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I'm being asked the following question:

If $A(t)$ is the amount of the investment at time $t$ for the case of continuous compounding, write a differential equation satisfied by $A(t)$.

(The initial information given is that the initial value is \$2000 and the interest rate is 4%. I was previously asked to find the value of the investment at the end of 10 years if the interest was compounded at a varying number of times each year, which I did with no problem.)

When I completed the first part of the exercise, in the case of continuously compounded interest, I used the formula $A(t)=(2000)\cdot e^{(0.04)(t)}$

Referencing my textbook, I found a few notes that I think are relevant.

$y'(t) = C(ke^{kt}) = k(Ce^{kt}) = ky(t)$

Using this information, I tried writing the equation: $\frac{dA}{dt}=(0.04)(2000e^{0.04t})$ since the interest rate is k = 0.04, and C = 2000. This answer was not accepted however. What am I doing wrong?

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You wrote it yourself, the differential equation is $$A'(t) = k A(t),$$ and all that remains is to pick out $k$. Of course, your solution is also a d.e. for $A$, it's just not interesting in that it doesn't express $A'$ in terms of $A$ alone. Indeed, the above d.e. shows that the relationship between $A'$ and $A$ is independent of time.

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  • $\begingroup$ Wait, then why wasn't my answer accepted. Are you saying that I should leave k as a variable, instead of plugging in the value of 0.04? $\endgroup$
    – Bassinator
    Oct 19, 2015 at 21:33
  • $\begingroup$ No, the point is that (presumably) the desired d.e. is $A'(t) = k A(t)$ (for an appropriate value $k$). $\endgroup$
    – Travis Willse
    Oct 19, 2015 at 21:35
  • $\begingroup$ An appropriate value k being 0.04, correct? So the answer would be $A'(t) = (0.04)A(t)$ $\endgroup$
    – Bassinator
    Oct 19, 2015 at 21:37
  • $\begingroup$ Yes, in fact, since the interest rate is $4\%$, $k$ is exactly $0.04$. $\endgroup$
    – Travis Willse
    Oct 19, 2015 at 21:41
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    $\begingroup$ I think you'll have to take that up with the relevant programmer. $\endgroup$
    – Travis Willse
    Oct 19, 2015 at 22:01

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