Simple differential equation

  • Thread starter zezima1
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  • #1
zezima1
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So I have the differential equation:

dV/dT = V/T

I solve it with separation and get:

ln(V) = ln(T) + c

where c has to be figured out from initial conditions.

Now this is how I am used to solving the equation. My book though does it differently. It simply integrate both sides from Vi to Vf or Ti to Tf. Why are these approaches the same? And how do you show that from the method I use?
 

Answers and Replies

  • #2
SadScholar
36
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When they put definite boundary conditions on the integral, the constant was naturally solved for. Here's how you would go about it from the other method you mentioned. I'm going to assume the exercise in the book is using T for temperature and V for volume, and I will refer to them as such, but replace my words as necessary.
$$
\int \frac{\mathrm{d}V}{V}=\int \frac{\mathrm{d}T}{T}\\
\Rightarrow \ln(V)=\ln(T)+C_0\text{, where }C_0\text{ is just some arbitrary constant.}\\
\Rightarrow V=e^{\ln(T)+C_0}=Te^{C_0}\equiv C_1 T\text{, where }C_1\text{ is some new arbitrary constant.}
$$
Now we have
$$
V=C_1T\text{.}
$$
We assume that at some initial temperature, there will be some initial volume.
$$
\Rightarrow V_i=C_1 T_i
\Rightarrow C_1=\frac{V_i}{T_i}
\Rightarrow V=\frac{V_i}{T_i}T
$$
This equation will have to hold for the final temperature and volume as well,
$$
\Rightarrow \frac{V_f}{T_f}=\frac{V_i}{T_i}
$$
 

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