Let’s consider the first order autonomous differential equation \(y^{\,\prime }=\,-\,\dfrac {(y^2-2y)(y^2-1)^2}{\sqrt {1+y^8}}\) and part of its (approximate) direction field below.

A direction field.

Figure 1: A direction field
(a)
Let’s determine all the equilibrium solutions.
(b)
Let’s classify each equilibrium solution as asymptotically stable, semistable, or unstable.
(c)
Suppose \(y(t)\) is a solution to \(y^{\,\prime }=-\dfrac {(y^2-2y)(y^2-1)^2}{\sqrt {1+y^8}}\) satisfying that \(y(0)=0.5\). Let’s determine \(\displaystyle \lim _{t\rightarrow \infty } y(t)\).
(d)
Let’s determine the possible limits as \(t\rightarrow \infty \) for non-constant solutions to
\(y^{\,\prime }=-\dfrac {(y^2-2y)(y^2-1)^2}{\sqrt {1+y^8}}\).

A second order autonomous equation is a differential equation of the form
\(y^{\,\prime \prime }=f(y,y^{\,\prime })\).

Let’s consider a special case: \(y^{\,\prime \prime }=f(y)\).

(a)
By multiplying both sides by \(y^{\,\prime }\) we get \(y^{\,\prime }y^{\,\prime \prime }=f(y)y^{\,\prime }\). Then we can integrate both sides (with respect to the independent variable). Let’s determine what we get when we do this, Using that \(y^{\,\prime \prime } \mathrm {d}t=\mathrm {d}(y^{\,\prime })\) and \(y^{\,\prime } \mathrm {d}t=\mathrm {d}y\). The result is a separable differential equation.
(b)
Let’s solve the IVP \(y^3y^{\,\prime \prime }=-1\), \(y(1)=1\), \(y^{\,\prime }(1)=-1\) using this method.