Jtdylktuy

I have a radioactive decay system to solve for $x(t)$ and $y(t)$ (no need for $z(t)$):
$$begin{cases}x'=-lambda x\
y'=lambda x-mu y\
z'=mu yend{cases}$$
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?

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) $lambdanot=mu$ 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?

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

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

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

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

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