Differential equations question I can't get my head around (1 Viewer)

[SDES]

I don't even do HSC, so I'll just try posting this here. Hope it's not too big of a question to ask here:

Consider the elements A, B and C. Initially, there is 800g of A, 100g of B and nothing for C. Each second:

• 4% of A turns into B
• 2% of B turns into C

Hence, the following is true (where a=A, b=B):

• da/dt = -0.04a
• db/dt = 0.04a - 0.02b

In finding the equation for the value for b, we are given this hint:

(db/dt) + 0.02b = (1/e^[0.02t])*(d/dt)(be^[0.02t])

​

From this, I have so far found the following:

a=800 @ t=0; a=768 @ t=1

da/dt α a --> da/dt = ak ---working out---> a = 800e^(-0.04t)

​

Given that db/dt = 0.04a - 0.02b, db/dt + 0.02b = 0.04a, therefore:

db/dt + 0.02b = 0.04(800e^-0.04t) = 32e^(-0.04t)
​

Meaning that the right hand side of the hint is essentially equal to 32e^(-0.04t).Now I'm kinda stuck at a dead end.
Thanks in advance for any help.
I'm pretty sure there's some equation typer built into forum posts - sorry I couldn't find it.

 

[khorne]

That's the LaTeX for it

 

[Trebla]

db/dt + 0.02b = 0.04(800e^-0.04t) = 32e^(-0.04t)
Meaning that the right hand side of the hint is essentially equal to 32e^(-0.04t).Now I'm kinda stuck at a dead end.
Thanks in advance for any help.
I'm pretty sure there's some equation typer built into forum posts - sorry I couldn't find it.

Assuming everything you've done so far is correct:
db/dt + 0.02b = 32e^-0.04t
We use the technique involving an "integrating factor". Which is e^{∫0.02 dt} = e^0.02t. Multiplying the equation with this factor gives:
e^0.02tdb/dt + 0.02e^0.02tb = 32e^-0.02t
Now the LHS is can actually be written using the reverse product rule.
Observe that d[e^0.02tb]/dt = e^0.02tdb/dt + 0.02e^0.02tb
i.e. d[e^0.02tb]/dt = 32e^-0.02t
=> e^0.02tb = -1600e^-0.02t + c
Using the initial conditions to evaluate c and then make b the subject

 

[SDES]

3unitz: I understand every single step you have posted, but what I am having problems with is the integration.

32e^(-0.02t) = (d/dt)[be^(0.02t)]

integrate both sides:
(32e^-0.02t)dt = d(be^0.02t)
∫(32e^-0.02t)dt = ∫(be^0.02t)d
(32e^-0.02t)/(-0.02) + C = be^0.02t
-1600e^-0.02t + C = be^0.02t


This is mainly because I'm not that familiar with 'd/dt' and its implication.
The bolded steps I have added, are they correct? Integrating be^0.02t results in itself, with no +C? Or would the integral of be^0.02t instead be something like (be^0.02t)/0.02?
Very confused right now, as what you show me is correct at the end (agrees with known values), but the working out puzzles me.


Trebla: from your final line,b = -1600e^-0.02t + c, and from the initial conditions of 'b', I get the equation b = -1600e^-0.04t+ 1700. I know from using a spreadhseet that b is about 198 at t=100, and using this equation for 'b', I get something like 1670.69.

 

[SDES]

Ah, my mistake, on both accounts. I should have realised d/dt just referred to the deriving of a function. And also the 'combining' of the two Cs. The last situation was just silly for me, since we were taught to just leave C alone and not do anything to it in most situations. Oh well.

Thanks for the very speedy response guys, the help is much appreciated.