Solve ODE: Integrating Factor Solution w/ Mike's Query

[bsodmike]

Homework Statement

This is a solution detailed in a past paper,

http://stuff.bsodmike.com/sensors_pastpaper.png

Homework Equations

Utilise an integrating factor to solve as detailed by myself https://www.physicsforums.com/showthread.php?t=283610".

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

 

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

 

[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,

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