Differential equations with applications and historical notes 3rd edition chapter 4 problem 3 (a) Problem 3.
A bright young executive with foresight but no initial capital makes
constant investments of D dollars per year at an annual interest rate of
100k percent. Assume that the investments are made continuously and
that interest is compounded continuously.
(a) Find the accumulated amount A at any time t.
Answer in the book :
$A = D(e^{kt}-1)/k$
I fail to get this answer with my calculation.
I attempted to solve this problem with A (accumulated investment result)
as a function of x ( money invested per year. $dx/dt=D$ ) and t (time as year)
Within the textbook, Example 1 Continuously compounded interest (page 20),
Quote,

More generally, if P is money deposited in a bank and if the interest
rate is 100k percent (k = 0.06 for 6 percent),
and if this interest is
compounded n times a year, then after t years the accumulated amount
is
$A=P(1+k/n)^{nt}$ Since this is continuous investment, n approaches to infinity.
Hence $A=Pe^{kt}$

I attempted A as a function of x and t.
$A=f(x,t)$
and change of money invested per year $dx/xt=D$
And since $A=xe^{kt}$ ,
$A=Dte^{kt}$
My answer satisfies initial condition. ($A=0$ when $x=0$ and $A=0$ when $t=0$)
But it doesn't match the answer in the textbook.
 A: As per user request, I'll post a complete solution here. Based on my comment, let's start with:
$$\frac{\partial A}{\partial t} = D + kA$$
$$\frac{\partial A}{\partial t} -kA= D $$
The theory behind an integrating factor is that we multiply each term by $\rho$ and get
$$\frac{\partial A}{\partial t} \rho(t) -kA \rho(t)= D \rho(t)$$
and then try to use the chain rule in reverse where
$$\frac{\partial A}{\partial t} \rho(t) -kA\rho(t) = \frac{\partial} {\partial t}(A *\rho(t)) $$ By expanding the right hand side with the chain rule, we get
$$\frac{\partial A}{\partial t} \rho(t) -kA\rho(t) = \frac{\partial A}{\partial t} \rho(t)+\rho^{'}(t)A $$
Thus,
$$\rho^{'}(t) A = -k A \rho(t)$$
$$\frac{\rho^{'}(t)}{\rho(t)}=-k $$
The solution to this equation is just
$$\rho(t) = e^{-kt}$$
Going back to our initial equation, we get
$$\frac{\partial A}{\partial t} e^{-kt} -kA e^{-kt}= D e^{-kt}$$
Integrating each side, we get
$$A e^{-kt} = \frac{-De^{-kt}}{k} + C$$
$$A = \frac{-D}{k} + C e^{kt}$$
To meet the initial condition, we set $C=\frac{D}{k}$ and get
$$A = \frac{De^{kt}}{k} -\frac{D}{k} = \frac{D(e^{kt}-1)}{k}$$
