Answer:
[tex]\bf x(t)=10(1-e^{-0.005*t})[/tex]
Step-by-step explanation:
The differential equation
[tex]\bf \displaystyle\frac{dx}{dt}=0.005(10-x)[/tex]
can be solved by separation of variables. Write the equation as
[tex]\bf \displaystyle\frac{dx}{10-x}=0.005dt[/tex]
Integrate on both sides
[tex]\bf \int\displaystyle\frac{dx}{10-x}=\int0.005dt\Rightarrow -ln(10-x)=0.005t+C\Rightarrow\\\\\Rightarrow ln(10-x)^{-1}=0.005t+C\Rightarrow (10-x)^{-1}=e^{0.005t}e^C[/tex]
where C is a constant.
[tex]\bf e^C[/tex] is also a constant and we will keep calling it C, (there is no reason to change the letter). We have then
[tex]\bf (10-x)^{-1}=Ce^{0.005t}\Rightarrow \displaystyle\frac{1}{10-x}=Ce^{0.005t}\Rightarrow 10-x=\displaystyle\frac{1}{Ce^{0.005t}}\Rightarrow\\\\\Rightarrow x(t)=10-(1/C)e{-0.005t}[/tex]
(1/C) is a constant, and for the same reason we will keep calling it C. So the general solution is
[tex]\bf x(t)=10-Ce^{-0.005t}[/tex]
Now, we use the initial condition x(0)=0
[tex]\bf x(0)=10-Ce^{-0.005*0}=0\Rightarrow C=10[/tex]
and the particular solution is
[tex]\bf x(t)=10-10e^{-0.005*t}=10(1-e^{-0.005*t})\\\\\boxed{x(t)=10(1-e^{-0.005*t})}[/tex]
