# Write a differential equation for compound interest?

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? ## 1 Answer 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. • 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? Oct 19 '15 at 21:33 • No, the point is that (presumably) the desired d.e. is$A'(t) = k A(t)$(for an appropriate value$k$). Oct 19 '15 at 21:35 • An appropriate value k being 0.04, correct? So the answer would be$A'(t) = (0.04)A(t)$Oct 19 '15 at 21:37 • Yes, in fact, since the interest rate is$4\%$,$k$is exactly$0.04\$. Oct 19 '15 at 21:41
• I think you'll have to take that up with the relevant programmer. Oct 19 '15 at 22:01