\begin{equation} \dv{y}{t} = a(t)f(y) \end{equation}
are a class of functions are called seperable. We can solve them using the division method
division method
the division method involves solving autonomous ODEs by dividing and treating it normally:
\begin{equation} y’ = 8y \end{equation}
\begin{equation} \frac{y’}{8} = y \end{equation}
we now write something fishy:
\begin{equation} \frac{\dd{y}}{y} = 8 \dd{t} \end{equation}
we now take the antiderivative of this:
\begin{equation} \int \frac{1}{y} \dd{y} = \int 8 \dd{t} \end{equation}
We will get that:
\begin{equation} \ln |y| = 8t + C \end{equation}
we finally get:
\begin{equation} |y| = e^{C} e^{8t} \end{equation}
getting rid of that absolute value:
\begin{align} y &= \pm e^{C} e^{8t} \\ &= K e^{8t} \end{align}
places where this breaks down
- sometimes, \(\frac{1}{f(y)}\) may not have a nice antiderivative
- sometimes, \(G(y)\), the antidepressant, may not be nicely invertible
general solution to y’(t) = ry(t)
generally, for \(r \in \mathbb{R}\), the solution to \(y’(t) = ry(t)\) is at \(y(t)=y_0e^{rt}\), where \(y_0 = y(0)\).
for autonomous ODEs for which \(ry(t) = f(y)\), we have that:
\begin{equation} \dv{y}{x} = ry(x) \end{equation}
which means:
\begin{equation} \frac{1}{y(x)} \dd{y} = r\dd{x} \end{equation}
and so:
\begin{equation} \ln \qty| y(x) | = rx +C \end{equation}
and hence:
\begin{equation} y(x) = K e^{rx} \end{equation}
plugging in \(x=0\), yields \(y(0) = Ke^{0} = K\).