Question

Consider the autonomous DE dy/dx = y² - y - 6. Use your ideas from Problem 35 to find intervals on the y-axis for which solution curves are concave up and intervals for which solution curves are concave down. Discuss why each solution curve of an initial-value problem of the form

dy/dx = y² - y – 6,y(0) = y₀, where -2 < y₀ < 3, has a point of inflection with the same y-coordinate. What is that y-coordinate? Carefully sketch the solution curve for which y(0) = -1. Repeat for y(2) = 2.

Short answer

The possible point of inflection is at $y=\frac{1}{2}$.

For $y<-2$and $\frac{1}{2}<y<3$we have $y\text{'}\text{'}\left(x\right)<0$,concave down.

For $y>3$and$-2<y<\frac{1}{3}$we have $y\text{'}\text{'}\left(x\right)>0$,concave up.

Step-by-step solution

Critical points.

$f$The zeros of the function $f$ in$dy/dx=f\left(y\right)$ are of special importance. Wesay that a real number c is a critical point of the autonomous differential equation$dy/dx=f\left(y\right)$ if it is a zero of $f$—that is, $f\left(c\right)=0$. A critical point is also called an equilibrium point or stationary point.

Step 2:Points of inflection.

Let $\frac{dy}{dx}=0$, then

$y^2-y-6=0$

Factoring the equation, we get

$$\begin{gathered} (y-3)(y+2)=0 \\ y=3,y=-2 \end{gathered}$$

So, the critical points are and .

Now,

$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=(2y-1)\frac{dy}{dx} \\ =(2y-1)(y-3)(y-2) \end{gathered}$$

Therefore, the possible point of inflection is at $y=\frac{1}{2}$.

For$y<-2$ and$\frac{1}{2}<y<3$we have $y\text{'}\text{'}\left(x\right)<0$, concave down.

For$y>3$and$-2<y<\frac{1}{3}$we have $y\text{'}\text{'}\left(x\right)>0$, concave up.

Points of inflection of solution of autonomous differential equations will have the same y-coordinates, because between critical points they are horizontal translations of each other.