conflicting answer to indefinite integral

Question

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...

Answer 1

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... $$y(0)=1 \implies 1=100+C \implies C=-99 $$ $$\implies y=100-99e^{-0.03t}$$

Answer 2

the C(constant) can either be positive or negative . so both of them are right.