(c)

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

$\Rightarrow \quad N=\frac{c}{\lambda}\left(1-e^{-\lambda t}\right)+N_0 e^{-\lambda t}$