Radioactive Decay 
Problem:Nuclei of a radioactive element $\Bbb X$ having decay constant $\lambda$ ,  ( decays into another stable nuclei $\Bbb Y$ ) is being produced by some external process at a constant rate $\Lambda$.Calculate the number of nuclei of $\Bbb X$ and $\Bbb Y$ at $t_{1/2}$

I tried to create an equation for rate of change of the number of nuclei a:
$$\dfrac{dN_{X}}{dt}=\Lambda-N_X\lambda $$
I did that because in simple decay $\dfrac{dN}{dt}=-\lambda N$ holds and here it's also being produced by rate.  But after integration should we write $$ln\Bigg(\dfrac{\lambda N_X-\Lambda}{\lambda N_0-\Lambda}\Bigg)=-\lambda t$$
or $$ln\Bigg(\dfrac{\lambda N_X-\Lambda}{N_0}\Bigg)=-\lambda t$$ First one because limit was on $N: (N_0\to N)$
And next what to substitute for $t$ (ie. what is $t_{1/2}$? $ln2/\lambda$ or something else?)
Also how to do it for $\Bbb Y$?
Just write $$\dfrac{dN_Y}{dt}=\lambda N_x $$?
 A: The first of your equations is correct.  You can see this in two ways.  First, just look at the dimensions.  In general, the argument of a logarithm should be dimensionless; only your first option is.  Second, and maybe more convincingly, look at what you get when you take $\Lambda \to0$.  You should be able to reproduce the standard decay equation:
\begin{equation}
N_X(t) = N_0\, e^{-\lambda\, t}~.
\end{equation}
In your first equation, the factors of $\lambda$ on the left-hand side cancel, and you get this result.  With your second equation, you would get $N_X(t) = \frac{N_0}{\lambda}\, e^{-\lambda\, t}$.  So that must be wrong.
As for what $t_{1/2}$ is, surely it must just be the half-life of $\mathbb{X}$ (with no creation).  In particular, if $\Lambda$ is large enough, $N_X$ will actually grow, so there is no time at which half of the material is left.  Since $\mathbb{Y}$ is stable, you can assume there's no relevant half-life there.
Also, your expression for $N_Y$ is correct.  It's a slightly harder integration, but not too bad.
A: So here's what I did, I discretized the problem and deduced $N_x$ and $N_y$ at some $t_n$. These happened to include sums that could easily be turned into integrals. Namely:
$$N_x(t_n)=N_0e^{-\lambda t_n}+\Lambda\sum_{i=0}^n\Delta t_ie^{-\lambda(t_n-t_i)}$$
$$N_y(t_n)=N_0(1-e^{-\lambda t_n})+\Lambda\sum_{i=0}^n\Delta t_i(1-e^{-\lambda(t_n-t_i)})$$
As a result, I found the following:
$$N_x(t)~=~N_0e^{-\lambda t}+{\Lambda\over\lambda}(1-e^{-\lambda t})$$
$$N_y(t)~=~N_0(1-e^{-\lambda t})-{\Lambda\over\lambda}(1-e^{-\lambda t})+\Lambda t$$
At $t_{1\over2}={ln(2)\over\lambda}$, $e^{-\lambda t}={1\over2}$, therefore:
$$N_x(t_{1\over2})~=~{N_0+{\Lambda\over\lambda}\over2}$$
$$N_y(t_{1\over2})~=~{N_0\over2}+{\Lambda\over\lambda}(ln(2)-{1\over2})$$
