Maths help please! (+REP) C4

[Matthew]

I'm sorta part of the way there with this question but I can't seem to get what is the correct answer

I'm probably just missing something really obvious so if I'm being a moron just say (A)

(BTW, take # to be 'theta')

d#/dt = -k(#+20)

express # in terms of t, k and an arbitary constant.


Okay so I've assumed that this is a 'differential equations' question.

This what I've done:

(Take | to mean 'the integral of')

| (1/#+20) d# = | (-k) dt

so ln(#+20) = -kt + c

#+20 = e^(-kt + c)

# = e^(-kt + c) - 20

-------------------

Now this is NEARLY the correct answer. The correct answer is # = -20 + ce^(-kt)

... so where have I gone wrong?

(If anyone can make sense of that, its hard to type this stuff out)
@Kardan; @Chippiewill; as I know you're good at maths (A)

[Kardan]

Ok, your final step is correct, we have:

theta = e^(-kt+c) - 20

Now, we use the exponential rules, where e^(a+b) = e^a x e^b, so we use this here to get:

theta = (e^-kt x e^c) - 20

Now, e^c is simply a constant, so we can call this constant A, so we have:

theta = Ae^-kt - 20, as required

[Matthew]

Ok, your final step is correct, we have:

theta = e^(-kt+c) - 20

Now, we use the exponential rules, where e^(a+b) = e^a x e^b, so we use this here to get:

theta = (e^-kt x e^c) - 20

Now, e^c is simply a constant, so we can call this constant A, so we have:

theta = Ae^-kt - 20, as required

Oh I see... Can't believe I missed that

Thanks for the quick reply, +REP