Definition 2.4.1.
A separable equation is a DE which has the following form, for some expressions \(f\) and \(g\text{.}\)
\begin{equation*}
\frac{dy}{dt} = f(y) g(t)
\end{equation*}
Section 2.4 Separable Equations
After some sections on qualitative methods, stability, and linearization, I am now going to try to actually solve some first order DEs. The easiest type of first order DE to solve is a separable equation.
Example 2.4.2.
The equation
\begin{equation*}
\frac{dy}{dt} = \left( y^2 + 2y - 1 \right) \left(
\frac{1}{t^2 +1} \right)
\end{equation*}
is a separable equation. On the right side, there are two expressions, one only involving \(y\) and one only involving \(t\text{,}\) which are multiplied together.
Example 2.4.3.
The equation
\begin{equation*}
\frac{dy}{dt} = \sin y \left( e^{t^2 + 1} - t^2 \right)
\end{equation*}
is a separable equation. On the right side, there are two expressions, one only involving \(y\) and one only involving \(t\text{,}\) which are multiplied together.
Example 2.4.4.
\begin{equation*}
\frac{dy}{dt} = ty + t^2 y^2 + t^3 y^3
\end{equation*}
is not a separable equation, since there is no way to factor the right side into two multiplied expressions \(f(y)
g(t)\text{.}\) Even though each piece of the sum on right side has this form, the entire right side needs to be factorable this way to satisfy the definition.
Separable equations are the first type of differential equation that I will actually solve in this course. They have a universal method (at least, a long as we can make sense of the resulting integrals.) First, I will take the expression involving the dependent variable and bring it to the left side. (I have to worry a bit about values of \(y\) that result in \(g(y) = 0\text{,}\) but I’ll worry about that issue in a later section.)
\begin{equation*}
\frac{1}{f(y)} \frac{dy}{dt} = g(t)
\end{equation*}
\(y\) is the dependent variable: the DE is built upon the assumption that \(y\) is some function of \(t\text{.}\) This means that both sides of the equations are functions of \(t\text{;}\) the left side implicity and the right side explicitly. Therefore, I’ll integrate both sides in \(t\text{.}\)
\begin{equation*}
\int \frac{1}{f(y)} \frac{dy}{dt} dt = \int g(t) dt
\end{equation*}
Now I use a strange looking substitution \(y = y(t)\) on the left side of the integral. The differential piece of the substitution is \(dy = \frac{dy}{dt} dt\text{,}\) which conveniently replaces a big piece of the left side integral.
\begin{equation*}
\int \frac{1}{f(y)} dy = \int g(t) dt
\end{equation*}
To finish, I do both the integrals. The result is some expression in \(y\) on the left and some function of \(t\) on the right. If I can, I simplify to solve for \(y\) on the left, which gives \(y\) as a function of \(t\) as desired. The function will include a constant of integration that arises from the integration step as well. If I cannot solve for \(y\text{,}\) I can leave the solution implicit. Even if I can solve for \(y\text{,}\) often the implicit for is preferred.
I should make one note about the constant on integration that arises in this solution method (and in many other as well). Since the constant of integraiton will eventually be determined by an initial value, mathematicians are often quite careless with the artihmetic of this constant. For example, if the following expression shows up \(2(t+c)\text{,}\) I would often simplify this to \(2t+c\) instead of \(2t + 2c\text{,}\) since whether I eventually figure out the constant from \(c\) or from \(2c\) later, its value is still determined by the initial condition. This is a strange carelessness after all the care mathematicians have always preached about arithmetic and algebra, but it is, in fact, justified. Let me now move on to some examples of solving separable equations.
Example 2.4.5.
Here is a separable DE, which I solve with the method described above, separating the variables, integrating both sides, and solving for the dependent variable. Note in the integration step that I only need to add the constant of integration on one side; if there were two constants, I could bring them over to one side and combine them into one.
\begin{align*}
\frac{dy}{dt} \amp = \frac{\sin t}{y}\\
y \frac{dy}{dt} \amp = \sin t \\
\int y \frac{dy}{dt} dt \amp = \int \sin t dt\\
\int y dy \amp = \int \sin t dt\\
\frac{y^2}{2} \amp = - \cos t + c\\
y \amp = \pm \sqrt{c - 2\cos t}
\end{align*}
Note that the constant of integration ends up incorporated into the expression. I need to add it in the integration step and them move it around with the algebra; I can’t calculate \(y\) and then just add the \(+ c\) at the very end.
If I impose an initial condition of \(y(0) = 1\text{,}\) I can determined the value of the constant of integration and thus a unique solution.
\begin{align*}
1 \amp = \sqrt{c - 2 \cos (0)} = \sqrt{c-2}\\
1 \amp = c-2\\
c \amp = 3\\
y \amp = \sqrt{3 - 2 \cos t}
\end{align*}
Figure 2.4.6 shows the direction field and for this example and Figure 2.4.7 shows several solution graphs. Notice the strange domain issues with this implicit plot. When \(|c| \leq 2\text{,}\) the solutions have a restricted domain, represented by the closed curves. There are no solutions at all when \(c \lt -2\text{.}\) There are solutions with domain \(\RR\) only for \(c \gt 2\text{.}\) When \(c =
2\text{,}\) the solution is the crossed graph, which is not always differentiable. When \(c=-2\text{.}\) the ‘solution’ is only defined at discrete points
Example 2.4.8.
Here is a separable DE, which I solve with the method described above, separating the variables, integrating both sides, and solving for the dependent variable. This is an autonomous example, so the right side integral is trivial.
\begin{align*}
\frac{dy}{dt} \amp = y^2 - 4\\
\int \frac{1}{y^2-4} dy \amp = \int 1 dt\\
\frac{-1}{2} \arctanh \left( \frac{y}{2} \right) \amp = t
+ c\\
\arctanh \left( \frac{y}{2} \right) \amp = -2t + c\\
\frac{y}{2} \amp = \tanh (-2t + c)\\
y \amp = 2 \tanh (-2t + c)
\end{align*}
Since this is an autonomous equation, I should look for singular solutions when the right side of the equation vanishes — these will be the steady states of the autonomous equation. On the graph, the steady state solutions should be horizontal lines, since the function never changes. Here, \(y=2\) and \(y=-2\) are steady states. Moreover, \(y=2\) is stable and \(y=-2\) is unstable. This can be seen in the direction field in Figure 2.4.9 and the graphs of some solution curves in Figure 2.4.10. Near the stable steady state, the solutions go towards the horizontal line. Near the unstable steady state, the solutions go away from the horizontal line.
It is also interesting to note that the range of \(\tanh\) is only the interval \((-1,1)\text{,}\) so it is impossible to get \(y \leq -2\) or \(y \geq 2\text{.}\) I might wonder if there are solutions in this range at all. In the implicit plot, I could draw curves with these \(y\) values, since there are slopes in that part of the cartesian plane. These other curves can be found by doing the integral differently, since both hyperbolic tangent and hyperbolic cotangent have the same anti-derivative. \(y=2 \coth (-2t + c)\) is also a solution and the range of \(coth\) covers exactly what is missing from \(\tanh\text{.}\)
