Solving a Simple Differential Equation: Comparing Different Approaches

Click For Summary
SUMMARY

The discussion centers on solving the differential equation dV/dT = V/T using two distinct methods: separation of variables and direct integration. The separation method leads to the equation ln(V) = ln(T) + c, while the integration method results in V = C_1 T, where C_1 is derived from initial conditions. Both approaches yield the same relationship between volume and temperature, demonstrating their equivalence through the derivation of constants from boundary conditions. The final relationship established is V_f/T_f = V_i/T_i, confirming the consistency of both methods.

PREREQUISITES
  • Understanding of differential equations
  • Familiarity with separation of variables technique
  • Knowledge of integration and logarithmic functions
  • Basic concepts of initial and boundary conditions in mathematical modeling
NEXT STEPS
  • Study the method of separation of variables in differential equations
  • Explore the application of definite integrals in solving differential equations
  • Learn about initial and boundary value problems in mathematical analysis
  • Investigate the implications of the ideal gas law in thermodynamic equations
USEFUL FOR

Students and professionals in mathematics, physics, and engineering who are solving differential equations and seeking to understand different methods of solution, particularly in relation to thermodynamic systems.

zezima1
Messages
119
Reaction score
0
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?
 
Physics news on Phys.org
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,
$$
\Rightarrow \frac{V_f}{T_f}=\frac{V_i}{T_i}
$$
 

Similar threads

Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
10
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
7
Views
2K