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A variable $a$ changes over time according to:

$\frac{da}{dt}=\frac{I}{\phi}a^{\gamma}$

where $I$ and $\phi$ are constants. $\gamma<0$ is also a constant.

I now want to find out the value of $a$ at $t=1$ if I know that $a_{t=0}=a_{0}$.

With the following two sources, Link 1 and Link2, I came up with this solution:

$a_{t}=\left[\left(\frac{I}{\phi}t+C\right)(1-\gamma)\right]^{\frac{1}{1-\gamma}}$

where the integration constant $C$ can be pinned down at $a_{0}$ such that: $a_{1}=\left[\left(\frac{I}{\phi}+\underbrace{\frac{(a_{0})^{1-\gamma}}{(1-\gamma)}}_{C\, at\, a_{0}}\right)(1-\gamma)\right]^{\frac{1}{1-\gamma}}$

I'd be really grateful if anyone can tell me whether this is right or completely wrong?

Fabian
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  • not sure if this is relativity or unrrlated to the title. –  Jan 21 '20 at 01:13
  • @RoddyMacPhee changed the title – Fabian Jan 21 '20 at 01:16
  • but acceleration is a function of velocity namely the slope of the tangent to the change in velocity in a velocity time diagram. –  Jan 21 '20 at 01:20
  • @RoddyMacPhee Actually, this is a problem I have in econ, ... It's just that the various "velocity/acceleration" problems on StackExchange seem to be closest to mine ... I'm sorry for the confusion. – Fabian Jan 21 '20 at 01:23
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    Broadly, this is a nonlinear, 1-dimensional first order differential equation. Many have explicit solutions, but in general they don't. As for whether that solution is correct, differentiate with respect to $t$ to see if the differential equation is satisfied. Plug in $0$ to see if you chose the right constant. – Charles Hudgins Jan 21 '20 at 01:24
  • @CharlesHudgins Thanks a lot for your answer. I think $C$ is right. However, I'm not so sure about the derivative with respect to time. I get something like $\frac{da}{dt}=\frac{I}{\phi}a_{0}^{\gamma}$. But is this right if it is $a_{0}$? – Fabian Jan 21 '20 at 01:56
  • If you get $a_0$ then something is wrong. This would be saying that $a(t) = \frac{I}{\phi} a_0^{\gamma} t + C$ for some constant $C$. My guess is that you substituted something incorrectly. Note that we aren't free to choose $C = a_0$, since $a_0$ is defined by $a_0 = \left[\left(\frac{I}{\phi} (0) + C\right)(1 - \gamma)\right]^{\frac{1}{1 - \gamma}}$. Indeed, $C$ will be uniquely determined by the value of $a_0$. – Charles Hudgins Jan 21 '20 at 02:40

1 Answers1

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So, as you mentioned physics in your title, I will solve it in a physicist way, i.e I will not be much rigorous in my steps. Hopefully this will help. First of all, let us call $I/\phi = \alpha$ and, because $\gamma <0$, we can write $\gamma = -|\gamma|$. With these notations, your equation becomes: \begin{eqnarray} \frac{da}{dt} = \alpha a^{-|\gamma|} =\alpha \frac{1}{a^{|\gamma|}} \tag{1}\label{1} \end{eqnarray} To simplify even more my notation, I will write write $|\gamma|$ as $\gamma$ from now on. Here's where the physicist point of view takes place: write (not rigorously) equation \ref{1} as: \begin{eqnarray} a^{\gamma} da = \alpha dt \Rightarrow \int a^{\gamma}da = \alpha\int dt \Rightarrow \frac{1}{\gamma+1}a^{\gamma+1}= \alpha t + C \tag{2}\label{2} \end{eqnarray} Thus, from (\ref{2}) we obtain: \begin{eqnarray} a = a(t) = \bigg{[}\bigg{(}\frac{I}{\phi}t+C\bigg{)}(\gamma+1)\bigg{]}^{\frac{1}{\gamma+1}} \end{eqnarray} where $C$ is completely defined by $a(t=0)$ by equation (\ref{2}).

IamWill
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  • There's nothing non-rigorous about this. Any mathematician would accept this as a proof. The only room for error is that we have to be sure $\gamma \neq -1$, since then the given antiderivative wouldn't be defined. But this isn't a problem since we've assumed (after redefining) that $\gamma > 0$. – Charles Hudgins Jan 21 '20 at 08:06
  • Thanks a lot! :-) – Fabian Jan 21 '20 at 11:05
  • @CharlesHudgins you are completely right! I assumed $\gamma \neq -1$ because of his first answer, but I should have mentioned this assumption. Thanks a lot for your comment! – IamWill Jan 21 '20 at 11:24