Answer
(a)$12$
(b) $\begin{cases}
12&y(0)=12\\
\infty&y(0)>12\\
-\infty&y(0)<12
\end{cases}$
(c) $ -3$
Work Step by Step
(a) Since
\begin{align*}
\frac{d y}{d t}&=-4(y-12)\\
\int \frac{dy}{y-12} &= \int -4dt\\
\ln|y-12|&= -4t +c\\
y-12&= e^{-4t+c}\\
y&=12+ C e^{-4t}
\end{align*}
Then
\begin{align*}
\lim _{t\to \infty }y&=\lim _{t\to \infty }(12+ C e^{-4t})\\
&=12
\end{align*}
(b) Since
\begin{align*}
\frac{d y}{d t}&=4(y-12)\\
\int \frac{dy}{y-12} &= \int 4dt\\
\ln|y-12|&= 4t +c\\
y-12&= e^{4t+c}\\
y&=12+ C e^{4t}
\end{align*}
Then
\begin{align*}
\lim _{t\to \infty }y&=\lim _{t\to \infty }(12+ C e^{4t})\\
&=\begin{cases}
12&y(0)=12\\
\infty&y(0)>12\\
-\infty&y(0)<12
\end{cases}
\end{align*}
(c) Since
\begin{align*}
\frac{d y}{d t}&=-4y-12\\
\int \frac{dy}{y+3} &= \int -4dt\\
\ln|y+3|&= -4t +c\\
y+3&= e^{-4t+c}\\
y&=-3+ C e^{-4t}
\end{align*}
Then
\begin{align*}
\lim _{t\to \infty }y&=\lim _{t\to \infty }(-3+ C e^{-4t} )\\
&=-3
\end{align*}
