I am currently studying the textbook Ordinary Differential Equations by Morris Tenenbaum and Harry Pollard. The section LESSON 1. How Differential Equations Originate. has the following exercise:
- The radium in a piece of lead decomposes at a rate which is proportional to the amount present. If 10 percent of the radium decomposes in 200 years, what percent of the original amount of radium will be present in a piece of lead after 1000 years?
The answer is said to be 59.05 percent.
Let $x$ be the amount of radium. Since the amount of radium in a piece of lead decomposes at a rate which is proportional to the amount present, we have the differential equation $\dfrac{dx}{dt} = -kx$, where $k > 0$ is a constant of proportionality.
Using the usual algebraic manipulations and then integration, we get that $x = Ae^{-kt}$, where $A$ is a constant.
So we now need to find the values of the constants, $A$ and $k$. Since we are told that 10 percent of the radium decomposes in 200 years, it seems to me that we have $0.9x = Ae^{-200k}$; but this still doesn't allow us to find $A$ and $k$. So what am I misunderstanding here? And what is the correct way to solve this?
EDIT
When $t = 0$, we have that $x = A$.
If I'm not mistaken, By plugging in $t = 0$ and concluding that $x = A$, this means that $x = A$ for all $t$, since $A$ is constant.
So we have that $$0.9x = xe^{-200k} \Rightarrow \dfrac{\ln(0.9)}{-200} = k = 0.00053$$
If we say that $C$ is the percent of the original amount of radium after $t$ years, then we have that $$Cx = xe^{-1000(0.00053)} \Rightarrow C = 0.5886,$$ which is 58.86% – not 59.05, as the textbook states.
Best Answer
Plug in $t=0$ into the solution you get, you will have $x=Ae^0=A$. So you have value for $A$. For $k$, as you have shown, you have $0.9x=xe^{-200k}$. So you have $$-200k=\ln(0.9).$$ When you consider the case after 1000 years, you have $$-1000k=5\ln(0.9)=\ln(0.9^5).$$ So you will see $$e^{-1000k}=0.9^5.$$ This is the percent you are looking for.