Direction Fields — Question 7

Question 7

For the equation $$y'=x^2-y,$$ consider the supplied candidate through the origin $$w(x)=x^2-2x+2-2e^{-x}.$$ You may use the calculus inequality $e^t>1+t$ for $t\ne 0$.

Tasks

1. Find the zero-slope isocline and describe the slope signs above and below it.

2. Verify that $w$ solves the equation and that $w(0)=w'(0)=0$.

3. Decide whether $w$ has a local extremum at $0$. Determine its concavity on either side of $0$ and classify the horizontal tangent there.

4. Draw the field, the isocline, and $w$ near the origin in your solution. Explain why observing a horizontal field segment is insufficient to conclude that a solution has a maximum or minimum.

Question 7 – Solution

Strategy. A zero first derivative identifies a stationary point, not its type. Inspect nearby derivative signs and concavity.

Step 1: Isocline and verification. Zero slopes occur on $\boxed{y=x^2}$. Slopes are positive below this parabola and negative above it. For the candidate, $$w'=2x-2+2e^{-x}=x^2-w,\qquad w(0)=w'(0)=0.$$ It is therefore a solution on $\mathbb R$ with a horizontal tangent at the origin.

Step 2: No local extremum. For $x\ne 0$, the supplied inequality gives $e^{-x}>1-x$, so $$w'=2(x-1+e^{-x})>0.$$ The derivative is positive on both sides of $0$. Thus $w$ increases through the origin and has neither a local maximum nor a local minimum there. Moreover, $x^2-w=w'>0$ for $x\ne 0$, so the curve lies below its zero-slope isocline on both sides.

Step 3: Concavity. Since $w''=2-2e^{-x}$, the curve is concave down for $x<0$ and concave up for $x>0$. Hence $$\boxed{(0,0)\text{ is an inflection point with a horizontal tangent.}}$$ The field gives the tangent slope at a point; the surrounding behavior is needed to classify that point.