Given a varying force of interest where $\delta=kt$, where an unknown amount of money will double itself, find an expression for K

Question

The book says the answer is $\frac{ln(2)}{50}$ but I do not see how unless this a typo or I am missing something very fundamental

It is known that an amount of money will double itself in 10 years at a varying force of interest of $\delta=kt$. Find an expression for k.

Well the double of $A(t)=pe^{kt}$ the principals cancel each other out and since $e^{5k}$ is half of $e^{10k}$:

$2e^{5k}=e^{10k} \to ln(2)+5k=10k \to ln(2)=5k \to k=\frac{ln(2)}{5}$

Answer 1

Hints: Actually the general formula is $$A(t)=A(0)e^{\int_0^t \delta(\tau)\, d\tau}.$$

In this case, $\delta(\tau)=k\tau$, so $\int_0^t \delta(\tau)\, d\tau = \frac{1}{2}kt^2$. Hence in this case, we have $$A(t)=A(0)e^{\frac{1}{2}kt^2}.$$ Now solve for $k$ if $A(10)=2A(0)$.