Direction Fields — Question 5

Question 5 – Solution

Strategy. Respect the equation’s domain before drawing. Use a constant quantity along a solution to identify the curve through the given point.

Step 1: Signs and missing points. Slopes are positive in Quadrants I and III, and negative in II and IV. Horizontal segments occur exactly where $x=0$ and $y\ne 0$. The expression $x/y$ is undefined on all of $y=0$; a vertical mark at $(1,0)$ would falsely assign a direction to a point outside the domain.

Step 2: Identify and verify the branch. Wherever a solution is defined, $$\frac{d}{dx}(y^2-x^2)=2yy'-2x=2y\frac{x}{y}-2x=0.$$ Thus $y^2-x^2$ is constant on its interval. At $(0,1)$ it equals $1$. Continuity and the positive initial value select $$\boxed{y=\sqrt{1+x^2},\qquad -\infty<x<\infty.}$$ Indeed $y'=x/\sqrt{1+x^2}=x/y$ and $y(0)=1$.

Step 3: Interval and geometry. This branch is smooth and satisfies $y\ge 1$ for every real $x$, so it never reaches the excluded axis. The entire real line is therefore its largest open interval. It decreases for $x<0$ and increases for $x>0$, with a minimum at $(0,1)$, matching the field.