Dog chases his tail ! - "parametric differential/Integral equation"..?

Question

I have the following situation where I am interested in the function $m(t)$

$$ \frac{dm}{dt}=4T(t)^{3}+T(t)^{2} $$
$$ T(\tau)=T_{0}-\int_{0}^{\tau}(\frac{dm}{dt})dt*Q_{S} $$

Is there a way to solve it analytically using Mathematica?
Thanks for your help in advance
Andi

ps.: I guess I just don't know what to look for

Take a look a DSolve - and maybe get familiar with Mathematica a bit. — gwr, Commented May 10, 2016 at 9:46
The notation is certainly confusing me. If I interpret $\frac{dm}{dt}(T)$ as the derivative of $m(T)$ with respect to $t$, then that's $0$; if that's the product of $\frac{dm}{dt}$ and $T$, then the right hand side is effectively a constant. So, what then? — J. M.'s missing motivation, Commented May 10, 2016 at 9:53
I hope it is more clear after that edit ... - m is the mass, T the temperature - mass changes depending on the Temperature. The temperature drops by losing mass (Sublimation) — G.Andi, Commented May 10, 2016 at 9:58
What do you get if you differentiate your second equation with respect to $t$? — J. M.'s missing motivation, Commented May 10, 2016 at 10:02
something arbitrary- but I just realized, that the Integral is written wrong - next edit ... — G.Andi, Commented May 10, 2016 at 10:06

kglr · Accepted Answer · 2016-05-10 11:22:16Z

5

Following up on @J.M.'s observations in the comments, differentiate

$$m(t)=m(0)-\frac{T(t)-T_0}{Q_S}$$

to get $$\frac{dm}{dt}=-\frac{T'(t)}{Q_S}$$

Combine with

$$\frac{dm}{dt}=4 \, T(t)^{3}+T(t)^{2}$$

to get a differential equation in T[t]:

$$ T'(t)=-\text{Qs} \left(4 \, T(t)^3+T(t)^2\right)$$

Use DSolve with initial value T[0] == t0:

DSolve[{T'[t] == -Qs ( 4 T[t]^3 + T[t]^2), T[0] == t0}, T, t]

Mathematica graphics

answered May 10, 2016 at 11:14

kglr

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$\begingroup$ Thank you very much @kglr, obviously it was a mathematics/approach problem :-) - and my bachelor studies were too long ago, ... maybe then I could have handled it better $\endgroup$
– G.Andi
Commented May 10, 2016 at 11:32
$\begingroup$ @G.Andi, my pleasure. You were almost there:) Thank you for the accept. $\endgroup$
– kglr
Commented May 10, 2016 at 11:35
$\begingroup$ now i will do this approach with my real dm/dt function which is a hell compared to $\frac{dm}{dt}=4 \, T(t)^{3}+T(t)^{2}$ :-) $\endgroup$
– G.Andi
Commented May 10, 2016 at 11:36

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