and a $v$.
In the unknown solution to the differential equation you're trying to discover something about (say, a formula for it), $t$ and $v$ depend on each other in some particular way -- they're not independent of each other. But when you evaluate the above integral you're deciding to forget about that and treat $t$ as a constant, independent of $v$. Therefore you have no reason to expect that the outcome of the integral will match the integral to its right, because you're pretending that the $t$ everywhere on the curve is equal to the final $t$ instead of varying with $v$.