Solve a differential equation in a radioactive decay system

Question

I have a radioactive decay system to solve for $x(t)$ and $y(t)$ (no need for $z(t)$): $$\begin{cases}x'=-\lambda x\\ y'=\lambda x-\mu y\\ z'=\mu y\end{cases}$$ with the initial conditions $x(0)=1,y(0)=0,z(0)=0$.

I found $x(t) = e^{-\lambda t}$, but $y(t)$ is proving to be difficult. I have tried subbing in $x(t) = e^{-\lambda t}$ so that we have a linear DE in $y$: $$y'+\mu y=\lambda e^{-\lambda t}$$ but ended up getting a giant solution which seems very wrong.

Is this the right approach or no?

Answer 1

The differential equation $$y'+\mu y=\lambda e^{-\lambda t}$$ is linear non-homogeneous with constant coefficients. The general solution is $$y'(t)=Ce^{-\mu t}+y_p(t)$$ where $C$ is an arbitrary constant and $y_p$ is a particular solution. As regards $y_p$ we are supposed to distinguish two cases 1) $\lambda\not=\mu$ and 2) $\lambda=\mu$. As a reference see the method of undetermined coefficients.

Once we have $y$, we can solve $z'=\mu y$ by a straightforward integration.

Can you take it from here?

Answer 2

$$y'+\mu y=\lambda e^{-\lambda t}$$

$$d(e^{\mu t}y)=\lambda e^{(\mu - \lambda)t} dt \text{ }[\text{I.F.} = e^{\mu t}]$$

$$\text{Integrating, } e^{\mu t}y=\frac{\lambda e^{(\mu - \lambda)t}}{\mu - \lambda} + C$$

$$\text{Putting } y(0)=0, 0=\frac{\lambda }{\mu - \lambda} + C$$

$$\text{Hence, } y(t)=\frac{\lambda }{\mu - \lambda}(e^{(\mu - \lambda)t}-1)e^{-\mu t}$$