Question

Help with modelling using DE. Will ask on next post.

Answer 1

If \[ (k_1 + k_2t)\frac{dq}{dt} + \frac{q}{C} = E(t) \]for some known positive constants k1 and k2 and if E(t) = Eo and q(0) = q0, where Eo and q0 are constants, show that: \[ q(t) = E_{0}C + (q_{0} - E_{0}C) (\frac{k1}{k_1+k_2t})^{\frac{1}{Ck_2}} \]

Answer 2

I solved it making use of the fact that it's separable, but my solution is: \[q(t) = c_1 e^{\frac{t}{CR}} + E_{0}C\]With \( R = k_1 + k_2t \).
I can't get that solution to be like the above :-(

Answer 3

Typo, should be \( \LARGE e^{\frac{{-t}}{CR}} \)

Answer 4

I think that maybe solving for c2 from the original solution: \[\Large q(t) = -e^{\frac{-t + c_2}{CR}} + E_0 C \]is the way, but I didn't find a way to solve it like the question.

Answer 5

How do you integrate \(E(t)\)?

Answer 6

Ohh \(E(t)\) is given \(E_0\). Sorry.

Answer 7

\[\frac{Cdq}{CE_0-q} = \frac{dt}{k_1 + tk_2}\]\[\implies \int_0^q \frac{Cdq}{CE_0-q} = \int_0^t\frac{dt}{k_1 + tk_2}\]
\[-C \ln \frac{CE_0 - q}{CE_0 - q_0} = \frac1{k_2}\ln\frac{k_1 + tk_2}{k_1}\]
\[\implies \frac{CE_0 - q}{CE_0 - q_0} = \left(\frac{k_1 + tk_2}{k_1}\right)^{-\frac1{Ck_2}}\]

Answer 8

\[q(t) = CE_0 - \left(CE_0 - q_0\right) \left(\frac{k_1 + tk_2}{k_1}\right)^{-\frac1{Ck_2}}\]

Answer 9

@Ishaan94 Thanks mate :-) Makes sense. I was doing the wrong integral then.

Answer 10

You're welcome :-)