Radioactive decay of an element A into an element B It's well known that the rate of decay of an element is proportional to its amount. Suppose we have a radioactive element $A$ which decays into a radioactive element $B$ ($B$ also decays). If the initial amount of $A$ is $A_0$ and there is no element $B$ at the beginning find a formula for the amount of $B$ at time $t$, $B(t)$.
I figured that the amount of $A$ that turned into $B$ in time $t$ is $t\dfrac{dA(t)}{dt}$. So the rate of change of $B$ is $\dfrac{d}{dt}\left(t\dfrac{dA(t)}{dt}\right)=\dfrac{dA(t)}{dt}+t\dfrac{d^2A(t)}{dt^2}=\dfrac{dB(t)}{dt}=bB(t)$ and therefore $$bB(t)=aA(t)+ta\dfrac{dA(t)}{dt}$$ where $a,b$ are the decay constants.I don't know if my thinking is correct, probably not because when I solve this DE I get a wrong answer which says that at time $t=0$ $B(t)\not=0$.
 A: The first thing to do is to note that the rate of change of $A$ is governed by the equation:
$$\frac{\mathrm{d}A}{\mathrm{d}t}=-\lambda A \implies A(t)=A_{0}e^{-\lambda t}$$
The total rate of change of $B$ is given by:
$$\frac{\mathrm{d}B}{\mathrm{d}t}=-\frac{\mathrm{d}A}{\mathrm{d}t}-\mu B = \lambda A - \mu B$$
But we know analytically the formula for $A(t)$, so we have:
$$\frac{\mathrm{d}B}{\mathrm{d}t}=\lambda A_{0}e^{-\lambda t}-\mu B$$
This is an inhomogeneous, linear ODE in $B$, and so we note that the homogeneous solution is given by:
$$B_{\mathrm{H}}=B_{0}e^{-\mu t}$$
And the particular solution is given by trying $B_{\text{PS}}=\alpha e^{-\lambda t}$, giving:
$$-\alpha \lambda e^{-\lambda t}=\lambda A_{0}e^{-\lambda t}-\mu \lambda e^{-\lambda t}$$
Thus we have: $\alpha \lambda =\lambda A_{0}-\mu\lambda$, thus $\alpha = A_{0}-\mu$, so we have:
$$B(t)=B_{0}e^{-\mu t}+(A_{0}-\mu)e^{-\lambda t}$$
A: We have $$\tag1\frac {d A}{d t}=-aA$$ because $A$ decays into $B$. And we have $$\tag2\frac{dB}{dt}=-\frac {d A}{d t}-bB$$ because the decay of $A$ produces $B$ and $B$ itself decays.
From $(1)$ you should find that $A(t)=A_0e^{-at}$ so that $(2)$ turns into 
$$ \frac{dB}{dt}=aA_0e^{-at}-bB,$$
which is slightly different than your DE.
