a country accepts A people a year from other countries, we know that : $x'(t) = −0.03x(t) + A$ while $x(0)=16m$ and $x(-10)=15m$ find A I have the following question :
In order to have population increasing, a country accepts A people a year
from other counntries, and the number x(t) of people in a country
changes by equation $$x'(t) = −0.03x(t) + A$$
where t is time in years. while $x(0)=16m$ and $x(-10)=15m$ find A.
This is what I did :
from the equation I know that each year the population of the country decrease in $0.03$
I tried to use the data, since $t$ is in years
$$x(t_0)*1.03=x(t_0+1)$$
Since we wish in 10 years the total popluation increase in million people therefore :
$$x(t_0)*1.03=x(t_0+1)+100,000$$
I get the feeling my solution is wrong, But I don't really find other ways to approach this.
Any ideas? I'll be glad if someone could tell me why my solution fails.
Thanks in advance!
 A: The provided information is
$$x'(t) = −0.03x(t) + A \tag{1}\label{eq1}$$
$$x(0) = 16m \; \; \text{ and } \; \; x(-10) = 15m \tag{2}\label{eq2}$$
Your attempted solution implies that the value of $x(t)$ only changes once per year. If so, I would assume something like a linear difference equation would have been provided. However, since \eqref{eq1} involves a derivative, it seems $x(t)$ is meant to be considered a continuous function instead. If this assumption is correct, then note that \eqref{eq1} shows an exponential decay along with (I assume) a fixed increase value. Note the equation implies using something where the derivative is a multiple of itself, which the exponential function satisfies. A general solution would be of the form
$$x(t) = ae^{bt} + c \tag{3}\label{eq3}$$
for real constants $a, b$ and $c$. Substituting \eqref{eq3} into \eqref{eq1} gives
$$abe^{bt} = -0.03ae^{bt} - 0.03c + A \tag{4}\label{eq4}$$
Since this must hold for all $t$, this means the coefficients of the $e^{bt}$ and constant terms must be the same on both sides of the equation, i.e.,
$$ab = -0.03a \; \; \Rightarrow \; \; a = 0 \; \text{ or } \; b = -0.03 \tag{5}\label{eq5}$$
$$0 = -0.03c + A \; \; \Rightarrow \; \; c = \frac{A}{0.03} \tag{6}\label{eq6}$$
Note that if $a = 0$, then \eqref{eq3} shows that $x(t)$ is a constant function, but \eqref{eq2} shows it's not, so $b = -0.03$ must be the case instead in \eqref{eq5}. Thus, \eqref{eq3} becomes
$$x(t) = ae^{-0.03t} + \frac{A}{0.03} \tag{7}\label{eq7}$$
Using the provided values, you can now substitute $t = 0$ and $t = -10$ into \eqref{eq7} to get $2$ equations in the $2$ unknowns of $a$ and $A$. You can then solve these equations to get $A$. I trust you can do these remaining calculations yourself.
