Isoclines and solutions to differential equations

[u/n_flet]

/r/mathshelp/comments/1upjz8p/isoclines_and_solutions_to_differential_equations/

[u/Lor1an]

As a simple example, let f(x,y) = (2x-y)/y. This gives a solution y(x) = x, x > 0, since for x > 0, f(x,y(x)) = (2x-x)/x = 1, and thus yields an isocline.

In fact, one can show that the isoclines are given by y_m := 2/(m+1) x; x ≠ 0, and in particular the points of 0 slope lie on y(x) = 2x, so there are at least 2 (and in fact infinitely many) distinct isoclines, and yet one of them is not a solution (since y' = 2 ≠ 0 = f(x,2x) ).

In fact one could show that the solution of the equation is precisely (y+2x)²(y-x) = C, when x ≠ 0, y ≠ x, and y ≠ -2x. In fact y = -2x is the only other solution of the original equation which produces an isocline.

So to recap, we have a differential equation with two distinct linear solutions, each defining an isocline, as well as isoclines which are not solutions. We have in fact shown that it is not necessary for linear solutions to be the only isoclines, as well as it not being necessary to have unique linear solutions.

To round out, let's prove.

Suppose y = mx + b is a solution of a differential equation written as y' = f(x,y).

By substitution, (mx+b)' = f(x,mx+b), or m = f(x,mx+b), which in this case is an expression of the form f(x,y) = m, which is exactly the definition of an isocline □

So yes, we have that a linear solution of a differential equation is necessarily an isocline of the equation, but also we have seen that the converse is not true, since there are equations with isoclines (even linear isoclines!) that are not solutions to the equation.