Question

I was wondering if I could solve a differential equation from my textbook in Chemical kinetics but I wasn't sure exactly how to tackle it - and if it's easy to solve on paper at all. Here is my equation:

Answer 1

\[\frac{ dB }{ dt }=k1*A0*\exp(-k1*t) - k2*B\]

Answer 2

I can't separate the variables, so I guess there is some other trick to get the result.

Answer 3

This is a linear differential equation in B and t

Answer 4

Do you know how to solve them?

Answer 5

The only method we've used so far in our Chemistry classes is to separate the variables.

Answer 6

today u will be learning another trick to solve it then :)

Answer 7

familiar wid 'product rule' of differentiation right ?

Answer 8

oh, thank you :)

Answer 9

A linear differential equation is of the form

y' + k(t)y = m(t)

this corresponds to that form doesn't it?

Answer 10

I am, yeah

Answer 11

We will be using the product rule to solve this equation

Answer 12

okay, listening :)

Answer 13

So your equation is of the form I just wrote above isn't it?

B' + k2*B = a0 * k1 * exp(-k1*t)

Answer 14

yes

Answer 15

To solve this kind of equation we multiply it with something we call an integrating factor. After that we can use the product rule simplify the equation.

In the form y' + k(t)y = m(y)

the integrating factor is

exp(integral(-k(t)dt))

okay?

Answer 16

okay

Answer 17

In your case the factor is

exp(integral(k2dt)) = exp(k2*t)

yes?

Answer 18

ah, yeah, sorry I'm that slow

Answer 19

Hey never mind..

Now please multiply the equation by that integrating factor and write it here

Answer 20

just a question - so, in the equation we have k(t)y to correspond to k2B... and in the integrating factor we have the integral of k(t)... when k2 is just a constant?

Answer 21

in this case -k2 is your k(t) and B is your y

so IF is exp(k2*t)

Answer 22

get it?

Answer 23

ah, right - we assume that the k(t) has t to the power 0, so only a constant remains

Answer 24

so, let me write the whole equation now

Answer 25

we have


\[\frac{ dB }{ dt }*e ^{-k2t} + k2Be ^{-k2t}=k1A0.e ^{-(k1+k2)t}\]

Answer 26

The IF is exp(k2*t)
correct that

and then see if you can recognize the LHS as

d(B*exp(k2*t))/dt

Answer 27

oops, yeah, without the minus in the exponent

Answer 28

\[\frac{ dB }{ dt }*e ^{k2t} + k2Be ^{k2t}=k1A0.e ^{(k2-k1)t}\]

Answer 29

get the dt to the other side now

you have

B*exp(k2*t) = k1*a0*exp((k2-k1)t)dt
\
isnt it?

Answer 30

yes, but how about the t on the left side? or it's okay for it to stay there?

Answer 31

can you solve it from here?

Answer 32

let me write it on paper to see better

Answer 33

Ugh, I thought I'd got it right but it turned out that there's some extra k2 in the answer I got compared to the one in the textbook. I'm attaching my solution. Help highly appreciated!

Answer 34

when you integrated the lhs of the equation you pulled out a k_2 unnecessarily. Use the fundamental theorem of calculus; you are integrating a derivative here

Answer 35

@TuringTest so in the left hand side, we assume e^k2*t is a constant and just take it out like this?

Answer 36

you don't assume it's a constant, the integration is implicit in using the integrating factor

Answer 37

lhs\[B'e^{k_2t}+K_2Be^{k_2t}=(Be^{k_2t})'\]so when you integrate it you just get\[Be^{k_2t}+C\]

Answer 38

oh, got it now, thank you very much!

Answer 39

welcome :)