occasionally, you can’t really get a specific solution.
\begin{equation} \dv{y}{t} = e^{t}\cos y \end{equation}
after doing the , you get:
\begin{equation} \ln (\sec y + \tan y) - e^{t} = C \end{equation}
you get sets of this function \(F(t,y)\) which shifts it up and down, by any constant C.
But at any given \((t,y)\), you get a slope \(e^{t}\cos y\).