conflicting answer to indefinite integral In modelling a mixing problem of a single tank using first order ODE , i ended up with
            $$\frac{dy}{dt}=3-0.03y$$
Case 1 : if I do $\frac{dy}{dt}=-0.03(y-100)$ which leads to $\ln|y-100|=-0.03t+c$ and ultimately 
$$y=100+Ce^{(-0.03t)}$$
Case 2 : if I do $\frac{dy}{dt}=0.03(100-y)$ which leads to $\ln|100-y|=0.03t+c$ and ultimately
$$y=100-Ce^{(0.03t)}$$
$$y(0)=1$$ which makes case 1 :$$y=100-99e^{-0.03t}$$
and case 2 :$$y=100-99e^{0.03t}$$ 
which one is apt...am i doing mistakes ?
edited after 11 mins. i got the answer guys...im unable to close the question my apologies...

this is not correct
Case 2 : if I do $\frac{dy}{dt}=0.03(100-y)$ which leads to $\ln|100-y|=0.03t+c$ and ultimately
Substitute $z=100-y$ before to integrate
Second case
$$\frac{dy}{dt}=0.03(100-y)$$
Substitute $z=100-y \implies dz=-dy$
$$-\int \frac{dz}{z}=0.03t+K$$
$$\ln|z|=-0.03t+K$$
$$100-y=e^{-0.03t}K$$
$$y=100-Ke^{-0.03t}$$
$$y=100+Ce^{-0.03t}$$
C is just a constant...
$$y(0)=1 \implies 1=100+C \implies C=-99 $$
$$\implies y=100-99e^{-0.03t}$$
