In a directional slope field, how can a straight line be a solution to a differential equation?

Question

I think I'm having a serious misunderstanding about differential equations, solutions to them, and what a directional slope field is.

From what I understand, if you have a differential equation like $\frac{dy}{dx} = y\cos{y}$, you can fill a graph with tiny slopes for each $(x,y)$ point. If you follow any path of slopes, you'll get a line. The equation of that line is a solution to the differential equation $\frac{dy}{dx} = y\cos{y}$.

One of the equilibrium points for this equation is $\frac{\pi}{2}$ (since when $y=\frac{\pi}{2}$, $y\cos{y}=0$). On the directional field, there will be a completely horizontal path of slopes at $y=\frac{\pi}{2}$. Therefore, from my understanding, if I trace that line, I'll get a solution to the differential equation. Tracing that line will get me the line represented by $y=\frac{\pi}{2}$.

How is $y=\frac{\pi}{2}$ a solution to $\frac{dy}{dx} = y\cos{y}$?! How?!

Isn't the derivate of $\frac{\pi}{2}$ just $0$?! What is happening?! I imagined that a solution to this equation would be something more complicated, so that when you took its derivative, it would give $y\cos{y}$. Help!

Answer 1

The equation $dy/dx = y \cos y$ means that if you have a solution $z(x)$, then substituting it to the equation will produce "true", or in other words it becomes equivalence.

So, if $z(x) = \pi/2$, $z'(x) = 0$ and $z \cos z =0$, so it is the solution.

Think of that in the same way how you think about algebraic equations, say $ax^2+bx +c=0$. Substitution of the solution into the equation does not produce a fancy expression, but rather just $0=0$.

Answer 2

I think your problem is understanding that a solution to a differential equation is much simpler than you expect at this point. A solution to $y'=f(x,y)$ is any function so that $y'(x)=f(x,y(x))$ for all $x$ from the domain of $y$. The constant function $y(x)=\frac\pi2$ satisfies this. As you observed, the left and right sides are zero for all $x$ and thus equal.

Of course, the knowledge of this solution might be useless if you are solving an initial value problem and the initial point is not on that solution.

You might also want to read up on the flow of differential equations, as that assembles the individual solutions into some kind of global solution object.

Answer 3

Restating the equation,

$$\frac{dy}{dx}=y\cdot \cos(y)$$

$$\frac{dy}{dx}=\frac{\pi}2\cdot \cos(\frac{\pi}2)$$

$y$ is a constant function, so its derivative is $0$. Substituting,

$$0=\frac{\pi}2\cdot \cos(\frac{\pi}2)$$.

Because $\cos(\frac{\pi}2)=0$, so $0=\frac{\pi}2\cdot0\implies0=0$, so the statement holds.