9. On the planet Lemniscate, the people use the elliptic table of elements, a far more advanced version of our periodic table. They're not very good at calculus, though, so they've asked for your help. They know that $\mathrm{Kr}$ is somewhat radioactive and deteriorates into $\mathrm{Pl}$, a very unstable element that deteriorates to form the stable element As. They started with a block of $\mathrm{Kr}$ of size 10 and nothing else. (Their units don't translate into English, sorry.) and nothing else. At time $t$, they let $x(t)$ be the amount of $\mathrm{Kr}, y(t)$ the amount of $\mathrm{Pl}$, and $z(t)$ the amount of As. They know that $x^{\prime}(t)=-x$, and that, in the absence of $\operatorname{Kr}, y^{\prime}(t)=-2 y$. Your job is to find at what time $t$ the quantity of $\mathrm{Pl}$ will be largest. You should assume that the entire amount of $\mathrm{Kr}$ that deteriorates has turned into $\mathrm{Pl}$.
Solution: This problem is long-winded since it's giving an autonomous linear system of differential equations without using any such language (and it includes a number of subtle references). The system we have is $x^{\prime}=-x, y^{\prime}=x-2 y$. It's not hard to see that $x=10 e^{-t}$ satisfies the first equation and the initial condition. Plugging this into the second equation and using the integrating factor $e^{2 t}$ (or using eigenvalues and eigenvectors to solve the system directly, though I don't want to begin to explain what that means) lets us solve for $y$. More precisely, we want to solve $y^{\prime}+2 y=10 e^{-t}$. Multiply by $e^{2 t}$ and simplify the left hand side to get $\left(y e^{2 t}\right)^{\prime}=10 e^{t}$. Integrating both sides with respect to $t$ then yields $y e^{2 t}=10 e^{t}+C$, or $y=10 e^{-t}+C e^{-2 t}$. Since $y(0)=0$, we find $C=-10$. Now to maximize $y$, we solve $y^{\prime}(t)=0$, or $-10 e^{-t}+20 e^{-2 t}=0$, or $t=\ln 2$.