Solving a system of three linear differential equations 
Can anyone help me check if the solution to this differentiation equation is (A).
Thanks
 A: Here's how to find the concentration of $A$ or $[A]$
$$
-\frac{d}{dt}[A]=k[A]
$$
$$
\frac{1}{[A]}d[A]=-k\ dt
$$
$$
\int \frac{1}{[A]}d[A]=-k\int dt
$$
$$
\ln |[A]|+K_1=-kt+K_2
$$
$$
\ln |[A]|=-kt+K_2-K_1=-kt+K
$$
$$
e^{\ln |[A]|}=e^{-kt+K}
$$
$$
[A]=e^{K-kt}=e^Ke^{-kt}=\frac{e^K}{e^{kt}}
$$
Note that
$$
[A]_0=\frac{e^K}{e^{k\cdot 0}}= \frac{e^K}{1}= e^K\neq 0 
$$ 
Therefore
$$
[A]=e^Ke^{-kt}=[A]_0e^{-kt}
$$
To find $[B]$, we use the fact that
$$
\frac{1}{2}\frac{d}{dt}[B]=k[A]=k[A]_0e^{-kt}
$$
$$
d[B]=2k[A]_0e^{-kt}\ dt
$$
$$
\int d[B]=2k[A]_0\int e^{-kt}\ dt
$$
$$
[B]+K_1=-2k[A]_0\left(\frac{e^{-kt}}{k}+K_2\right)
$$
$$
[B]=-2[A]_0e^{-kt}-2k[A]_0K_2-K_1=-2[A]_0e^{-kt}+K
$$
Note that
$$
[B]_0=-2[A]_0e^{-k\cdot 0}+K=-2[A]_0+K=-2[A]_0+2[A]_0=0
$$
Therefore
$$
[B]=-2[A]_0e^{-kt}+K=-2[A]_0e^{-kt}+2[A]_0=2[A]_0(1-e^{-kt})
$$
And now we can find $[C]$ using
$$
\frac{d}{dt}[C]=\frac{1}{2}\frac{d}{dt}[B]
$$
$$
d[C]=\frac{1}{2}d[B]
$$
$$
\int d[C]=\frac{1}{2}\int d[B]
$$
$$
[C]+K_1=\frac{1}{2}[B]+K_2
$$
$$
[C]=\frac{1}{2}[B]+K_2-K_1=\frac{1}{2}[B]+K=[A]_0(1-e^{-kt})+K
$$
Note that
$$
[C]_0=[A]_0(1-e^{-k\cdot 0})+K=[A]_0(1-1)+K=K=\frac{[A]_0}{2}
$$
Therefore
$$
[C]=[A]_0(1-e^{-kt})+K=[A]_0(1-e^{-kt})+\frac{[A]_0}{2}=[A]_0\left(\frac{3}{2}-e^{-kt}\right)
$$
