Maths - Differential Equations

[Mansize_tissue]

Any Mathematicians feel like giving me a hand?

Because I'm lazy, let's say theta = x

A cold liquid at temperature x degrees Celsius, where x < 20, is standing in a warm room. The temperature of the liquid obeys the differential equation

dx/dt = 2(20 - x)

where t is measured in hours.

Find the general solution of this differential equation.

Here are my workings:

--> dx/dt = 2(20 - x)

--> Int(1/[40 - 2x]) dx = Int(1) dt

--> 1/2 ln(40 - 2x) = -t

--> 40 - 2x = e^-2t

--> x = 20 - (e^-2t)/2


Now the actual answer apparently is x = 20 - Ae^-2t, where I'm assuming A is a constant.

Can anyone see where I've slipped up and where I should've added in the constant?

Thank you in advance.

 

[Mansize_tissue]

Thanks guys. I knew the constant had to go in somewhere, I just didn't know how to incorporate it really.

--> dx/dt = 2(20 - x)

--> Int(1/[40 - 2x]) dx = Int(1) dt

--> 1/2 ln(40 - 2x) + ln(c) = -t

--> 1/2 ln(c[40 - 2x]) = -t

--> c(40 - 2x) = e^-2t

--> x = 20 - (e^-2t)/2c

I still can't do it!

Maybe if it was -ln(c) I'd be a bit closer to the answer, but there's still the rogue 1/2 in there at the end.

 

[Mansize_tissue]

Thanks some more!

--> dx/dt = 2(20 - x)

--> Int(1/[40 - 2x]) dx = Int(1) dt

--> 1/2 ln(40 - 2x) = -t + c

--> ln(40 - 2x) = -2t + c

--> 40 - 2x = e^(-2t + c)

--> x = 20 - [e^(-2t + c)]/2

--> x = 20 - [e^(-2t) * e^c]/2

--> x = 20 - [Ae^(-2t)]/2

Is that right, Ricochet J? Now what about the 1/2?

 

[Mansize_tissue]

You missed the negative sign when integrating:

1/(20-x) dx = 2 dt

It should be:

-ln|20-x| = 2t + c


Thank you very much! I still can't see why my method doesn't work. The only thing I did differently was expanding the brackets at the beginning. :S

 

[Mansize_tissue]

Your method did work. As I said before integration constants are arbitrary. You can relabel them all you like. Look back at your method and relabel the constant. You say yourself "Maybe if it was -ln(c) I'd be a bit closer to the answer, but there's still the rogue 1/2 in there at the end." Well...make it -ln(c) and stick a 1/2 in there. You can do that because it is arbitrary.

I suppose that would work, thanks. But if I can just disregard the 1/2 I'm changing the equation. I mean, Ae^-2t is not the same as (Ae^-2t)/2, is it? Had you not have said anything I wouldn't have known which numbers I could just get rid of and replace with the constant, A, and which to leave.

 

[Mansize_tissue]

Ahhh, thank you very much guys. I really appreciate your help.

I didn't realise they'd both reach the same result, Visage, thank you for explaining that to me. The only problem with doing it my was is the value for A would differ from what they're looking for, despite the equation being identical. Oh wait, I'm asked to find the particular solution anyway and not just right down what A is, so they'll see that the equation is the same anyway.