# ODE Solution Query

1. Jan 12, 2009

### bsodmike

1. The problem statement, all variables and given/known data

This is a solution detailed in a past paper,

2. Relevant equations
Utilise an integrating factor to solve as detailed by myself here.

3. The attempt at a solution

It can be said that,

$$\dfrac{d\theta}{dt}+\dfrac{\theta}{\beta}=\dfrac{\theta_m}{\beta}$$

Hence, the solution by employing an integrating factor would yield,

$$e^{t/\beta}\theta=\dfrac{\theta_m}{\beta}\int{e^{t/\beta}\,dt}$$

$$\theta=\dfrac{\theta_m}{\beta}+Ce^{-t/\beta}$$

The problem is that I'm having the extra 'beta' term as shown above. Any ideas?

Thanks!
Mike

2. Jan 12, 2009

### bsodmike

A more direct approach, solving the homogenous part:

$$\int{\dfrac{1}{\theta}\,d\theta}=\int{-\dfrac{1}{\beta}\,dt}$$

$$ln(\theta)=-\dfrac{t}{\beta}+C$$

$$\theta=e^{-t/\beta+C}$$

Now, my memory here is fuzzy but $$e^{-t/\beta+C} = e^{-t/\beta}+e^C$$ ??

3. Jan 12, 2009

### bsodmike

Ah, another small blunder, of course!

Now, $$\int{e^{t/\beta}\,dt}=\beta e^{t/\beta}+C$$. thus, the 'beta's cancel,

$$e^{t/\beta}\theta=\theta_m e^{t/\beta}+C$$

Dividing by $$e^{t/\beta}$$, yields,

$$\theta=Ce^{-t/\beta}+\theta_m$$