Net decay rate of active nuclii is
$\frac{d N}{d t}=c-\lambda N$
$\Rightarrow \quad \frac{d N}{c-\lambda N}=d t$
Integrating both sides, we get
$\int \frac{d N}{c-\lambda N}=\int d t \quad \dots(i)$
Now, $\quad c-\lambda N=u$
Differentiating,
$-\lambda d N=d u$
above equation, we have
$d N=\frac{d x}{-\lambda} \quad \dots(i)$
Substituting is Eq. $(i)$, we get
$\frac{-1}{\lambda} \int \frac{d u}{u}=\int d t$
$\Rightarrow -\frac{1}{\lambda} \log u=t+k$
where, $k$ is constant of integration.
$\Rightarrow -\frac{1}{\lambda} \log (c-\lambda N)=t+k \quad \dots(ii)$
Now, at $t=0$ and $N=N_0$,
So, $\frac{-1}{\lambda} \log \left(c-\lambda N_{0}\right)=k$.
Substituting for $k$ in Eq. $(ii)$, we get
$-\frac{1}{\lambda} \log (c-\lambda N)=t-\frac{1}{\lambda} \log$
$\left(c-\lambda N_0\right)$
$\log \left(\frac{c-\lambda N}{c-\lambda N_{0}}\right)=-\lambda t$
Taking antilog we get,
$\frac{c-\lambda N}{c-\lambda N_0}=e^{-\lambda t}$
$\Rightarrow c-\lambda N=\left(c-\lambda N_0\right) e^{-\lambda t}$
Solving for $N$, we get
$\Rightarrow \quad N=\frac{c}{\lambda}\left(1-e^{-\lambda t}\right)+N_0 e^{-\lambda t}$
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