Exercise · ZW-DIRECTION-FIELD-009 · ★★★ Advanced
For \(y'=x-y\), the line \(y=x-1\) is the isocline of slope \(1\). Is it also a solution of the equation?
Choose one option.
Show answer
- B. Yes: its slope is \(1\) and \(x-y=1\) at all its points
Explanation
On \(y=x-1\) we have \(x-y=1\), the field slope there; the line has slope \(1\), so it is tangent to the field at every point and hence a solution. It is a special case: in general an isocline is not a solution.
Review the concept: Direction field
A direction field draws, at many points of the plane, a small segment with the slope that the differential equation assigns to that point.
\[\text{slope at }(x,y)=f(x,y)\]
More practice problems: direction field
- ZW-DIRECTION-FIELD-001 For y'=x+y, what slope does the field segment have at the point (1,2)?
- ZW-DIRECTION-FIELD-002 For y'=xy, what slope does the segment have at (2,-1)?
- ZW-DIRECTION-FIELD-003 For y'=y-x, at which points are the segments horizontal?
- ZW-DIRECTION-FIELD-004 In the field of y'=x^2, how are the segments at (1,0) and (1,5)?
- ZW-DIRECTION-FIELD-005 Which equation produces a field with all segments parallel?
- ZW-DIRECTION-FIELD-006 For y'=x^2+y^2, the segments have slope 1 along:
- ZW-DIRECTION-FIELD-007 The solutions of y'=-x/y are circles x^2+y^2=C. What slope does the field segment have at (3,4)?
- ZW-DIRECTION-FIELD-008 In the field of y'=y^2-4, which horizontal lines are constant solutions?